7. Chain Rule & Implicit Differentiation
Exercises
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Consider the functions \(f(x)=3x\) and \(g(x)=2x\). Compute \(g(4)\) and \(f(g(4))\).
\(g(4)=8\)
\(f(g(4))=24\)We replace the \(x\) in \(g(x)=2x\) by \(4\): \[ g(4)=2\cdot4=8 \] Next we replace the \(x\) in \(f(x)=3x\) by \(g(4)=8\): \[ f(g(4))=3\cdot8=24 \]
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Consider the functions \(f(x)=3x^2+2\) and \(g(x)=4x+1\). Compute \(g(2)\) and \(f(g(2))\).
\(g(2)=9\)
\(f(g(2))=245\)We replace the \(x\) in \(g(x)=4x+1\) by \(2\): \[ g(2)=4\cdot2+1=8+1=9 \] Next we replace the \(x\) in \(f(x)=3x^2+2\) by \(g(2)=9\): \[ f(g(2))=3\cdot9^2+2=3\cdot81+2=243+2=245 \]
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Consider the functions \(f(x)=3\cos(\pi x)\) and \(g(x)=\sin(\pi x)\). Compute \(f\left(\dfrac{1}{3}\right)\) and \(g\left(f\left(\dfrac{1}{3}\right)\right)\).
\(f\left(\dfrac{1}{3}\right)=\dfrac{3}{2}\)
\(g\left(f\left(\dfrac{1}{3}\right)\right)=-1\)We replace the \(x\) in \(f(x)=3\cos(\pi x)\) by \(\dfrac{1}{3}\): \[ f\left(\dfrac{1}{3}\right)=3\cos\left(\dfrac{\pi}{3}\right) =3\cdot\dfrac{1}{2}=\dfrac{3}{2} \] Next we replace the \(x\) in \(g(x)=\sin(\pi x)\) by \(f\left(\dfrac{1}{3}\right)=\dfrac{3}{2}\): \[ g\left(f\left(\dfrac{1}{3}\right)\right)=\sin\left(\dfrac{3\pi}{2}\right) =-1 \]
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Consider the functions \(f(x)=3x^2+5\) and \(g(x)=4x^2+1\).
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Compute \(g(1)\) and \(f(g(1))\).
\(g(1)=5\)
\(f(g(1))=80\)We replace the \(x\) in \(g(x)=4x^2+1\) by \(1\): \[ g(1)=4\cdot1^2+1=4+1=5 \] Next we replace the \(x\) in \(f(x)=3x^2+5\) by \(g(1)=5\): \[ f(g(1))=3\cdot5^2+5=3\cdot25+5=75+5=80 \]
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Compute \(f(1)\) and \(g(f(1))\).
\(f(1)=8\)
\(g(f(1))=257\)We replace the \(x\) in \(f(x)=3x^2+5\) by \(1\): \[ f(1)=3\cdot1^2+5=3+5=8 \] Next we replace the \(x\) in \(g(x)=4x^2+1\) by \(f(1)=8\): \[ g(f(1))=4\cdot8^2+1=4\cdot64+1=256+1=257 \]
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Notice that \(g(f(1))\) is not the same as \(f(g(1))\).
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Compute \(f(g(x))\).
\(f(g(x))=48x^4+24x^2+8\)
We replace the \(x\) in \(f(x)=3x^2+5\) by \(g(x)=4x^2+1\): \[\begin{aligned} f(g(x))&=f(4x^2+1) \\ &=3(4x^2+1)^2+5 \\ &=3(16x^4+8x^2+1)+5 \\ &=48x^4+24x^2+3+5 \\ &=48x^4+24x^2+8 \end{aligned}\]
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Since \(f(g(x))=48x^4+24x^2+8\), we compute: \[\begin{aligned} f(g(1))&=48\cdot1^4+24\cdot1^2+8 \\ &=48+24+8=80 \end{aligned}\] which agrees with the answer in part (a).
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Compute \(g(f(x))\).
\(g(f(x))=36x^4+120x^2+101\)
We replace the \(x\) in \(g(x)=4x^2+1\) by \(f(x)=3x^2+5\): \[\begin{aligned} g(f(x))&=g(3x^2+5) \\ &=4(3x^2+5)^2+1 \\ &=4(9x^4+30x^2+25)+1 \\ &=36x^4+120x^2+100+1 \\ &=36x^4+120x^2+101 \end{aligned}\]
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Since \(g(f(x))=36x^4+120x^2+101\), we compute: \[\begin{aligned} g(f(1))&=36\cdot1^4+120\cdot1^2+101 \\ &=36+120+101=257 \end{aligned}\] which agreees with the answer in part (b).
Notice that \(g(f(x))\) is not the same as \(f(g(x))\).
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Consider the functions \(f(x)=x^2+2x\) and \(g(x)=x^3-4x\).
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Compute \(g(3)\) and \(f(g(3))\).
\(g(3)=15\)
\(f(g(3))=255\)We replace the \(x\) in \(g(x)=x^3-4x\) by \(3\): \[ g(3)=3^3-4\cdot3=27-12=15 \] Next we replace the \(x\) in \(f(x)=x^2+2x\) by \(g(3)=15\): \[ f(g(3))=(15)^2+2(15)=225+30=255 \]
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Compute \(f(3)\) and \(g(f(3))\).
\(f(3)=15\)
\(g(f(3))=3315\)We replace the \(x\) in \(f(x)=x^2+2x\) by \(3\): \[ f(3)=3^2+2\cdot3=9+6=15 \] Next we replace the \(x\) in \(g(x)=x^3-4x\) by \(f(3)=15\): \[ g(f(3))=(15)^3-4(15)=3375-60=3315 \]
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Notice that \(g(f(3))\) is not the same as \(f(g(3))\).
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Compute \(f(g(x))\).
\(f(g(x))=x^6-8x^4+2x^3+16x^2-8x\)
We replace the \(x\) in \(f(x)=x^2+2x\) by \(g(x)=x^3-4x\): \[\begin{aligned} f(g(x))&=f(x^3-4x) \\ &=(x^3-4x)^2+2(x^3-4x) \\ &=x^6-8x^4+16x^2+2x^3-8x \\ &=x^6-8x^4+2x^3+16x^2-8x \end{aligned}\]
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If \(f(g(x))=x^6-8x^4+2x^3+16x^2-8x\), then \[\begin{aligned} f(g(3))&=3^6-8\cdot3^4+2\cdot3^3+16\cdot3^2-8\cdot3 \\ &=729-648+54+144-24 =255 \end{aligned}\] which agrees with the answer in part (a).
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Compute \(g(f(x))\).
\(g(f(x))=x^6+6x^5+12x^4+8x^3-4x^2-8x\)
We replace the \(x\) in \(g(x)=x^3-4x\) by \(f(x)=x^2+2x\): \[\begin{aligned} g(f(x))&=g(x^2+2x) \\ &=(x^2+2x)^3-4(x^2+2x) \\ &=x^6+6x^5+12x^4+8x^3-4x^2-8x \end{aligned}\]
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If \(g(f(x))=x^6+6x^5+12x^4+8x^3-4x^2-8x\), then \[\begin{aligned} g(f(3))&=3^6+6\cdot3^5+12\cdot3^4+8\cdot3^3-4\cdot3^2-8\cdot3 \\ &=729+1458+972+216-36-24 =3315 \end{aligned}\] which agreees with the answer in part (b).
Notice that \(g(f(x))\) is not the same as \(f(g(x))\).
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Consider the functions \(f(u)=\cos(\pi u)\) and
\(g(u)=\sin(\pi u)\).
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Compute \(g\left(\dfrac{1}{6}\right)\) and \(f\left(g\left(\dfrac{1}{6}\right)\right)\).
\(g\left(\dfrac{1}{6}\right)=\dfrac{1}{2}\)
\(f\left(g\left(\dfrac{1}{6}\right)\right)=0\)We replace the \(u\) in \(g(u)=\sin(\pi u)\) by \(\dfrac{1}{6}\): \[ g\left(\dfrac{1}{6}\right)=\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2} \] Next we replace the \(u\) in \(f(u)=\cos(\pi u)\) by \(g\left(\dfrac{1}{6}\right)=\dfrac{1}{2}\): \[ f\left(g\left(\dfrac{1}{6}\right)\right)=\cos\left(\frac{\pi}{2}\right)=0 \]
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Compute \(f\left(\dfrac{1}{6}\right)\) and \(g\left(f\left(\dfrac{1}{6}\right)\right)\).
\(f\left(\dfrac{1}{6}\right)=\dfrac{\sqrt{3}}{2}\)
\(g\left(f\left(\dfrac{1}{6}\right)\right)=\sin\left({\dfrac{\pi\sqrt{3}}{2}}\right)\)We replace the \(u\) in \(f(u)=\cos(\pi u)\) by \(\dfrac{1}{6}\): \[ f\left(\dfrac{1}{6}\right)=\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2} \] Next we replace the \(u\) in \(g(u)=\sin(\pi u)\) by \(f\left(\dfrac{1}{6}\right)=\dfrac{\sqrt{3}}{2}\): \[ g\left(f\left(\dfrac{1}{6}\right)\right)=\sin\left(\dfrac{\pi\sqrt{3}}{2}\right) \]
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Notice that \(g\left(f\left(\dfrac{1}{6}\right)\right)\) is not the same as \(f\left(g\left(\dfrac{1}{6}\right)\right)\).
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Compute \(f(g(u))\).
\(f(g(u))=\cos(\pi\sin(\pi u))\)
We replace the \(u\) in \(f(u)=\cos(\pi u)\) by \(g(u)=\sin(\pi u)\): \[\begin{aligned} f(g(u))=\cos(\pi\sin(\pi u)) \end{aligned}\]
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If \(f(g(u))=\cos(\pi\sin(\pi u))\), then \[\begin{aligned} f\left(g\left(\dfrac{1}{6}\right)\right)&=\cos\left(\pi\sin\left(\dfrac{\pi}{6}\right)\right) \\ &=\cos\left(\dfrac{\pi}{2}\right) \\ &=0\end{aligned}\] which agrees with the answer in part (a).
