10. Derivatives and Tangent Lines

c. Derivative as a Function

1. Definition

So far we have computed the derivative of a function, \(f(x)\), at \(x=a\) where \(a\) is an actual number, obtaining a number, \(f'(a)\). However, what happens if we let \(a\) be a variable? Then we obtain \(f'(a)\) as a function of \(a\). If we replace the \(a\) by an \(x\), then we obtain the following definition of the derivative of \(f(x)\) as a function:

We can also write the derivative, \(f'(x)\), using the limit of a difference quotient, but then we also need to replace the limit variable since it was previously \(x\):

The derivative of the function \(f(x)\) can also be written as the limit: \[ f'(x)=\lim_{z\rightarrow x} \dfrac{f(z)-f(x)}{z-x} \]

The derivative of a function \(y=f(x)\) can be written in either prime notation or differential notation: \[ f'(x)=\dfrac{df}{dx}=\dfrac{dy}{dx} \] The differential notation reminds us of the difference quotient definition: \[ \dfrac{df}{dx}=\lim_{\Delta x\to0}\dfrac{\Delta f}{\Delta x} \]

If \(f(x)=x^2\), find its derivative \(\dfrac{df}{dx}\). Use it to find \(\dfrac{df}{dx}(3)\).

We write out and compute the limit definition: \[\begin{aligned} \dfrac{df}{dx}&=\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h} =\lim_{h\rightarrow 0} \dfrac{(x+h)^2-(x)^2}{h} \\ &=\lim_{h\rightarrow 0} \dfrac{x^2+2xh+h^2-x^2}{h} =\lim_{h\rightarrow 0} \dfrac{2xh+h^2}{h} \\ &=\lim_{h\rightarrow 0} (2x+h) =2x \end{aligned}\]

This time we use the alternate formula: \[\begin{aligned} \dfrac{df}{dx}&=\lim_{z\rightarrow x} \dfrac{f(z)-f(x)}{z-x} =\lim_{z\rightarrow x} \dfrac{z^2-x^2}{z-x} \\ &=\lim_{z\rightarrow x} (z+x) =2x \end{aligned}\]

Using \(\dfrac{df}{dx}=2x\), we find \(\dfrac{df}{dx}(3)=6\). This is the same answer we found in a previous example.