13. Chain Rule & Implicit Differentiation

Implicit Functions

An explicit function definition is an equation in which one variable is explicitly defined in terms of the other. For example, \(y=x^4\) is explicit because \(y\) is defined in terms of \(x\).

An implicit function definition is an equation relating two variables in which one variable is regarded as a function of the other. In fact, one equation can define several implicit (continuous) functions which may be distinguished by giving a point of the graph. It is usually impossible to solve an implicit definition to give an explicit definition of the function.

As an example, the graph of the equation \(x^4+x^2y^2+y^6=21\) is shown at the right. The equation implicitly defines \(y\) as many different continuous functions of \(x\). The upper half of the graph is the function passing through \((x,y)=(2,1)\) while the lower half is the function passing through \((x,y)=(2,-1)\). (Check \((2,1)\) and \((2,-1)\) are solutions to the equation.)

eg_implicit1

d. Implicit Differentiation

We would like to compute the derivatives of implicitly defined functions so that we can construct tangent lines. We do this by example.

The equation \(x^4+x^2y^2+y^6=21\) implicitly defines \(y\) as a function of \(x\) which passes through the point \((x,y)=(2,1)\). Compute \(\dfrac{dy}{dx}\) at \((2,1)\) and find the tangent line.

eg_implicit1_tan

We first check that \((x,y)=(2,1)\) is a solution to the equation: \[ x^4+x^2y^2+y^6=(2)^4+(2)^2(1)^2+(1)^6=16+4+1=21 \] We start by rewriting the equation indicating that we regard \(y\) as a function of \(x\): \[ x^4+x^2y_{(x)}^{\enspace2}+y_{(x)}^{\enspace6}=21 \] Notice that we wrote the \((x)\) small because we don't usually write it; we just think it. Now we take the \(x\) derivative of both sides, i.e. we apply the operator \(\dfrac{d}{dx}\) to both sides. We include many more steps than necessary, to make it clear how we are applying the Sum, Product, Power and Chain Rules: \[\begin{aligned} \dfrac{d}{dx}(x^4)+\dfrac{d}{dx}(x^2y^2)+\dfrac{d}{dx}(y^6) =\dfrac{d}{dx}(21)& \qquad \text{Sum Rule} \\ \dfrac{d}{dx}(x^4)+\dfrac{d}{dx}(x^2)y^2+x^2\dfrac{d}{dx}(y^2) +\dfrac{d}{dx}(y^6) =0& \qquad \text{Product Rule} \\ 4x^3+2xy^2+x^22y\dfrac{dy}{dx}+6y^5\dfrac{dy}{dx} =0& \qquad \text{Power and Chain Rules (*)} \end{aligned}\] Notice that \(y\) is a function of \(x\). So the derivatives \(\dfrac{d}{dx}(y^2)\) and \(\dfrac{d}{dx}(y^6)\) are done by Chain Rule and we write \(\dfrac{dy}{dx}\) for the derivative of the inner function. We now solve for \(\dfrac{dy}{dx}\) by grouping terms. \[ (2x^2y+6y^5)\dfrac{dy}{dx}=-4x^3-2xy^2 \] \[ \dfrac{dy}{dx}=\dfrac{-4x^3-2xy^2}{2x^2y+6y^5} \qquad \qquad \text{(**)} \] Finally, we evaluate the derivative at \((x,y)=(2,1)\). \[ \left.\dfrac{dy}{dx}\right|_{(2,1)} =\dfrac{-32-4}{8+6} =-\,\dfrac{18}{7} \] We use the point-slope formula to find the equation of the tangent line at \((x_o,y_o)=(2,1)\): \[ y=y_o+m(x-x_o) =1-\dfrac{18}{7}(x-2) \]

Notice that the derivative given in equation (**) depends on both \(x\) and \(y\). You need to know both coordinates to be able to find the derivative at a point on the graph.

If you are only interested in finding the derivative at a single point, then it is usually simpler to plug numbers into equation (*) rather than equation (**). This avoids algebra mistakes. At \((x,y)=(2,1)\), equation (*) becomes: \[ 32+4+8\dfrac{dy}{dx}+6\dfrac{dy}{dx}=0 \quad \Rightarrow \quad \left.\dfrac{dy}{dx}\right|_{(2,1)}=-\,\dfrac{18}{7} \]

Here's an alternate way to think about the tangent line.
The original equation, \(x^4+x^2y^2+y^6=21\), implicitly defines \(y\) as a function of \(x\) which passes through the point \((x,y)=(2,1)\). This means \(f(2)=1\) and \(f'(2)=\left.\dfrac{dy}{dx}\right|_{(2,1)}\). We can use these in the general equation of the tangent line: \[ y=f(2)+f'(2)(x-2)=1-\dfrac{18}{7}(x-2) \]

The equation \(x^2y^3+xy^2=12\) implicitly defines \(y\) as a function of \(x\) which passes through the point \((x,y)=(1,2)\). (Check \((x,y)=(1,2)\) is a solution to the equation.) Compute \(\dfrac{dy}{dx}\) at \((1,2)\) and find the tangent line.

ex_implicit2_tan

\(\left.\dfrac{dy}{dx}\right|_{(1,2)}=-\,\dfrac{5}{4}\)
\(y=2-\dfrac{\rule{0pt}{10pt}5}{4}(x-1)=-\,\dfrac{5}{4}x+\dfrac{13}{4}\)

We first check that \((x,y)=(1,2)\) is a solution to the equation. \[ x^2y^3+xy^2=(1)^2(2)^3+(1)(2)^2=8+4=12 \] We apply the operator \(\dfrac{d}{dx}\) to both sides, using the Product and Chain Rules: \[ (2x)y^3+x^2 3y^2\dfrac{dy}{dx} +(1)y^2+x 2y\dfrac{dy}{dx} =0 \] Now we evaluate at \((x,y)=(1,2)\) and solve for the derivative: \[ 16+12\dfrac{dy}{dx}+4+4\dfrac{dy}{dx}=0 \] \[ f'(1)=\left.\dfrac{dy}{dx}\right|_{(1,2)}=-\,\dfrac{20}{16}=-\,\dfrac{5}{4} \] Since \(f(1)=2\), the tangent line is: \[\begin{aligned} y&=f(1)+f'(1)(x-x_o) =2-\dfrac{5}{4}(x-1) \\ &=-\,\dfrac{5}{4}x+\dfrac{13}{4} \end{aligned}\]

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