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Compute \(g(f(u))\).
\(g(f(u))=\sin(\pi\cos(\pi u))\)
We replace the \(u\) in \(g(u)=\sin(\pi u)\) by \(f(u)=\cos(\pi u)\): \[\begin{aligned} g(f(u))&=\sin(\pi\cos(\pi u)) \end{aligned}\]
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If \(g(f(u))=\sin(\pi\cos(\pi u))\), then \[\begin{aligned} g\left(f\left(\dfrac{1}{6}\right)\right)&=\sin\left(\pi\cos\left(\dfrac{\pi}{6}\right)\right) \\ &=\sin\left(\dfrac{\pi\sqrt{3}}{2}\right) \end{aligned}\] which agreees with the answer in part (b).
Notice that \(g(f(x))\) is not the same as \(f(g(x))\).
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Consider the functions \(f(x)\) and \(g(x)\) whose graphs are shown below.
\(f(x)\) 
\(g(x)\) To enlarge the plot, press Ctrl-+. To return, press Ctrl-- or Ctrl-0.
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Approximate \(g(2)\) and \(f(g(2))\).
\(g(2)=5\)
\(f(g(2))=21\)We look at the graph of \(g(x)\) at \(x=2\) and read off \(g(2)=5\).
Next, we look at the graph of \(f(x)\) at \(x=g(2)=5\) and read off \(f(g(2))=f(5)=21\). -
Approximate \(f(2)\) and \(g(f(2))\).
\(f(2)=0\)
\(g(f(2))=9\)We look at the graph of \(f(x)\) at \(x=2\) and read off \(f(2)=0\).
Next, we look at the graph of \(g(x)\) at \(x=f(2)=0\) and read off \(g(f(2))=g(0)=9\).Notice that \(g(f(2))=9\) is not the same as \(f(g(2))=21\).
Consider the functions \(f(x)\) and \(g(x)\) whose graphs are shown below.
\(f(x)\) 
\(g(x)\) To enlarge the plot, press Ctrl-+. To return, press Ctrl-- or Ctrl-0.
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Approximate \(g(-2)\) and \(f(g(-2))\).
\(g(-2)=1\)
\(f(g(-2))=24\)We look at the graph of \(g(x)\) at \(x=-2\) and read off \(g(-2)=1\).
Next, we look at the graph of \(f(x)\) at \(x=g(-2)=1\) and read off \(f(g(-2))=f(1)=24\). -
Approximate \(f(-2)\) and \(g(f(-2))\).
\(f(-2)=3\)
\(g(f(-2))=-4\)We look at the graph of \(f(x)\) at \(x=-2\) and read off \(f(-2)=3\).
Next, we look at the graph of \(g(x)\) at \(x=f(-2)=3\) and read off \(g(f(-2))=g(3)=-4\).Notice that \(g(f(-2))\) is not the same as \(f(g(-2))\).
Find the derivative of \(y\) with respect to \(t\) for \(y=4x\) and \(x=7t\).
\(\dfrac{dy}{dt}=28\)
The chain rule gives \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}\).
First, we find the information we need for the formula: \[ \dfrac{dy}{dx}=4\qquad\dfrac{dx}{dt}=7 \]
We then plug these values into the formula for the chain rule: \[\begin{aligned} \dfrac{dy}{dt}&=4\cdot7=28 \end{aligned}\]ad
Since we have complete information, we check by computing the composition and taking its derivative directly: \[\begin{aligned} y(x(t))&=4(7t)=28t \\ \dfrac{dy}{dt}&=28 \end{aligned}\]
Find the derivative of \(y\) with respect to \(t\) for \(y=5x^2+2x+3\) and \(x=3t^2+4\).
\(\dfrac{dy}{dt}=180t^3+252t\)
The chain rule gives \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}\).
First, we find the information we need for the formula: \[ \dfrac{dy}{dx}=10x+2\qquad\dfrac{dx}{dt}=6t \]
We then plug these values into the formula for the chain rule: \[ \dfrac{dy}{dt}=(10x+2)6t=10x(6t)+2(6t) \]
Then we remember that the derivative of the outer function needs to be evaluated at the inner function. \[\begin{aligned} \dfrac{dy}{dt}&=(10(3t^2+4)+2)(6t)=(30t^2+40+2)(6t) \\ &=6t(30t^2+40+2) \\ &=180t^3+240t+12t \\ &=180t^3+252t \end{aligned}\]ad
Since we have complete information, we check by computing the composition and taking its derivative directly: \[\begin{aligned} y(x(t))&=5(3t^2+4)^2+2(3t^2+4)+3 \\ &=5(9t^4+24t^2+16)+6t^2+8+3 \\ &=45t^4+120t^2+80+6t^2+11 \\ &=45t^4+126t^2+91 \\ \dfrac{dy}{dt}&=180t^3+252t \end{aligned}\]
Consider the functions \(f(x)\) and \(g(x)\) whose graphs are shown below. Estimate \((f\circ g)'(2)\).
\(f(x)\)
\(g(x)\) To enlarge the plot, press Ctrl-+. To return, press Ctrl-- or Ctrl-0.
Approximately, what are \(g(2)\) and \(g'(2)\)? Then what is \(f'(g(2))\)?
\(\displaystyle f'(g(2))\approx-40\)
Looking at the graph of \(g\), we see \(g(2)\approx5\). Drawing a tangent line to \(g\) at \(x=2\), we see \(g'(2)\approx-4\).
\(f(x) \,\text{and}\, f_{\tan}(x)\) 
\(g(x) \,\text{and}\, g_{\tan}(x)\) Looking at the graph of \(f\), we draw a tangent line at \(x=g(2)=5\). Its slope is approximately \(f'(g(2))=f'(5)\approx10\). So \[ (f\circ g)'(2)=f'(g(2))g'(2)\approx(10)(-4)=-40 \]
Consider the functions \(f(x)\) and \(g(x)\) whose graphs are shown below. Estimate \((f\circ g)'(-2)\).
\(f(x)\)
\(g(x)\) To enlarge the plot, press Ctrl-+. To return, press Ctrl-- or Ctrl-0.
Approximately, what are \(g(-2)\) and \(g'(-2)\)? Then what is \(f'(g(-2))\)?
\(\displaystyle (f\circ g)'(-2)\approx40\)
Looking at the graph of \(g\), we see \(g(-2)\approx1\). Drawing a tangent line to \(g\) at \(x=-2\), we see \(g'(-2)\approx4\).
\(f(x) \,\text{and}\, f_{\tan}(x)\) 
\(g(x) \,\text{and}\, g_{\tan}(x)\) Looking at the graph of \(f\), we draw a tangent line at \(x=g(-2)=1\). Its slope is approximately \(f'(g(-2))=f'(1)\approx10\). So \[ (f\circ g)'(-2)=f'(g(-2))g'(-2)\approx(10)(4)=40 \]
Find the derivative of \(z\) with respect to \(x\) for \(z=\cos(y)\) and \(y=\sin(x)\).
\(\dfrac{dz}{dx}=-\sin(\sin(x))\cos(x)\)
The chain rule gives \(\dfrac{dz}{dx}=\dfrac{dz}{dy}\dfrac{dy}{dx}\).
First, we find the information we need for the formula: \[ \dfrac{dz}{dy}=-\sin(y)\qquad\dfrac{dy}{dx}=\cos(x) \]
We then plug these values into the formula for the chain rule: \[ \dfrac{dz}{dx}=-\sin(y)\cos(x) \]
Then we remember that the derivative of the outer function needs to be evaluated at the inner function. \[ \dfrac{dz}{dx}=-\sin(\sin(x))\cos(x) \]ad
Find \(\dfrac{dy}{dt}\) if \(y=3x^2+11x\) and \(x=4t^2+3t+6\).
\(\dfrac{dy}{dt}=192t^3+216t^2+430t+141\)
The chain rule gives \(\dfrac{dy}{dt}=\left.\dfrac{dy}{dx}\right|_{x(t)}\dfrac{dx}{dt}\).
First, we find the two derivatives: \[ \dfrac{dy}{dx}=6x+11\qquad\dfrac{dx}{dt}=8t+3 \] and evaluate the outer derivative at the inner function: \[ \left.\dfrac{dy}{dx}\right|_{x(t)}=6(4t^2+3t+6)+11 =24t^2+18t+47 \] We then plug these values into the formula for the chain rule: \[\begin{aligned} \dfrac{dy}{dt}&=(24t^2+18t+47)(8t+3) \\ &=192t^3+216t^2+430t+141 \end{aligned}\]ad
Since we have complete information, we check by computing the composition and taking its derivative directly: \[\begin{aligned} y(x(t))&=3(4t^2+3t+6)^2+11(4t^2+3t+6) \\ &=3(16t^4+9t^2+36+24t^3+48t^2+36t)+44t^2+33t+66 \\ &=48t^4+27t^2+108+72t^3+144t^2+108t+44t^2+33t+66 \\ &=48t^4+72t^3+215t^2+141t+174 \\ \dfrac{dy}{dt}&=192t^3+216t^2+430t+141 \end{aligned}\]
Find \(\dfrac{dy}{dt}\) if \(y=x^3+2x^2+x+2\) and \(x=2t^2+5t\).
\(\dfrac{dy}{dt}=48t^5+300t^4+632t^3+495t^2+104t+5\)
The chain rule gives \(\dfrac{dy}{dt}= \left.\dfrac{dy}{dx}\right|_{x(t)}\dfrac{dx}{dt}\).
