This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.

# Derivative as a function

### From Mathematics Is A Science

## Contents

## 1 Motion and the derivative

Recall, if $x$ is time, $f(x)$ is the location at time $x$, then $f^{\prime}(a)$ is the *instantaneous velocity* of the motion at $a$. (Unlike speed, velocity has direction).

The motion can be horizontal or vertical.

However, we always make the time axis horizontal.

**Example.** The graph shows the positions of two runners as functions of time. Describe what happens.

Here's what happened: they started together, then were apart from each other, at all times $A$ was ahead, then finished together.

More details:

- $A$ maintains the same speed,
- $B$ starts slow and speeds up.

Estimate the speed from the graph. Speed of $B$ at $t=9$: $$g^{\prime}(9) = \frac{60}{10} = 6 \text{ m/s}$$ Speed of $B$ at $t=16$: $$g^{\prime}(16) = \frac{100}{3} = 33 \text{ m/s}$$ Naturally, $$g^{\prime}(0) = 0.$$

**Important observation:** Here we have three values of $g^{\prime}$. That a new function! It is called the *derivative function*.

Rough sketch here:

**Review exercise.** Compute $f^{\prime}(2)$ from the definition for
$$ f(x) = -x^{2} - x $$
Definition:
$$f^{\prime}(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} $$
Algebra first. Substitution...

What do we substitute?

1. Replace $x$ with $2 + h$. $$\begin{aligned}f(x) &= -x^{2} - x \\ f(2 + h) &= -(2 + h)^{2} - (2 + h) \end{aligned},$$ $$f(2) = -2^{2} - 2.$$

2. Replace $x$ with 2.

Substitute now: $$\begin{aligned} f^{\prime}(2) &= \lim_{h \to 0} \frac{\left[ -(2 - h)^{2} - (2-h) \right] - \left[ -2^{2} - 2 \right]}{h} \\ &= \lim_{h \to 0} \frac{-4 - 4h - h^{2} - 2 - h + 4 + 2}{h} \\ &= \lim_{h \to 0} \frac{-5h - h^{2}}{h} \\ &= \lim_{h \to 0} (-5 -h ) \\ &= -5 - 0 \\ &= 5 \end{aligned}$$

*Review substitution:*

Given $f(x) = \sin x + x^{3}$, consider $$\begin{gathered} f(1) = \sin 1 + 1^{3}, \\ f(-1) = \sin(-1) + (-1)^{3}, \\ f(t) = \sin t + t^{3}, \\ f(-x) = \sin(-x) + (-x)^{3}, \\ f(x^{2}) = \sin(x^{2}) +(x^{2})^{3}. \end{gathered} $$

## 2 Derivative as a function

Until now, the derivative has been a *number*:
$$f^{\prime}(1)=0, \quad f^{\prime}(3)=5. $$
Indeed, it's defined as a limit and, when it exists, it's a number.

And if the limit does not exist, the derivative does not exist: $$f^{\prime}(0) \text{ d.n.e}$$

Next, let's think of the derivative as a function.

Given the graph of $f$, we can find $f'(a)$, for each point $a$:

**Graphical procedure:**

- Find $f^{\prime}(a)$ at each of these $a$'s a s the slope of the graph.
- Plot these points, i.e., $(a,f^{\prime}(a))$, on the axes below the graph of $f$.
- Based on these points guess the shape of $y = f^{\prime}(x)$.

**Numerical procedure:**

So, differentiation serves as a function of functions:

Next, with $f$ given by its formula, find $f^{\prime}$.

**Algebraic procedure:**

Use the definition, as before, but with $a$ is unspecified.

**Example.** Compute $f^{\prime}$ as a function for $f(x) = x^{2}$.

Definition: $$f^{\prime}(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},$$ for each $a$.

Substitute $x^{2}$
$$ \lim_{h \to 0} \frac{(a + h)^{2} - a^{2}}{h} .$$
We want $f^{\prime}(a)$ for *all* $a$ in one computation!
$$\lim_{h \to 0} \frac{a^{2} + 2ah + h^{2} - a^{2}}{h}.$$
Divide by $h$.
$$\lim_{h \to 0} \frac{2ah + h^{2}}{h}.$$
Crucial observation: $a$ is independent of $h$, so, as far as the limit is concerned, it's treated as a constant.
$$ \lim_{h \to 0} (2a + h) $$
Where does this go as $h$ goes to 0?
$$ 2a + 0 = 2a. $$

Last point: I want $f^{\prime}$ to be a function of $x$, not $a$, just as the formula of $f$. We can plug in $a = \text{ anything }$ to get the derivative as a number. So, we just replace $a$ with $x$.

Answer $f^{\prime}(x) = 2x$.

The two functions, $f$ and $f'$, must have the *same input*, why?

Consider again: $x$ is time, $f(x)$ is the position as a function of time, then $f^{\prime}(a)$ is the velocity at time $x = a$. Then $f^{\prime}$, as a function, should be a function of time, $x$: $$\begin{aligned} \text{feet, } y &= f(x), \quad \text{position} \\ \text{feet/sec, } y &= f^{\prime}(x), \quad \text{velocity} \end{aligned}$$

## 3 Derivative exists, so what?

Some limits don't exist. Some for $f^{\prime}$ So, when it does, what does it tell us? Suppose $$ f^{\prime}(a) = \lim_{h \to 0} \frac{f(a + h) + f(a)}{h} $$ exists.

