The derivative as a function

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Here we study the derivative of a function, as a function, in its own right.

The derivative of a function, as a function

First, we have to find an alternate definition for $f'(a)$ , the derivative of a function $f$ at $a$ .

Let’s start with the average rate of change of the function $f$ as the input changes from $a$ to $x$ . We will introduce a new variable, $h$ , to denote the difference between $x$ and $a$ . That is $x-a=h$ or $x=a+h$ . Take a look at the figure below.

Now we can write ${\text {average rate of change }}=\frac {f(a+h)-f(a)}{(a+h)-a}=\frac {f(a+h)-f(a)}{h}$ What happens if $h\to 0$ ? In other words, what is the meaning of the limit $\lim _{h\to 0} \frac {f(a+h)-f(a)}{h}?$ Obviously, this limit represents $f'(a)$ , the instantaneous rate of change of $f$ at $a$ ! Therefore, we have an alternate way of writing the definition of the derivative at the point $a$ , namely $f'(a) = \lim _{h\to 0} \frac {f(a+h)-f(a)}{h}.$

Let $f(x) = x^2-2x$ . Using the alternate expression for the derivative, find the slope of the tangent line to the curve $y=f(x)$ at the point $(2,f(2))$ .

The slope of the tangent line is given by the derivative, $f'(2)$ . $f'(2) = \lim _{h\to 0}\frac {f\left (\answer [given]{2+h}\right )-f\left (2\right )}{h}.$ Now substitute in for the function we know, $f'(2) = \lim _{h\to 0}\frac {(2+h)^2-2(2+h) -0}{h}.$ Now expand the numerator of the fraction, $f'(2) =\lim _{h\to 0} \frac {4+4h+h^2-4-2h }{h}.$ Now combine like-terms, $f'(2) = \lim _{h\to 0} \frac {2h+h^2}{h}.$ Factor an $h$ from every term in the numerator, $f'(2) = \lim _{h\to 0}\frac {\cancel {h}\left (2+h\right )}{\cancel {h}}.$ Compute the limit, $f'(2) = \lim _{h\to 0}(2+h)=\answer [given]{2}.$

This alternate definition of the derivative of $f$ at $a$ , namely,

$f'(a) = \lim _{h\to 0}\frac {f(a+h)-f(a)}{h},$ (provided that the limit exists), allows us to define $f'(x)$ for any value of $x$ ,

$f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(x)}{h},$ (provided that the limit exists).

And this is how we define a new function, $f'$ , the derivative of $f$ . The domain of $f'$ consists of all points in the domain of $f$ where the function $f$ is differentiable. $f'(x)$ gives us the instantaneous rate of change of $f$ at any point $x$ in the domain of $f'$ .
Given a function $f$ from some set of real numbers to the real numbers, the derivative $f'$ is also a function from some set of real numbers to the real numbers. Understanding the relationship between the functions $f$ and $f'$ helps us understand any situation (real or imagined) involving changing values.

Given the function $f(x) = 3x+2$ , find $f'(x)$ .

Start with the definition of $f'(x)$ $f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(\answer [given]{x})}{h}.$ Replace $f$ with its formula, $f'(x) = \lim _{h\to 0}\frac {3(x+h)+2-\left (\answer [given]{3x+2}\right )}{h}.$ Simplify the top, $f'(x) = \lim _{h\to 0}\frac {3\cancel {h}}{\cancel {h}}.$ Evaluate the limit. $f'(x) = \lim _{h\to 0}{3}=\answer [given]{3}.$

Given the function $f(x) = |x|$ , find $f'(x)$ .