First, we find the two derivatives: \[ \dfrac{dy}{dx}=3x^2+4x+1\qquad\dfrac{dx}{dt}=4t+5 \]
and evaluate the outer derivative at the inner function: \[\begin{aligned} \left.\dfrac{dy}{dx}\right|_{x(t)}&=3(2t^2+5t)^2+4(2t^2+5t)+1 \\ &=3(4t^4+20t^3+25t^2)+8t^2+20t+1 \\ &=12t^4+60t^3+75t^2+8t^2+20t+1 \\ &=12t^4+60t^3+83t^2+20t+1 \end{aligned}\]
We then plug these values into the formula for the chain rule: \[\begin{aligned} \dfrac{dy}{dt}&=(12t^4+60t^3+83t^2+20t+1)(4t+5) \\ &=48t^5+240t^4+332t^3+80t^2+4t+60t^4+300t^3+415t^2+100t+5 \\ &=48t^5+300t^4+632t^3+495t^2+104t+5 \end{aligned}\]ad
Since we have complete information, we check by computing the composition and taking its derivative directly: \[\begin{aligned} y(x(t))&=(2t^2+5t)^3+2(2t^2+5t)^2+2t^2+5t+2 \\ &=8t^6+60t^5+150t^4+125t^3+2(4t^4+20t^3+25t^2)+2t^2+5t+2 \\ &=8t^6+60t^5+150t^4+125t^3+8t^4+40t^3+50t^2+2t^2+5t+2 \\ &=8t^6+60t^5+158t^4+165t^3+52t^2+5t+2 \\ \dfrac{dy}{dt}&=48t^5+300t^4+632t^3+495t^2+104t+5 \end{aligned}\]
Find the derivative of \(y=\cos(\pi\sin(\pi x))\).
What are the outer and inner functions?
\(\dfrac{dy}{dx}=-\pi^2\sin(\pi \sin(\pi x))\cos(\pi x)\)
We identify the outer function as \(y(u)=\cos(\pi u)\) and the inner function as \(u(x)=\sin(\pi x)\). Then the chain rule gives: \[ \dfrac{dy}{dx}=\left.\dfrac{dy}{du}\right|_{u(x)}\dfrac{du}{dx} \] First, we find the derivatives: \[ \dfrac{dy}{du}=-\pi\sin(\pi u) \qquad \dfrac{du}{dx}=\pi\cos(\pi x) \] and evaluate the derivative of the outer function at the inner function: \[ \left.\dfrac{dy}{du}\right|_{u(x)}=-\pi\sin(\pi \sin(\pi x)) \] We use the chain rule: \[\begin{aligned} \dfrac{dy}{dx}&=-\pi\sin(\pi \sin(\pi x))\pi\cos(\pi x) \\ &=-\pi^2\sin(\pi \sin(\pi x))\cos(\pi x) \end{aligned}\]
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Find the derivative of \(y\) with respect to \(t\) for \(y=e^x\) and \(x=5t\).
\(\dfrac{dy}{dt}=5e^{5t}\)
The chain rule gives \(\dfrac{dy}{dt}=\left.\dfrac{dy}{dx}\right|_{x(t)} \dfrac{dx}{dt}\).
First, we find the derivatives: \[ \dfrac{dy}{dx}=e^x\qquad\dfrac{dx}{dt}=5 \] and evaluate the outer derivative at the inner function: \[ \left.\dfrac{dy}{dx}\right|_{x(t)}=e^{5t} \] We then plug these values into the formula for the chain rule: \[ \dfrac{dy}{dt}=e^{5t}5=5e^{5t} \]ad
Find the derivative of \(y\) with respect to \(t\) for \(y=7^x\) and \(x=5t^2+4t\).
\(\dfrac{dy}{dt}=\ln(7)7^{5t^2+4t}(10t+4)\)
The chain rule gives \(\dfrac{dy}{dt}=\left.\dfrac{dy}{dx}\right|_{x(t)}\dfrac{dx}{dt}\).
First, we find the derivatives: \[ \dfrac{dy}{dx}=\ln(7)7^x\qquad\dfrac{dx}{dt}=10t+4 \] and evaluate the outer derivative at the inner function: \[ \left.\dfrac{dy}{dx}\right|_{x(t)}=\ln(7)7^{5t^2+4t} \] We then plug these values into the formula for the chain rule: \[ \dfrac{dy}{dt}=\ln(7)7^{5t^2+4t}(10t+4) \]ad
Find the derivative of \(y\) with respect to \(t\) for \(y=4^x\) and \(x=\sin(t)\).
\(\dfrac{dy}{dt}=\cos(t)\ln(4)4^{\sin(t)}\)
The chain rule gives \(\dfrac{dy}{dt}=\left.\dfrac{dy}{dx}\right|_{x(t)} \dfrac{dx}{dt}\).
First, we find the derivatives: \[ \dfrac{dy}{dx}=\ln(4)4^x\qquad\dfrac{dx}{dt}=\cos(t) \]
and evaluate the outer derivative at the inner function: \[ \left.\dfrac{dy}{dx}\right|_{x(t)}=\ln(4)4^{\sin(t)} \]
We then plug these values into the formula for the chain rule: \[ \dfrac{dy}{dt}=\cos(t)\ln(4)4^{\sin(t)} \]ad
Differentiate \(f(x)=(x^3-4x+7)^{15}\).
Think of two functions, neither of them identical to \(f\), that when you take their composite, it makes \(f\). Use those functions to find the answer.
\(\dfrac{df}{dx}=15(x^3-4x+7)^{14}(3x^2-4)\)
We identify the outer function as \(f(y)=y^{15}\) and the inner function as \(y(x)=x^3-4x+7\). Then the chain rule gives \[ \dfrac{df}{dx}=\left.\dfrac{df}{dy}\right|_{y(x)}\dfrac{dy}{dx} \] First, we find the derivatives: \[ \dfrac{df}{dy}=15y^{14}\qquad\dfrac{dy}{dx}=3x^2-4 \] and evaluate the derivative of the outer function at the inner function: \[ \left.\dfrac{df}{dy}\right|_{y(x)}=15(x^3-4x+7)^{14} \] We use the chain rule: \[ \dfrac{df}{dx}=15(x^3-4x+7)^{14}(3x^2-4) \]
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Differentiate \(g(y)=\cos(y^2)\).
\(g\) is a composition of two functions, namely an outer function and an inner function. Figure out those two functions.
\(g'(y)=-2y\sin(y^2)\)
We identify the outer function as \(g(u)=\cos(u)\) and the inner function as \(u(y)=y^2\). Their derivatives are \[ g'(u)=-\sin(u) \qquad \text{and} \qquad u'(y)=2y \] Then the chain rule gives: \[\begin{aligned} (g\circ u)'(y)&=g'(u(y))u'(y) \\ &=-\sin(y^2)2y =-2y\sin(y^2) \end{aligned}\]
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Differentiate \(g(y)=\cos^2(y)\).
\(g'(y)=-2\sin(y)\cos(y)=-\sin(2y)\)
We identify the outer function as \(g(u)=u^2\) and the inner function as \(u(y)=\cos(y)\). Their derivatives are \[ g'(u)=2u \qquad \text{and} \qquad u'(y)=-\sin(u) \] Then the chain rule gives: \[\begin{aligned} (g\circ u)'(y)&=g'(u(y))u'(y) =2\cos(y)(-\sin(y)) \\ &=-2\sin(y)\cos(y) =-\sin(2y) \end{aligned}\] where in the last step we used the trig identity \(\sin(2y)=2\sin(y)\cos(y)\).
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Differentiate \(g(y)=\cos^2(y^2)\).
This is a double chain rule with an outer function, an intermediate function and an inner function.
\(\dfrac{dg}{dy}=-4y\cos(y^2)\sin(y^2) =-2y\sin(2y^2)\)
This is a double chain rule. The outer function is \(g(u)=u^2\). The intermediate function is \(u(z)=\cos(z)\) and the inner function as \(z(y)=y^2\). Their derivatives are: \[ g'(u)=2u \qquad u'(z)=-\sin(z) \qquad z'(y)=2y \] Then the chain rule gives \[\begin{aligned} (g\circ u\circ z)'(y)&=g'(u(z(y)))u'(z(y))z'(y) \\ &=2(\cos(y^2))(-\sin(y^2))2y \\ &=-4y\cos(y^2)\sin(y^2) =-2y\sin(2y^2) \end{aligned}\]
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Differentiate \(F(\theta)=\tan^2(7\theta)\).
This is a double chain rule with an outer function, an intermediate function and an inner function.
\(F'(\theta)=14\tan(7\theta)\sec^2(7\theta)\)
This is a double chain rule. The outer function is \(F(u)=u^2\). The intermediate function is \(u(j)=\tan(j)\) and the inner function as \(j(\theta)=7\theta\). So that \[ (F\circ u\circ j)(y)=(\tan(7\theta))^2 \] Then their derivatives are: \[ F'(u)=2u \qquad u'(j)=\sec^2(j) \qquad j'(\theta)=7 \] And the chain rule gives: \[\begin{aligned} (F\circ u\circ j)'(\theta)&=F'(u(j(\theta)))u'(j(\theta))j'(\theta) \\ &=2(\tan(7\theta))(\sec^2(7\theta))7 \\ &=14\tan(7\theta)\sec^2(7\theta) \end{aligned}\]
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Differentiate \(w(\theta)=\csc^2(\cot(\theta))\).
This is a double chain rule with an outer function, an intermediate function and an inner function.
\(\dfrac{dw}{d\theta}=2\cot(\cot(\theta))\csc^2(\cot(\theta))\csc^2(\theta)\)
This is a double chain rule. The outer function is squaring and the intermediate function is \(\csc\) and the inner function is \(\cot\). To apply the chain rule we differentiate the outer function and evaluate it at the intermediate function composed with the inner function and then multiply by the derivative of the intermediate function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[\begin{aligned} w'(\theta)&=2\csc(\cot\theta)\cdot(-\csc(\cot\theta)\cot(\cot\theta))\cdot (-\csc^2\theta) \\ &=2\cot(\cot\theta)\csc^2(\cot\theta)\csc^2\theta \end{aligned}\]
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Differentiate \(f(x)=\sin^2(x)\).
