The derivative as a function

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

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

1 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.

\[ f'(x) = \lim _{h\to 0}{3}=\answer [given]{3}. \]

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 $\mathbb R$. 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 positiveincreasingnegativedecreasing. 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 $y = f(x)$?

2025-01-06 19:56:27