Recall, the domain of $f$ is $\RR$ and $f$ is in fact a piecewise defined function, since $f(x) = |x|= \begin {cases} \answer [given]{-x} &\text {if $x<0$,}\\ \answer [given]{x} &\text {if $x\ge 0$}. \end {cases}$ We will first compute $f'(x)$ when $x>0$ . Start with the definition of $f'(x)$ $f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(\answer [given]{x})}{h}.$ Replace $f$ with its formula, $f'(x) = \lim _{h\to 0}\frac {|x+h|-\answer [given]{|x|}}{h}.$ $f'(x) = \lim _{h\to 0}\frac {x+h-\answer [given]{x}}{h}.$ Note: When $x>0$ , then for all small enough values of $h$ it follows that $x+h>0$ . Therefore, $|x+h|=x+h$ . Now we have $f'(x) = \lim _{h\to 0}\frac {h}{h}= \lim _{h\to 0}\answer [given]{1}=\answer [given]{1}.$ Now, we can compute the derivative $f'(x)$ when $x<0$ . $f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(\answer [given]{x})}{h}.$ Replace $f$ with its formula, $f'(x) = \lim _{h\to 0}\frac {|x+h|-\answer [given]{|x|}}{h}.$ $f'(x) = \lim _{h\to 0}\frac {-x-h-\answer [given]{-x}}{h}.$ Note: When $x<0$ , then for all small enough values of $h$ it follows that $x+h<0$ . Therefore, $|x+h|=-(x+h)=-x-h$ . Now we have $f'(x) = \lim _{h\to 0}\frac {-h}{h}= \lim _{h\to 0}\answer [given]{-1}=\answer [given]{-1}.$ What remains to be done is to check whether the derivative $f'(0)$ exists. Start with the definition of $f'(0)$ $f'(0) = \lim _{h\to 0}\frac {f(0+h)-f(\answer [given]{0})}{h}.$ Replace $f$ with its formula, $f'(0) = \lim _{h\to 0}\frac {|0+h|-\answer [given]{|0|}}{h}.$ $f'(0) = \lim _{h\to 0}\frac {|h|}{h}.$ Note: When $h>0$ , then $|h|=h$ , but when $h<0$ , then $|h|=-h$ . Therefore, instead of computing the limit above, we will compute the two one-sided limits and compare them.

$\lim _{h\to 0^+}\frac {|h|}{h}= \lim _{h\to 0^+}\frac {h}{h}=\lim _{h\to 0^+}\answer [given]{1}=\answer [given]{1};$ $\lim _{h\to 0^-}\frac {|h|}{h}= \lim _{h\to 0^-}\frac {-h}{h}=\lim _{h\to 0^-}\answer [given]{-1}=\answer [given]{-1};$ Since the two one-sided limits are not equal it follows that $\lim _{h\to 0}\frac {|h|}{h} {\text { DOES NOT EXIST!}}$ Therefore, $f'(0)$ DOES NOT EXIST, which means that $f$ is NOT DIFFERENTIABLE at $x=0$ ! To summarize

$f'(x) = \begin {cases} \answer [given]{-1} &\text {if $x<0$,}\\ \answer [given]{1} &\text {if $x> 0$}. \end {cases}$

Is it true that for any function $f$ the domain of $f'$ is equal to the domain of $f$ ?

yes no

This is not true. Consider the function $f(x)=|x|$ . The domain of $f$ is $\RR$ and the domain of $f'$ is $(-\infty ,0)\cup (0,\infty )$

This example demonstrates that a function $f$ and its derivative, $f'$ , may have different domains.

Can two different functions, say, $f$ and $g$ , have the same derivative?

yes no

Many different functions can share the same derivatives. Consider two different functions, $f$ and $g$ , defined by
$f(x)=x$ and $g(x)=x+5$ . Then, $f'(x)=1$ , and $g'(x)=1$ , for all real numbers $x$ .
So, the derivatives of these two different functions are equal.

Let’s compare the graphs of $f$ and $f'$ for the derivatives we’ve computed so far:

$f(x)=x^2, f'(x)=2x$

$f(x)=3x+2, f'(x)=3$

$f(x)=|x|, f'(x)=\begin {cases} 1 & \text {for } x>0 \\ -1 & \text {for } x<0 \end {cases}$

For each of the three pairs of functions, describe $y=f(x)$ when $f'$ is positive, and when $f'$ is negative.

When $f'$ is positive, $y=f(x)$ is . When $f'$ is negative, $y=f(x)$ is positiveincreasingnegativedecreasing

Here we see the graph of $f'$ , the derivative of some function $f$ .

Which of the following graphs could be $y = f(x)$ ?