\(f'(x)=2\sin(x)\cos(x)=\sin(2x)\)
The outer function is squaring and the inner function is \(\sin\). To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ f'(x)=2\sin(x)\cos(x) =\sin(2x) \]
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Differentiate \(z(x)=\sin^2(\cos^2(x))\).
\(\begin{aligned} z'(x)&=-4\sin(\cos^2(x))\cos(\cos^2(x))\cos(x)\sin(x) \\ &=-\sin(2\cos^2(x))\sin(2x) \end{aligned}\)
It is useful to rewrite the function as: \[ z(x)=[\sin([\cos(x)]^2)]^2 \] There are \(4\) levels of functions: squaring, \(\sin\), squaring and \(\cos\). To apply the chain rule we differentiate each level of function and multiply them: \[\begin{aligned} z'(x)&=2[\sin([\cos(x)]^2)]\cdot[\cos([\cos(x)]^2)]\cdot2[\cos(x)]\cdot[-\sin(x)] \\ &=-4\sin(\cos^2(x))\cos(\cos^2(x))\cos(x)\sin(x) \\ &=-\sin(2\cos^2(x))\sin(2x) \end{aligned}\]
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Differentiate \(g(y)=\cot^2(y)\).
\(g'(y)=-2\cot(y)\csc^2(y)\)
The outer function is squaring and the inner function is \(\cot\). To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ g'(y)=2\cot(y)\cdot(-\csc^2(y))=-2\cot(y)\csc^2(y) \]
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If \(z(t)=\cot^2(\cos^2(t))\), find \(\dfrac{dz}{dt}.\)
In previous problems we found: \[ \dfrac{d}{dy}(\cot^2(y))=-2\cot(y)\csc^2(y) \] \[ \dfrac{d}{dy}(\cos^2(y))=-\sin(2y) \] So use these as the outer and inner functions.
\(z'(t)=2\cot(\cos^2(t))\csc^2(\cos^2(t))\sin(2t)\)
In previous problems we found: \[ \dfrac{d}{dy}(\cot^2(y))=-2\cot(y)\csc^2(y) \] \[ \dfrac{d}{dy}(\cos^2(y))=-\sin(2y) \] We use these as the outer and inner functions. To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[\begin{aligned} z'(t)&=-2\cot(\cos^2(t))\csc^2(\cos^2(t))\cdot(-\sin(2t)) \\ &=2\cot(\cos^2(t))\csc^2(\cos^2(t))\sin(2t) \end{aligned}\]
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If \(f(x)=\sin(5^x)\), find \(\dfrac{df}{dx}\).
\(f'(x)=5^x\ln(5)\cos(5^x)\)
The outer function is \(\sin\) and the inner function is \(5^x\). To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ f'(x)=\cos(5^x)\ln(5)5^x=5^x\ln(5)\cos(5^x) \]
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If \(g(y)=\sec(\sin(y))\), find \(\dfrac{dg}{dy}\).
\(g'(y)=\sec(\sin(y))\tan(\sin(y))\cos(y)\)
The outer function is \(\sec\) and the inner function is \(\sin\). To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ g'(y)=\sec(\sin(y))\tan(\sin(y))\cos(y) \]
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If \(f(y)=\sin(y^2+2y+2)\), find \(\dfrac{df}{dy}\).
\(f'(y)=\cos(y^2+2y+2)(2y+2)\)
The outer function is \(\sin\) and the inner function is \(y^2+2y+2\). To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ f'(y)=\cos(y^2+2y+2)(2y+2) \]
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If \(g(x)=\cot(\sin(x))\), find \(\dfrac{dg}{dx}\).
\(g'(x)=-\csc^2(\sin(x))\cos(x)\)
The outer function is \(\cot\) and the inner function is \(\sin\). To apply the chain rule we differentiate the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ g'(x)=-\csc^2(\sin(x))\cos(x) \]
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If \(f(x)=\tan((x^2+2x+2)^{50})\), find \(\dfrac{df}{dx}\).
\(f'(x)=100(x+1)(x^2+2x+2)^{49}\sec^2((x^2+2x+2)^{50})\)
This is a double chain rule. The outer function is \(\tan\) and the intermediate function is \(u^{50}\) and the inner function is \(u=x^2+2x+2\). To apply the chain rule we differentiate the outer function and evaluate it at the intermediate function composed with the inner function and then multiply by the derivative of the intermediate function at the inner function and then multiply by the derivative of the inner function. \[\begin{aligned} f'(x)&=\sec^2((x^2+2x+2)^{50})50(x^2+2x+2)^{49}(2x+2) \\ &=100(x+1)(x^2+2x+2)^{49}\sec^2((x^2+2x+2)^{50}) \end{aligned}\]
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If \(r(\theta)=(\sin^3(\theta)+\cos(2\theta))^{12}\), find \(\dfrac{dr}{d\theta}\).
Use the chain rule, then the sum rule and the chain rule.
\(r'(\theta)=12[\sin^3(\theta)+\cos(2\theta)]^{11}[3\sin^2(\theta) \cos(\theta)-2\sin(2\theta)]\)
The outer function is \(x^{12}\) and the inner function is \(x=\sin^3(\theta)+\cos(2\theta)\). The derivative of the outer function is \(12x^{11}\). To calculate the derivative of the inner function we use the Sum Rule and then the Chain Rule: \[\begin{aligned} \dfrac{dx}{d\theta}&=\dfrac{d}{d\theta}(\sin^3(\theta)+\cos(2\theta)) \\ &=\dfrac{d}{d\theta}(\sin^3(\theta))+\dfrac{d}{d\theta}(\cos(2\theta)) \\ &=3\sin^2(\theta)\cos(\theta)-2\sin(2\theta) \end{aligned}\] To apply the chain rule to the function \(r\) we take the derivative of the outer function and evaluate it at the inner function and then multiply by the derivative of the inner function: \[ r'(\theta)=12[\sin^3(\theta)+\cos(2\theta)]^{11}[3\sin^2(\theta) \cos(\theta)-2\sin(2\theta)] \]
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If \(f(x)=e^{\tan(\sin(x))+\cos^4(x)}\), find \(\dfrac{df}{dx}\).
Use the chain rule, then the sum rule and the chain rule.
\(f'(x)=e^{\tan(\sin(x))+\cos^4(x)}(\sec^2(\sin(x))\cos(x)-4\cos^3(x) \sin(x))\)
The outer function is \(e^y\) and the inner function is \(y=\tan(\sin( x))+\cos^4(x)\). The derivative of the outer function is \(e^y\). To calculate the derivative of the inner function we use the Sum rule and then the Chain rule: \[\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}(\tan(\sin(x))+\cos^4(x)) \\ &=\dfrac{d}{dx}(\tan(\sin(x)))+\dfrac{d}{dx}(\cos^4(x)) \\ &=\sec^2(\sin(x))\cos(x)-4\cos^3(x)\sin(x) \end{aligned}\] To apply the chain rule to the function \(f\) we take the derivative of the outer function and evaluate it at the inner function and multiply by the derivative of the inner function: \[ f'(x)=e^{\tan(\sin(x))+\cos^4(x)}(\sec^2(\sin(x))\cos(x)-4\cos^3(x)\sin( x)) \]
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If \(f(x)=(5^{\sin(x)}+e^{\cot(x)})^{100}\), find \(f'(x)\).
\(f'(x)=100(5^{\sin(x)}+e^{\cot(x)})^{99}[5^{\sin(x)}\ln(5)\cos(x)-e^{ \cot(x)}\csc^2(x)]\)
The outer function is \(u^{100}\) and the inner function is \(u=5^{\sin( x)}+e^{\cot(x)}\). The derivative of the outer function is \(100u^{99}\). To calculate the derivative of the inner function we use the Sum rule and then the Chain rule: \[\begin{aligned} u'(x)&=\dfrac{d}{dx}(5^{\sin(x)}+e^{\cot(x)}) \\ &=\dfrac{d}{dx}(5^{\sin(x)})+\dfrac{d}{dx}(e^{\cot(x)}) \\ &=5^{\sin(x)}\ln(5)\cos(x)-e^{\cot(x)}\csc^2(x) \end{aligned}\] To apply the chain rule to the function \(f\) we take the derivative of the outer function and evaluate it at the inner function and multiply by the derivative of the inner function: \[ f'(x)=100(5^{\sin(x)}+e^{\cot(x)})^{99}[5^{\sin(x)}\ln(5)\cos(x) -e^{\cot(x)}\csc^2(x)] \]
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If \(f(x)=\sin(\sin(x)\tan(2x))\), find \(f'(x)\).
Use the chain rule, then the product rule and the chain rule.
\(f'(x)=\cos(\sin(x)\tan(2x))[2\sin(x)\sec^2(2x)+\tan(2x)\cos(x)]\)
The outer function is \(\sin\) and the inner function is \(\sin(x)\tan(2 x)\). The derivative of the outer function is \(\cos\). To calculate the derivative of the inner function we use the Product rule and then the Chain rule: \[\begin{aligned} \dfrac{d}{dx}(\sin(x)\tan(2x))&=\sin(x)\dfrac{d}{dx}(\tan(2x))+\tan(2x) \dfrac{d}{dx}(\sin(x)) \\ &=2\sin(x)\sec^2(2x)+\tan(2x)\cos(x) \end{aligned}\] To apply the chain rule to the function \(f\) we take the derivative of the outer function and evaluate it at the inner function and multiply by the derivative of the inner function: \[ f'(x)=\cos(\sin(x)\tan(2x))[2\sin(x)\sec^2(2x)+\tan(2x)\cos(x)] \]
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If \(f(x)=\sin(3^{\sin(x)}\cos(\cos(x)))\), find \(f'(x)\).
Use the chain rule, then the product rule and the chain rule.
