The derivative as a function

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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

\[ \lim _{h\to 0} \frac {f(a+h)-f(a)}{h}?\]

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).

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. The value $f'(x)$ 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 numerator,

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

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 $(-\infty , \infty )$ and $f$ can be considered 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

\[ f'(x) = \lim _{h\to 0}\frac {h}{h}= \lim _{h\to 0}\answer [given]{1}=\answer [given]{1}.\]

Next we 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, $|x+h|=-(x+h)=-x-h$. That gives

\[ f'(x) = \lim _{h\to 0}\frac {-h}{h}= \lim _{h\to 0}\answer [given]{-1}=\answer [given]{-1}.\]

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 $(-\infty , \infty )$ 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 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 $y = f(x)$?