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

We know that to find the derivative of a function at a point $x=a$ we write $f'(a) = \lim _{h\to 0}\frac {f(a+h)-f(a)}{h},$ (provided that the limit exists). However, if we replace the given number $a$ with a variable $x$ , we now have $f'(x) = \lim _{h\to 0}\frac {f(x+h)-f(x)}{h},$ (provided that the limit exists). This defines 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 $f'$ .

The notation:

$f'(a)$ means take the derivative of $f$ first, then evaluate at $x=a$ .

In other words, given $f$ a function of $x$ $f'(a) = \eval {\ddx f(x)}_{x=a}.$

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.

Let $f(x) = 3x+2$ . What is $f'(-1)$ ?

$f'(-1) = 0$ because $f'(3)$ is a number, and a number corresponds to a horizontal line, which has a slope of zero. $f'(-1) = 3$ because $y=f(x)$ is a line with slope $3$ . We cannot solve this problem yet.

Given the function $f(x) = 3x+2$ , find $f'(x)$ .
Then, use this this result to compute $f'(-1)$ in order to verify your answer in previous question.

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, $f'(x) = \lim _{h\to 0}\frac {3(x+h)+2-\answer [given]{(3x+2)}}{h}.$ Simplify the top, $f'(x) = \lim _{h\to 0}\frac {3\answer [given]{h}}{h}.$ Cancel the $h$ in the numerator and denominator, $f'(x) = \lim _{h\to 0}\answer [given]{3}.$

Evaluate the limit. $f'(x) = \answer [given]{3}.$ Therefore, $f'(-1) = \answer [given]{3}.$

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

yes no

Let’s consider the function $f(x)=|x|$ . The domain of $f$ is $\RR$ . Remember, $f$ is in fact a piecewise defined function, since $f(x)=|x| = \begin {cases} \answer [given]{ -x} &\text {if $x<0$ ,}\\ x &\text {if $x\geq 0$}. \end {cases}$

We can easily see that all nonzero real numbers are in the domain of $f'$ . Why?
Because, if $x>0$ , then the graph of $f$ near $x$ is a line with the slope $1$ . Therefore, $f'(x)=1$ , for $x>0$ . Similarly, if $x<0$ , then the graph of $f$ near $x$ is a line with the slope $-1$ . So, $f'(x)=-1$ , for $x<0$ . Therefore, $f'(x)$ is defined for all nonzero numbers. $f'(x) = \begin {cases} \answer [given]{ -1} &\text {if $x<0$ ,}\\ 1 &\text {if $x> 0$}. \end {cases}$ But, what about $x=0$ ? Is 0 in the domain of $f'$ ?
Let’s try to compute $f'(0)$ , and see what happens. $f'(0)= \lim _{h\to 0} \dfrac {f(0+h) - f(\answer [given]{0})}{h}= \lim _{h\to 0} \dfrac {|0+h|-|\answer [given]{0}| }{h} = \lim _{h\to 0} \dfrac {|\answer [given]{h}| }{h}$ The last limit does not exist. Recall $\lim _{h\to 0^+} \dfrac {|h| }{h}= \lim _{h\to 0^+} \dfrac {\answer [given]{h} }{h}=\lim _{h\to 0^+} \answer [given]{1}=\answer [given]{1}$ and $\lim _{h\to 0^-} \dfrac {|h| }{h}= \lim _{h\to 0^-} \dfrac {\answer [given]{-h} }{h}=\lim _{h\to 0^+} \answer [given]{-1}=\answer [given]{-1}$ Since $f'(0)$ is not defined, $f$ is not differentiable at $0$ , and , therefore, $0$ is not in the domain of $f'$ .
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 $y = f(x)$ ?