\(\begin{aligned} f'(x)&=\cos(3^{\sin(x)}\cos(\cos(x)))3^{\sin(x)} \\ &\quad[\sin(x)\sin(\cos(x))+\ln(3)\cos(x)\cos(\cos(x))] \end{aligned}\)
The outer function is \(\sin\) and the inner function is \(3^{\sin(x)} \cos(\cos(x))\). The derivative of the outer function is \(\cos\). To calculate the derivative of the inner function we use the Product rule and then the Chain rule: \[\begin{aligned} &\dfrac{d}{dx}(3^{\sin(x)}\cos(\cos(x))) \\ &=3^{\sin(x)}\dfrac{d}{dx}(\cos(\cos(x)))+\cos(\cos(x))\dfrac{d}{dx}(3^{ \sin(x)}) \\ &=3^{\sin(x)}\sin(\cos(x))\sin(x)+\cos(\cos(x))3^{\sin(x)}\ln(3)\cos(x) \\ &=3^{\sin(x)}[\sin(x)\sin(\cos(x))+\ln(3)\cos(x)\cos(\cos(x))] \end{aligned}\] To apply the chain rule to the function \(f\), we take the derivative of the outer function evaluated at the inner function and then multiply by the derivative of the inner function: \[\begin{aligned} f'(x)&=\cos(3^{\sin(x)}\cos(\cos(x)))3^{\sin(x)} \\ &\quad[\sin(x)\sin(\cos(x))+\ln(3)\cos(x)\cos(\cos(x))] \end{aligned}\]
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If \(f(x)=\cos\left(\dfrac{3^x}{\sin(x^5)}\right)\), find \(f'(x)\).
Use the chain rule, then the quotient rule and the chain rule.
\(f'(x)=-\sin\left(\dfrac{3^x}{\sin(x^5)}\right)\dfrac{3^x\ln(3)\tan(x^5 )-5x^43^x\cos(x^5)}{\sin^2(x^5)}\)
The outer function is \(\cos\) and the inner function is \(\dfrac{3^x}{ \sin(x^5)}\). The derivative of the outer function is \(-\sin\). To calculate the derivative of the inner function we use the Quotient rule and then the Chain rule: \[\begin{aligned} \dfrac{d}{dx}\left(\dfrac{3^x}{\sin(x^5)}\right) &=\dfrac{\sin(x^5)\dfrac{d}{dx}(3^x)-3^x\dfrac{d}{dx}(\sin(x^5))}{\sin^2(x^5)} \\ &=\dfrac{3^x\ln(3)\sin(x^5)-5x^43^x\cos(x^5)}{\sin^2(x^5)} \end{aligned}\] To apply the chain rule to the function \(f\) we take the derivative of the outer function evaluated at the inner function and multiply by the derivative of the inner function: \[ f'(x)=-\sin\left(\dfrac{3^x}{\sin(x^5)}\right)\dfrac{3^x\ln(3)\tan(x^5)- 5x^43^x\cos(x^5)}{\sin^2(x^5)} \]
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If \(u(v)=\sin^{23}\left(3^x+\dfrac{v^4}{\sin(2v)}\right)\), find \(u'(v )\).
Use the chain, sum, quotient, and chain rules. This is also a double chain rule.
\(\begin{aligned} u'(v)&=23\sin^{22}\left(3^x+\dfrac{v^4}{\sin(2v)}\right) \cos\left(3^x+\dfrac{v^4}{\sin(2v)}\right) \\ &\quad\left(3^x\ln(3)+\dfrac{4v^3\sin(2v)-2v^4\cos(2v)}{\sin^2(2v)}\right) \end{aligned}\)
This is a double chain rule. The outer function is \(s^{23}\) and the intermediate function is \(s=\sin(l)\) and the inner function is \(l=3^x +\dfrac{v^4}{\sin(2v)}\). The derivatives of the outer and intermediate functions are \(23s^{22}\) and \(\cos\), respectively. To calculate the derivative of the inner function we use the Sum Rule, Exponential Rule, Quotient Rule and then the Chain Rule: \[\begin{aligned} l'(v)&=\dfrac{d}{dv}\left(3^x+\dfrac{v^4}{\sin(2v)}\right) \\ &=\dfrac{d}{dv}(3^x)+\dfrac{d}{dv}\left(\dfrac{v^4}{\sin(2v)}\right) \\ &=3^x\ln(3)+\dfrac{\sin(2v)\dfrac{d}{dv}(v^4)-v^4\dfrac{d}{dv}(\sin(2v)) }{\sin^2(2v)} \\ &=3^x\ln(3)+\dfrac{4v^3\sin(2v)-2v^4\cos(2v)}{\sin^2(2v)} \end{aligned}\] To apply the chain rule to the function \(u\) we differentiate each level of function and multiply them: \[\begin{aligned} u'(v)&=23\sin^{22}\left(3^x+\dfrac{v^4}{\sin(2v)}\right) \cos\left(3^x+\dfrac{v^4}{\sin(2v)}\right) \\ &\quad\left(3^x\ln(3)+\dfrac{4v^3\sin(2v)-2v^4\cos(2v)}{\sin^2(2v)}\right) \end{aligned}\]
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Do the Tutorial until you feel confident.
Find the derivative of an implicitly defined function.
If \(x\) and \(y\) are related by \(x^2+2xy+y=13\), find the derivative of \(y\) with respect to \(x\) at \((1,4)\).
\(\left.\dfrac{dy}{dx}\right|_{(1,4)}=-\dfrac{10}{3}\)
We differentiate both sides with respect to \(x\), using the Sum, Power, and Product rules: \[ 2x+2x\dfrac{dy}{dx}+2y+\dfrac{dy}{dx}=0 \] Now we evaluate at \((x,y)=(1,4)\) and solve for the derivative: \[\begin{aligned} 2+2\dfrac{dy}{dx}+8+\dfrac{dy}{dx}&=0 \\ 10+3\dfrac{dy}{dx}&=0 \\ 3\dfrac{dy}{dx}&=-10 \\ \left.\dfrac{dy}{dx}\right|_{(1,4)}&=-\dfrac{10}{3} \end{aligned}\]
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If \(x\) and \(y\) are related by \(x^3+3x^2y+3xy+y^3=89\), find the derivative of \(x\) with respect to \(y\), \(\dfrac{dx}{dy}\), at \((x,y)=(2,3)\).
\(\left.\dfrac{dx}{dy}\right|_{(2,3)}=-\dfrac{15}{19}\)
We differentiate both sides with respect to \(y\), using the Sum, Power, Product and Chain rules: \[ 3x^2\dfrac{dx}{dy}+3x^2+6xy\dfrac{dx}{dy}+3x+3y\dfrac{dx}{dy}+3y^2=0 \] Now we evaluate at \((x,y)=(2,3)\) and solve for the derivative: \[\begin{aligned} 12\dfrac{dx}{dy}+12+36\dfrac{dx}{dy}&+6+9\dfrac{dx}{dy}+27=0 \\ 57\dfrac{dx}{dy}+45&=0 \\ \left.\dfrac{dx}{dy}\right|_{(2,3)} &=-\dfrac{45}{57}=-\dfrac{15}{19} \end{aligned}\]
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If \(y\) and \(z\) are related by \(y^2+2y\sin\left(\dfrac{\pi z}{2}\right)+\sin(\pi z)=3\), find the derivative of \(z\) with respect to \(y\) at \((y,z)=(3,3)\).
\(\left.\dfrac{dz}{dy}\right|_{(3,3)}=\dfrac{4}{\pi}\)
We apply \(\dfrac{d}{dy}\) to both sides, using the Sum, Power, Product and Chain rules: \[ 2y+\pi y\cos\left(\dfrac{\pi z}{2}\right)\dfrac{dz}{dy} +2\sin\left(\dfrac{\pi z}{2}\right)+\pi\cos(\pi z)\dfrac{dz}{dy}=0 \] Now we evaluate at \((y,z)=(3,3)\) and solve for the derivative: \[\begin{aligned} 6-2-\pi\dfrac{dz}{dy}&=0 \\ 4-\pi\dfrac{dz}{dy}&=0 \\ \left.\dfrac{dz}{dy}\right|_{(3,3)}&=\dfrac{4}{\pi} \end{aligned}\]
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If \(x\) and \(y\) are related by \(x^2y+y^2x=6\), find \(\dfrac{dy}{dx}\) at \((1,2)\).
\(\left.\dfrac{dy}{dx}\right|_{(1,2)}=-\,\dfrac{8}{5}\)
We differentiate both sides, using the Sum, Product, Power and Chain rules: \[ x^2\dfrac{dy}{dx}+2xy+y^2+2xy\dfrac{dy}{dx}=0 \] Now we evaluate at \((x,y)=(1,2)\) and solve for the derivative: \[\begin{aligned} \dfrac{dy}{dx}+4+4+4\dfrac{dy}{dx}&=0 \\ 5\dfrac{dy}{dx}+8&=0 \\ \left.\dfrac{dy}{dx}\right|_{(1,2)}&=-\,\dfrac{8}{5} \end{aligned}\]
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If \(y\) and \(z\) are related by \(y^2+z^2=24-y^2z^2\), find \(\dfrac{dz}{dy}\) at \((y,z)=(2,2)\).
\(\left.\dfrac{dz}{dy}\right|_{(2,2)}=-1\)
We apply \(\dfrac{d}{dy}\) to both sides, using the Sum, Power, Product and Chain rules: \[ 2y+2z\dfrac{dz}{dy}=-2y^2z\dfrac{dz}{dy}-2yz^2 \] Now we evaluate at \((y,z)=(2,2)\) and solve for the derivative: \[\begin{aligned} 4+4\dfrac{dz}{dy}&=-16\dfrac{dz}{dy}-16 \\ 20\dfrac{dz}{dy}&=-20 \\ \left.\dfrac{dz}{dy}\right|_{(2,2)}&=-1 \end{aligned}\]
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If \(x\) and \(z\) are related by \(x^2+z^2+\cos(\pi xz)=14\), find \(\dfrac{dz}{dx}\) at \((x,z)=(2,3)\).