First, clearly $\lim\limits_{h \to 0} h = 0$.

- Question
- What does it mean for the limit of the numerator?
- Answer
- Its limit is 0.

Why?

Consider $$\lim_{h \to 0} \left[f(a + h) - f(a) \right] = 0$$ Rewrite (DR) $$\begin{gathered} \lim_{h \to 0} f(a + h) - \lim_{h \to 0} \underbrace{f(a)}_{\text{No } h \text{ here}} = 0 \\ \lim_{h \to 0} f(a + h) - f(a) = 0 \end{gathered}$$ So, $$ \lim_{h \to 0} f(a + h) = f(a) $$ This is a familiar limit!

Indeed, the limit means that $f$ is continuous at $a$.

To summarize...

**Theorem.** If the derivative of $f$ exists at $a$, then $f$ is continuous at $a$.

Conversely, if $f$ is *not* continuous at $a$, then $f^{\prime}(a)$ does not exist, i.e., $f$ is *not differentiable*.

We know from algebra: if $f^{\prime}(a)$ exists then $f$ is continuous at $x=a$. Now geometrically, if it is not, the secant lines become more and more vertical, $$\text{slope } \to \infty$$ as it gets steeper. So our limit is $\infty$:

**Fact:** There are continuous functions that are not differentiable. Compare:

- Continuous : There is no break.
- Differentiable : There is a tangent line.

**Cusp.**

Let's try to use the geometric definition of the derivative -- via secant lines. Approach $a$ separately from the left and right, they turn and the end result is...

- Two lines, so not differentiable:

$$\lim_{h \to 0^{-}} \neq \lim_{h \to 0^{+}}$$

- Same line, so differentiable:

**Example.** Analyze algebraically,
$$f(x) = |x|$$
at 0.
$$\begin{aligned}
f^{\prime}(0) &= \lim_{h \to 0} \frac{|0 + h| - |0|}{h} \\
&= \lim_{h \to 0} \frac{|h|}{h} \\
\end{aligned}$$
Consider:
$$\lim_{h \to 0^{-}} \frac{|h|}{h} = \frac{-h}{h} = -1, \qquad \lim_{h \to 0^{+}} \frac{|h|}{h} = \frac{h}{h} = 1$$
Not equal, so the limit does not exist.

**Fact.** There are functions the graph of which look smooth that they are not differentiable.

**Example.**
$$f(x) = \sqrt[3]{x}$$
at $x=0$.

The limit is infinite: $$f^{\prime}(0) = \infty $$

**Review example.** Find $f^{\prime}(x)$ for $f(x) = x^{2} + 1$ for all $x$.
$$\begin{aligned}
f^{\prime} &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\
&= \lim_{h \to 0} \frac{\left[ (x + h)^{2} + 1\right] - \left[ x^{2} + 1\right]}{h}
\end{aligned}$$
The terms cancel
$$\begin{aligned}
=\lim_{h \to 0} &\frac{x^{2} + 2xh + h^{2} + 1 - x^{2} -1}{h} \\
&\quad = \lim_{h \to 0} (2x + h) \\
&\quad = \lim_{h \to 0} \frac{2xh + h^{2}}{h} \\
&\quad = 2x
\end{aligned}$$

Some analysis, we compare the graphs of the derivative and the function:

**Example.** Now how about plotting $f^{\prime}$ based on the graph of $f$ only.

How? First look at the points with a horizontal tangent, i.e, for those $x$'s where $f^{\prime}(x) = 0$. These will typically separate intervals on which $f^{\prime} > 0$ or $f^{\prime} < 0$. To see which one, look at the slopes of $f$.

**Notation:** Consider
$$f^{\prime}(a) = \lim_{h \to 0} \underbrace{\frac{f(a+h) - f(a)}{h}}_{\text{Rename them as: }\frac{\Delta y}{\Delta x}= \frac{\text{rise}}{\text{run}}} $$
Here, the fraction is called the "difference quotient":
$$\frac{\text{change of } y}{\text{change of } x}$$
Remember, $\Delta$ stands for "difference" (and $d$ in $dx$, $dy$ as well).

This suggests...

- New notation
- $$\frac{dy}{dx}$$

for the derivative.

It's called the *Leibniz Notation*:
$$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$
Here $dx$ and $dy$ can be thought of as "infinitely small" increments of $x,y$ but we prefer to understand them via *limits*.

## 4 Higher derivatives

**Examples** Consider
$$\left( x^{2} + 1 \right)^{\prime}=2x.$$
We did differentiation once. How about one more time?
$$ \left( 2x \right)^{\prime} = 2 $$
Twice!

**New notation:**

- $\left(x^{2} + 1 \right)^{\prime\prime} = 2$ is the
*second derivative*, - $\left(x^{2} + 1 \right)^{\prime\prime\prime} = 2^{\prime} = 0$ is the
*third derivative*.

*Leibniz notation:*

- First derivative : $$ \frac{d}{dx} \left( x^{2} + 1 \right) = 2x $$
- Second derivative : $$ \frac{d}{dx} 2x = 2 $$