# 3.2 The derivative as a function

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• Define the derivative function of a given function.
• Graph a derivative function from the graph of a given function.
• State the connection between derivatives and continuity.
• Describe three conditions for when a function does not have a derivative.
• Explain the meaning of a higher-order derivative.

As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it.

## Derivative functions

The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.

## Definition

Let $f$ be a function. The derivative function    , denoted by ${f}^{\prime },$ is the function whose domain consists of those values of $x$ such that the following limit exists:

${f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.$

A function $f\left(x\right)$ is said to be differentiable at $a$    if $f\left(a\right)$ exists. More generally, a function is said to be differentiable on $S$    if it is differentiable at every point in an open set $S,$ and a differentiable function    is one in which ${f}^{\prime }\left(x\right)$ exists on its domain.

In the next few examples we use [link] to find the derivative of a function.

## Finding the derivative of a square-root function

Find the derivative of $f\left(x\right)=\sqrt{x}.$

Start directly with the definition of the derivative function. Use [link] .

$\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}\hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x+h\right)=\sqrt{x+h}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=\sqrt{x}\hfill \\ \text{into}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}·\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\hfill & & & \begin{array}{c}\text{Multiply numerator and denominator by}\hfill \\ \sqrt{x+h}+\sqrt{x}\phantom{\rule{0.2em}{0ex}}\text{without distributing in the}\hfill \\ \text{denominator.}\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\hfill & & & \text{Multiply the numerators and simplify.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}\hfill & & & \text{Cancel the}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =\frac{1}{2\sqrt{x}}\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

## Finding the derivative of a quadratic function

Find the derivative of the function $f\left(x\right)={x}^{2}-2x.$

Follow the same procedure here, but without having to multiply by the conjugate.

$\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{\left({\left(x+h\right)}^{2}-2\left(x+h\right)\right)-\left({x}^{2}-2x\right)}{h}\hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x+h\right)={\left(x+h\right)}^{2}-2\left(x+h\right)\phantom{\rule{0.2em}{0ex}}\text{and}\hfill \\ f\left(x\right)={x}^{2}-2x\phantom{\rule{0.2em}{0ex}}\text{into}\hfill \\ {f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{2}+2xh+{h}^{2}-2x-2h-{x}^{2}+2x}{h}\hfill & & & \text{Expand}\phantom{\rule{0.2em}{0ex}}{\left(x+h\right)}^{2}-2\left(x+h\right).\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{2xh-2h+{h}^{2}}{h}\hfill & & & \text{Simplify.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h\left(2x-2+h\right)}{h}\hfill & & & \text{Factor out}\phantom{\rule{0.2em}{0ex}}h\phantom{\rule{0.2em}{0ex}}\text{from the numerator.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\left(2x-2+h\right)\hfill & & & \text{Cancel the common factor of}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =2x-2\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$

Find the derivative of $f\left(x\right)={x}^{2}.$

${f}^{\prime }\left(x\right)=2x$

We use a variety of different notations to express the derivative of a function. In [link] we showed that if $f\left(x\right)={x}^{2}-2x,$ then ${f}^{\prime }\left(x\right)=2x-2.$ If we had expressed this function in the form $y={x}^{2}-2x,$ we could have expressed the derivative as ${y}^{\prime }=2x-2$ or $\frac{dy}{dx}=2x-2.$ We could have conveyed the same information by writing $\frac{d}{dx}\left({x}^{2}-2x\right)=2x-2.$ Thus, for the function $y=f\left(x\right),$ each of the following notations represents the derivative of $f\left(x\right)\text{:}$

${f}^{\prime }\left(x\right),\text{}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx},\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime },\text{}\phantom{\rule{0.2em}{0ex}}\frac{d}{dx}\left(f\left(x\right)\right).$

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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