\(\left.\dfrac{dz}{dx}\right|_{(2,3)}=-\dfrac{2}{3}\)
We differentiate both sides, using the Sum, Power, Chain, and Product rules: \[ 2x+2z\dfrac{dz}{dx}-\pi\sin(\pi xz)\left(x\dfrac{dz}{dx}+z\right)=0 \] Now we evaluate at \((x,z)=(2,3)\) and solve for the derivative: \[\begin{aligned} 4+6\dfrac{dz}{dx}&=0 \\ \left.\dfrac{dz}{dx}\right|_{(2,3)}&=-\dfrac{4}{6}=-\dfrac{2}{3} \end{aligned}\]
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If \(x\) and \(z\) are related by \(2^x+2^z=6\), find \(\dfrac{dz}{dx}\) at \((x,z)=(1,2)\).
\(\left.\dfrac{dz}{dx}\right|_{(1,2)}=-\dfrac{1}{2}\)
We differentiate both sides, using the Sum and Exponential rules: \[ 2^x\ln(2)+2^z\ln(2)\dfrac{dz}{dx}=0 \] Now we evaluate at \((x,z)=(1,2)\) and solve for the derivative: \[\begin{aligned} 2\ln(2)+4\ln(2)\dfrac{dz}{dx}&=0 \\ 4\dfrac{dz}{dx}&=-2 \\ \left.\dfrac{dz}{dx}\right|_{(1,2)}&=-\dfrac{1}{2} \end{aligned}\]
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In this equation we can solve for \[ z=\log_2\left(6-2^x\right) \] So we can differentiate explicitly: \[ \dfrac{dz}{dx}=\dfrac{1}{\ln(2)}\dfrac{-2^x\ln(2)}{6-2^x} =\dfrac{-2^x}{6-2^x} \] \[ \left.\dfrac{dz}{dx}\right|_{x=1}=\dfrac{-2}{6-2}=-\dfrac{1}{2} \] But implicit differentiation was easier.
For an ideal gas the pressure \(P\) and the volume \(V\) are related by \(PV=K\) for some constant \(K\). Currently, the pressure of a sample in a box is \(P=1.1\,\text{atm}\) and the volume is \(V=15\,\text{m}^3\). Find the rate of change of the pressure with respect to the volume.
\(\left.\dfrac{dP}{dV}\right|_{(15\,\text{m}^3,1.1\,\text{atm})} =-\,\dfrac{1.1\,\text{atm}}{15\,\text{m}^3} =-.0733\dfrac{\text{atm}}{\text{m}^3}\)
We apply \(\dfrac{d}{dV}\) to both sides using the Product rule, and solve for \(\dfrac{dP}{dV}\): \[\begin{aligned} P+V\dfrac{dP}{dV}&=0 \\ \dfrac{dP}{dV}&=-\,\dfrac{P}{V} \end{aligned}\] Now we evaluate at \((V,P)=(15\,\text{m}^3,1.1\,\text{atm})\): \[ \left.\dfrac{dP}{dV}\right|_{(15\,\text{m}^3,1.1\,\text{atm})} =-\,\dfrac{1.1\,\text{atm}}{15\,\text{m}^3} =-.0733\dfrac{\text{atm}}{\text{m}^3} \]
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In this equation we can solve for \[ P=\dfrac{K}{V} \] So we can differentiate explicitly using the Power rule and substituting for \(P\): \[ \dfrac{dP}{dV}=-\dfrac{K}{V^2}=-\dfrac{P}{V} \]
A boy is running such that the relationship between the distance covered, \(S\), and time, \(t\), is \(S^3+t^3=K\) for some constant \(K\). Currently, the distance traveled is \(S=12\,\text{m}\) and the time is \(t =10\,\text{s}\). Find the rate of change of the distance traveled with respect to time.
\(\left.\dfrac{dS}{dt}\right|_{(10\,\text{m},12\,\text{s})} =-\,\dfrac{36\,\text{s}^2}{25\,\text{m}^2} =-1.44\dfrac{\text{s}^2}{\text{m}^2}\)
We apply \(\dfrac{d}{dt}\) to both sides, using the Sum rule, the Power rule, and the Chain rule, and solve for \(\dfrac{dS}{dt}\): \[\begin{aligned} 3S^2\dfrac{dS}{dt}+3t^2&=0 \\ \dfrac{dS}{dt}&=-\,\dfrac{t^2}{S^2} \end{aligned}\] Now we evaluate at \((t,S)=(10\,\text{m},12\,\text{s})\): \[ \left.\dfrac{dS}{dt}\right|_{(10\,\text{m},12\,\text{s})} =-\,\dfrac{144\,\text{s}^2}{100\,\text{m}^2} =-\,\dfrac{36\,\text{s}^2}{25\,\text{m}^2} =-1.44\dfrac{\text{s}^2}{\text{m}^2} \]
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In this equation we can solve for \[ S=\sqrt[\scriptstyle3]{K-t^3} \] So we can differentiate explicitly using the Chain and Difference and Power rules and substituting for \(S\): \[ \dfrac{dS}{dt} =-\,\dfrac{t^2}{\sqrt[\scriptstyle3]{K-t^3\,}^{\,2}} =-\,\dfrac{t^2}{S^2} \] But implicit differentiation was easier.
Two charges, \(q_1\), and \(q_2\), are repelling each other. The distance between them, \(r\), and the force, \(F\), are related by \(r^2F=kq_1q_2\) and where \(k\) is Coulomb's constant. When \(r=2\,\text{m}\) the force is \(F=15\,\text{N}\), what is the rate of change of \(r\) with respect to \(F\)?
\(\left.\dfrac{dr}{dF}\right|_{((15\,\text{N},2\,\text{m}))} =-\,\dfrac{1}{15}\dfrac{\text{m}}{\text{N}}\)
We apply \(\dfrac{d}{dF}\) to both sides, using the the Power rule and solve for \(\dfrac{dr}{dF}\): \[\begin{aligned} r^2&+F2r\dfrac{dr}{dF}=0 \\ &\dfrac{dr}{dF}=-\,\dfrac{r}{2F} \end{aligned}\] Now we evaluate at \((F,r)=(15\,\text{N},2\,\text{m})\): \[ \left.\dfrac{dr}{dF}\right|_{((15\,\text{N},2\,\text{m}))} =-\,\dfrac{1}{15}\dfrac{\text{m}}{\text{N}} \]
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The position on a rod is given by the position, \(x\). The total mass between \(x=0\) and \(x=a\) is \(m\). They are related by \(\dfrac{m^3}{a^2}=K\) for some constant \(K\). When \(a=3\,\text{m}\), we have \(m=0.02\,\text{kg}\). What is the rate of change of \(m\) with respect to \(a\) (the density of the rod at the point \(x=a\))?
\(\left.\dfrac{dm}{dx}\right|_{(3\,\text{m},0.02\,\text{kg})} =\dfrac{0.04\,\text{kg}}{9\,\text{m}} =.0044\overline{4}\,\dfrac{\text{kg}}{\text{m}}\)
We apply \(\dfrac{d}{da}\) to both sides using the Quotient rule and solve for \(\dfrac{dm}{da}\): \[\begin{aligned} \dfrac{a^2 3m^2\dfrac{dm}{da}-m^3 2a}{a^4}&=0 \\ 3a\dfrac{dm}{da}-2m&=0 \\ \dfrac{dm}{da}=\dfrac{2m}{3a}& \end{aligned}\] Now we evaluate at \((a,m)=(3\,\text{m},0.02\,\text{kg})\): \[ \left.\dfrac{dm}{da}\right|_{(3\,\text{m},0.02\,\text{kg})} =\dfrac{0.04}{9}\dfrac{\text{kg}}{\text{m}} =.0044\overline{4}\,\dfrac{\text{kg}}{\text{m}} \]
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For a pendulum, the period \(T\) and the acceleration of gravity \(g\) are related by \(T\sqrt{g}=2\pi\sqrt{L}\) where \(L\) is the length of the pendulum. When \(g=2\,\dfrac{\text{m}}{\text{sec}^2}\) and \(T=1\,\text{sec}\), what is \(\dfrac{dT}{dg}\)?
\(\left.\dfrac{dT}{dg}\right|_{(2,1)} =-\dfrac{1}{4}\,\dfrac{\text{sec}^3}{\text{m}}\)
We apply \(\dfrac{d}{dg}\) to both sides using the Product and Power rules and solve for \(\dfrac{dT}{dg}\): \[\begin{aligned} \dfrac{T}{2\sqrt{g}}+\sqrt{g}\dfrac{dT}{dg}&=0 \\ \dfrac{dT}{dg}&=-\,\dfrac{T}{2g} \end{aligned}\] Now we evaluate at \((g,T)= \left(2\,\dfrac{\text{m}}{\text{sec}^2},1\,\text{sec}\right)\): \[ \left.\dfrac{dT}{dg}\right|_{(2,1)} =-\dfrac{1}{4}\,\dfrac{\text{sec}^3}{\text{m}} \]
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In this equation we can solve for: \[ T=\dfrac{K}{\sqrt{g}} \] So we can differentiate explicitly: \[ \dfrac{dT}{dg}=-\dfrac{K}{2g\sqrt{g}}=-\dfrac{T}{2g} \]
A rod is heating up. The length \(L\) and the temperature \(T\) are related by \(\dfrac{L-L_0}{T-T_0}=\alpha L_0\) where \(\alpha\) is the coefficient of linear expansion and \(L_0\) is the original length \(2\,\text{m}\) and \(T_0\) is the original temperature \(25\,^{\circ}\text{C}\). If \(L=3\,\text{m}\) and \(T=50^\circ\text{C}\), what is \(\dfrac{dL}{dT}\)?
