The derivative as a function

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

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 the real numbers to the real numbers, the derivative $f'$ is also a function from the 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 $x<0$},\\ \answer [given]{x} &\text {if $x\ge 0$}. \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 $x<0$},\\ \answer [given]{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 $y = f(x)$ ?

The derivative as a function of functions

While writing $f'$ is viewing the derivative of $f$ as a function in its own right, the derivative itself $\ddx$ is in fact a function that maps functions to functions,

$\begin{align*} \ddx x^2 &= 2x\\ \ddx f(x) &= f'(x). \end{align*}$

As a function, is $\ddx$ one-to-one?

yes no

Many different functions share the same derivative since the derivative records only the slope of the tangent line and not the value, or height of the function.