\(\left.\dfrac{dL}{dT}\right|_{(50\,^{\circ}\text{C},3\,\text{m})} =\dfrac{1}{25}\,\dfrac{\text{m}}{^{\circ}\text{C}} =0.04\,\dfrac{\text{m}}{^{\circ}\text{C}}\)
We apply \(\dfrac{d}{dT}\) to both sides using the Quotient rule and solve for \(\dfrac{dL}{dT}\): \[\begin{aligned} \dfrac{(T-T_0)\dfrac{dL}{dT}-(L-L_0)}{(T-T_0)^2}&=0 \\ (T-T_0)\dfrac{dL}{dT}-(L-L_0)&=0 \\ \dfrac{dL}{dT}=\dfrac{L-L_0}{T-T_0} \end{aligned}\] Now we evaluate at \((T,L)=(50^\circ\text{C},3\,\text{m})\): \[ \left.\dfrac{dL}{dT}\right|_{(50,3)} =\dfrac{1}{25}\,\dfrac{\text{m}}{^{\circ}\text{C}} =0.04\,\dfrac{\text{m}}{^{\circ}\text{C}} \]
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In this equation we can solve for: \[ L=\alpha L_0(T-T_0)+L_0 \] So we can differentiate explicitly: \[ \dfrac{dL}{dT}=\alpha L_0=\dfrac{L-L_0}{T-T_0} \]
If \(x\) and \(z\) are related by \(3x^2+z^2=21\), find \(\dfrac{dz}{dx}\) at \((x,z)=(2,-3)\).
\(\left.\dfrac{dz}{dx}\right|_{(2,-3)}=2\)
We differentiate both sides, using the Sum and Power rules: \[ 6x+2z\dfrac{dz}{dx}=0 \] Now we evaluate at \((x,z)=(2,-3)\) and solve for the derivative: \[\begin{aligned} 12-6\dfrac{dz}{dx}&=0 \\ \left.\dfrac{dz}{dx}\right|_{(2,-3)}&=2 \end{aligned}\]
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In this equation we can solve for: \[ z=-\sqrt{21-3x^2} \] We take the negative square root because \(z=-3\) is negative. So we can differentiate explicitly using the Chain and Power rules and substituting for \(z\): \[\begin{aligned} \dfrac{dz}{dx}&=\dfrac{3x}{\sqrt{21-3x^2}} \\ \dfrac{dz}{dx}(2)&=\dfrac{6}{3}=2 \end{aligned}\]
If \(x\) and \(z\) are related by \(3xz^2=14-2x^2\), find \(\dfrac{dz}{dx}\) at \((x,z)=(2,1)\).
\(\left.\dfrac{dz}{dx}\right|_{(2,1)}=-\,\dfrac{11}{12}=-0.91\overline{6}\)
We differentiate both sides, using the Sum, Product, and Power rules: \[ 6xz\dfrac{dz}{dx}+3z^2=-4x \] Now we evaluate at \((x,z)=(2,1)\) and solve for the derivative: \[\begin{aligned} 12\dfrac{dz}{dx}+3&=-8 \\ \left.\dfrac{dz}{dx}\right|_{(2,1)}&=-\,\dfrac{11}{12} =-0.91\overline{6} \end{aligned}\]
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If \(x\) and \(y\) are related by \(x^2+y^3=9\), find the equation of the tangent line at \((1,2)\).
The equation for a straight line is \(y=mx+b\).
The equation of the tangent line at \((1,2)\) is:
\(y=-\,\dfrac{1}{6}x+\dfrac{13}{6}\)The equation for a straight line is \(y=mx+b\). We find the slope \(m=\dfrac{dy}{dx}\) by implicit differentiation: \[\begin{aligned} 2x+3y^2\dfrac{dy}{dx}&=0 \\ \dfrac{dy}{dx}&=-\dfrac{2x}{3y^2} \end{aligned}\] Now we evaluate at \((x,y)=(1,2)\): \[ m=\left.\dfrac{dy}{dx}\right|_{(1,2)}=-\,\dfrac{2}{12}=-\,\dfrac{1}{6} \] We plug in \(x=1\), \(y=2\) and \(m=-\,\dfrac{1}{6}\) into the equation for a line and solve for \(b\): \[\begin{aligned} y&=mx+b \\ 2&=-\,\dfrac{1}{6}(1)+b \\ b&=2+\dfrac{1}{6}=\dfrac{13}{6} \end{aligned}\] So the equation is: \[ y=-\,\dfrac{1}{6}x+\dfrac{13}{6} \]
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Review Exercises
These are Composition, Chain rule, and Implicit Differentiation problems.
Consider the functions \(f(x)=x+4\) and \(g(x)=2x\). Compute \(g(4)\) and \(f(g(4))\).
\(g(4)=8\)
\(f(g(4))=12\)We replace the \(x\) in \(g(x)=2x\) by \(4\): \[ g(4)=2\cdot4=8 \] Next we replace the \(x\) in \(f(x)=x+4\) by \(g(4)=8\): \[ f(g(4))=8+4=12 \]
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If \(g(x)=\left(\dfrac{x}{x^4+1}+5^x\right)^{500}\), find \(\dfrac{dg}{dx}\).
\(g'(x)=500\left(\dfrac{x}{x^4+1}+5^x\right)^{499} \left(\dfrac{1-3x^4}{(x^4+1)^2}+5^x\ln(5)\right)\)
The outer function is \(u^{500}\) and the inner function is \(u =\dfrac{x}{x^4+1}+5^x\). The derivative of the outer function is \( 500u^{499}\). To calculate the derivative of the inner function we use the Sum, Exponential, Quotient, Sum, and Power rules: \[\begin{aligned} &\dfrac{d}{dx}\left(\dfrac{x}{x^4+1}+5^x\right) \\ &=\dfrac{d}{dx}\left(\dfrac{x}{x^4+1}\right)+\dfrac{d}{dx}(5^x) \\ &=\dfrac{(x^4+1)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(x^4+1)}{(x^4+1)^2} +5^x\ln(5) \\ &=\dfrac{x^4+1-4x^4}{(x^4+1)^2}+5^x\ln(5) \\ &=\dfrac{1-3x^4}{(x^4+1)^2}+5^x\ln(5) \\ \end{aligned}\] To apply the chain rule to the function \(g\), we take the derivative of the outer function evaluated at the inner function and then multiply by the derivative of the inner function: \[ g'(x)=500\left(\dfrac{x}{x^4+1}+5^x\right)^{499} \left(\dfrac{1-3x^4}{(x^4+1)^2}+5^x\ln(5)\right) \]
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If \(x\) and \(y\) are related by \(x^2+2x+2y+y^2=11\), find \(\dfrac{dy}{dx}\) at \((1,2)\).
\(\left.\dfrac{dy}{dx}\right|_{(1,2)}=-\,\dfrac{2}{3}\)
We differentiate both sides, using the Sum and Power rules: \[ 2x+2+(2+2y)\dfrac{dy}{dx}=0 \] Now we evaluate at \((x,y)=(1,2)\) and solve for \(\dfrac{dy}{dx}\): \[\begin{aligned} 4+6\left.\dfrac{dy}{dx}\right|_{(1,2)}&=0 \\ \left.\dfrac{dy}{dx}\right|_{(1,2)}&=-\,\dfrac{2}{3} \end{aligned}\]
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- Consider the functions \(f(x)=x^2+2x+3\) and \(g(x)=2x+4\).
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Compute \(g(1)\) and \(f(g(1))\).
\(g(1)=6\)
\(f(g(1))=51\)We replace the \(x\) in \(g(x)=2x+4\) by \(1\): \[ g(1)=2\cdot1+4=6 \] Next we replace the \(x\) in \(f(x)=x^2+2x+3\) by \(g(1)=6\): \[ f(g(1))=6^2+2\cdot6+3=36+12+3=51 \]
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Compute \(f(1)\) and \(g(f(1))\).
\(f(1)=6\)
\(g(f(1))=16\)We replace the \(x\) in \(f(x)=x^2+2x+3\) by \(1\): \[ f(1)=1^2+2\cdot1+3=1+2+3=6 \] Next we replace the \(x\) in \(g(x)=2x+4\) by \(f(x)=6\): \[ f(g(1))=2\cdot6+4=12+4=16 \]
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Notice that \(g(f(1))\) is not the same as \(f(g(1))\).
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Compute \(f(g(x))\).
\(f(g(x))=4x^2+20x+27\)
We replace the \(x\) in \(f(x)=x^2+2x+3\) by \(g(x)=2x+4\): \[\begin{aligned} f(g(x))&=(2x+4)^2+2(2x+4)+3 \\ &=4x^2+16x+16+4x+8+3 \\ &=4x^2+20x+27 \end{aligned}\]
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If \(f(g(x))=4x^2+20x+27\), then \[\begin{aligned} f(g(1))&=4\cdot1^2+20\cdot1+27 \\ &=4+20+27=51 \end{aligned}\] which agrees with the answer from part (a).
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Compute \(g(f(x))\).
\(g(f(x))=2x^2+4x+10\)
We replace the \(x\) in \(g(x)=2x+4\) by \(f(x)=x^2+2x+3\): \[\begin{aligned} g(f(x))&=2(x^2+2x+3)+4 \\ &=2x^2+4x+6+4 \\ &=2x^2+4x+10 \end{aligned}\]
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If \(g(f(x))=2x^2+4x+10\), then: \[\begin{aligned} g(f(1))&=2\cdot1^2+4\cdot1+10 \\ &=2+4+10=16 \end{aligned}\] which agrees with the answer in part (b).
Notice that \(g(f(x))\) is not equal to \(f(g(x))\).
If \(f(x)=\sin^2(\cot^2(x^3+5))\), what is \( \dfrac{df}{dx}\)?
\(f'(x)=-3x^2\sin(2\cot^2(x^3+5))\cot(x^3+5)\csc^2(x^3+5)\)
It is useful to rewrite the function as: \[ f(x)=[\sin([\cot(x^3+5)]^2)]^2 \] There are \(5\) levels of functions: squaring, \(\sin\), squaring, \(\cot\), and \(x^3+5\). To apply the chain rule we differentiate each level of function and multiply them: \[\begin{aligned} f'(x)&=2\sin([\cot(x^3+5)]^2)\cdot\cos([\cot(x^3+5)]^2)\cdot2\cot(x^3+5) \cdot[-\csc^2(x^3+5)]\cdot3x^2 \\ &=-6x^2\sin([\cot(x^3+5)]^2)\cos([\cot(x^3+5)]^2)\cot(x^3+5)\csc^2(x^3+5) \\ &=-6x^2\sin(\cot^2(x^3+5))\cos(\cot^2(x^3+5))\cot(x^3+5)\csc^2(x^3+5) \\ &=-3x^2\sin(2\cot^2(x^3+5))\cot(x^3+5)\csc^2(x^3+5) \end{aligned}\]
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If \(x\) and \(y\) are related by \(xy^2+x^2y=30\), find the equation of the tangent line at \((2,3)\).
The equation of the tangent line at \((2,3)\) is:
\(y=-\,\dfrac{21}{16}x+\dfrac{45}{8}\)We differentiate both sides, using the Sum, Product, and Power rules and solve for \(\dfrac{dy}{dx}\): \[\begin{aligned} 2xy\dfrac{dy}{dx}+y^2+x^2\dfrac{dy}{dx}+2xy=0 \\ \dfrac{dy}{dx}=-\,\dfrac{y^2+2xy}{x^2+2xy} \end{aligned}\] Now we evaluate at \((x,y)=(2,3)\): \[ \left.\dfrac{dy}{dx}\right|_{(2,3)}=-\,\dfrac{21}{16} \] We plug in \(\dfrac{dy}{dx}\), \(x\), and \(y\) into \(y=mx+c\) and solve for \(c\): \[\begin{aligned} -\,\dfrac{21}{16}+c&=3 \\ c&=\dfrac{45}{8} \end{aligned}\] So the equation is: \[ y=-\,\dfrac{21}{8}x+\dfrac{45}{8} \]
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- Consider the functions \(f(x)=x^3+5x^2\) and \(g(x)=2x+3\).
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Compute \(g(1)\) and \(f(g(1))\).
\(g(1)=5\)
\(f(g(1))=250\)We replace the \(x\) in \(g(x)=2x+3\) by \(x=1\): \[ g(1)=2\cdot1+3=5 \] Next we replace the \(x\) in \(f(x)=x^3+5x^2\) by \(g(1)=5\): \[ f(g(1))=5^3+5\cdot5^2=125+125=250 \]
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Compute \(f(1)\) and \(g(f(1))\).
\(f(1)=6\)
\(g(f(1))=15\)We replace the \(x\) in \(f(x)=x^3+5x^2\) by \(x=1\): \[ f(1)=1^3+5\cdot1^2=1+5=6 \] Next we replace the \(x\) in \(g(x)=2x+3\) by \(f(x)=6\): \[ g(f(1))=2\cdot6+3=12+3=15 \]
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Notice that \(g(f(1))\) is not the same as \(f(g(1))\).
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Compute \(f(g(x))\).
\(f(g(x))=8x^3+56x^2+114x+72\)
We replace the \(x\) in \(f(x)=x^3+5x^2\) by \(g(x)=2x+3\): \[\begin{aligned} f(g(x))&=(2x+3)^3+5(2x+3)^2 \\ &=8x^3+36x^2+54x+27+20x^2+60x+45 \\ &=8x^3+56x^2+114x+72 \end{aligned}\]
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If \(f(g(x))=8x^3+56x^2+114x+72\), then \[\begin{aligned} f(g(1))&=8\cdot1^3+56\cdot1^2+114\cdot1+72 \\ &=8+56+114+72=250 \end{aligned}\] which agrees with the answer from part (a).
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Compute \(g(f(x))\).
\(g(f(x))=2x^3+10x^2+3\)
We replace the \(x\) in \(g(x)=2x+3\) by \(f(x)=x^3+5x^2\): \[\begin{aligned} g(f(x))&=2(x^3+5x^2)+3 \\ &=2x^3+10x^2+3 \end{aligned}\]
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If \(g(f(x))=2x^3+10x^2+3\), then: \[\begin{aligned} g(f(1))&=2\cdot1^3+10\cdot1^2+3 \\ &=2+10+3=15 \end{aligned}\] which agrees with the answer in part (b).
Notice that \(g(f(x))\) is not equal to \(f(g(x))\).
If \(f(x)=\sec^3\left(x+\sqrt{x^2+1}\right)\), what is \(\dfrac{df}{dx}\)?
\(f'(x)=3\sec^3\left(x+\sqrt{x^2+1}\right)\tan\left(x+\sqrt{x^2+1}\right) \left(1+\dfrac{x}{\sqrt{x^2+1}}\right)\)
The outer function is \(u^3\) and the intermediate function is \(u=\sec(r)\) and the inner function is \(r=x+\sqrt{x^2+1}\). The derivative of the outer function is \(3u^2\) and the derivative of the intermediate function is \(\sec(r)\tan(r)\). To calculate the derivative of the inner function, we use the Sum, Chain, Sum, and Power rules: \[\begin{aligned} &\dfrac{d}{dx}\left(x+\sqrt{x^2+1}\right) \\ &=\dfrac{d}{dx}(x)+\dfrac{d}{dx} \left(\sqrt{x^2+1}\right) \\ &=1+\dfrac{x}{\sqrt{x^2+1}} \end{aligned}\] To apply the chain rule to the function \(f\) we take the derivative of the outer function at the intermediate function composed with the inner function multiplied by the derivative of the intermediate function evaluated at the inner function multiplied by the derivative of the inner function: \[\begin{aligned} f'(x)&=\left[3\sec^2\left(x+\sqrt{x^2+1}\right)\right]\left[\sec\left(x+\sqrt{x^2+1}\right) \tan\left(x+\sqrt{x^2+1}\right)\right] \left[1+\dfrac{x}{\sqrt{x^2+1}}\right] \\ &=3\sec^3\left(x+\sqrt{x^2+1}\right)\tan\left(x+\sqrt{x^2+1}\right) \left(1+\dfrac{x}{\sqrt{x^2+1}}\right) \end{aligned}\]
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If \(x\) and \(y\) are related by \(x\sin(\pi y)-y^2\cos(\pi x)=\dfrac{1}{4}\left(1-\dfrac{1}{\sqrt{2}}\right)\), find the equation of the tangent line at \(\left(\dfrac{1}{4},\dfrac{1}{2}\right)\).
The equation of the tangent line at \(\left(\dfrac{1}{4},\dfrac{1}{2}\right)\) is: \[ y=-\left(\sqrt{2}+\dfrac{\pi}{4}\right)x+\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\pi}{16} \]
We differentiate both sides, using the Difference, Product, Power, and Chain rules and solve for \(\dfrac{dy}{dx}\): \[\begin{aligned} \pi x\cos(\pi y)&\dfrac{dy}{dx}+\sin(\pi y)+\pi y^2\sin(\pi x)+2y\cos(\pi x) \dfrac{dy}{dx}=0 \\ &\dfrac{dy}{dx}=-\,\dfrac{\sin(\pi y)+\pi y^2\sin(\pi x)} {\pi x\cos(\pi y)+2y\cos(\pi x)} \end{aligned}\] Now we evaluate at \((x, y)=\left(\dfrac{1}{4},\dfrac{1}{2}\right)\): \[\begin{aligned} \left.\dfrac{dy}{dx}\right|_{(\frac{1}{4},\frac{1}{2})} &=-\,\dfrac{1+\dfrac{\pi}{4}\dfrac{1}{\sqrt{2}}} {0+\dfrac{1}{\sqrt{2}}} \\ &=-\sqrt{2}-\dfrac{\pi}{4} \end{aligned}\] So the equation of the tangent line is: \[\begin{aligned} y&=y_0+m(x-x_0) \\ &=\dfrac{1}{2}-\left(\sqrt{2}+\dfrac{\pi}{4}\right)\left(x-\dfrac{1}{4}\right) \\ &=-\left(\sqrt{2}+\dfrac{\pi}{4}\right)x+\dfrac{1}{2}+\dfrac{\sqrt{2}}{4}+\dfrac{\pi}{16} \end{aligned}\]
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Differentiate \(p(x)=\sin(\cos[\tan(\cot[\sec(\csc[x])])])\). This is a challenge problem, make sure to keep track of your work.
Apply the Chain Rule \(5\) times. Be careful with the last two steps.
\(\begin{aligned} p'(x)=&-\cos(\cos[\tan(\cot[\sec(\csc[x])])]) \\ &\cdot\sin[\tan(\cot[\sec(\csc[x])])] \\ &\cdot\sec^2(\cot[\sec(\csc[x])]) \\ &\cdot\csc^2[\sec(\csc[x])] \\ &\cdot\sec(\csc[x])\tan(\csc[x]) \\ &\cdot\csc[x]\cot[x] \\ \end{aligned}\)
Apply the Chain Rule \(5\) times: \[\begin{aligned} p'(x)= &\cos(\cos[\tan(\cot[\sec(\csc[x])])]) \\ &\cdot(-\sin[\tan(\cot[\sec(\csc[x])])]) \\ &\cdot\sec^2(\cot[\sec(\csc[x])]) \\ &\cdot(-\csc^2[\sec(\csc[x])]) \\ &\cdot\sec(\csc[x])\tan(\csc[x]) \\ &\cdot(-\csc[x]\cot[x]) \\ =&-\cos(\cos[\tan(\cot[\sec(\csc[x])])]) \\ &\cdot\sin[\tan(\cot[\sec(\csc[x])])] \\ &\cdot\sec^2(\cot[\sec(\csc[x])]) \\ &\cdot\csc^2[\sec(\csc[x])] \\ &\cdot\sec(\csc[x])\tan(\csc[x]) \\ &\cdot\csc[x]\cot[x] \end{aligned}\]
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Compute the composition of the pair of functions.