The Derivative as a Function

The slope of the tangent line at every point, collected into a brand-new function.

0 / 9 problems mastered

In §2.7 you found the slope at one point $a$ . Now let the point move. For every $x$ , the slope of the tangent line to $y=f(x)$ is a number, so the slopes themselves form a function, written $f'$ :

f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}

The fraction is the slope of a secant line through two points on the curve, a horizontal distance $h$ apart. As $h\to 0$ , the second point slides into the first and the secant becomes the tangent. Try it in the first lab.

Secant → tangent

Lab 1 · drag the sliders

point a gap h

f(x) = 0.25x³ − x + 1a = 1

secant slope [f(a+h) − f(a)] / h–

true slope f′(a) = 0.75a² − 1–

Watch the secant slope approach f′(a) as h shrinks, from either side (h > 0 or h < 0). That limit is the derivative.

Reading f′ off the graph of f

Exercises 1–14 ask you to sketch $f'$ from a picture of $f$ . You never need a formula. Just ask, at each $x$ : how steep is $f$ , and is it going up or down?

Where $f$ has a peak or valley (horizontal tangent), $f'=0$ : the $f'$ graph crosses the x-axis.
Where $f$ is increasing, $f'>0$ (above the axis). Where $f$ is decreasing, $f'\lt 0$ (below).
Where $f$ is steepest, $|f'|$ is largest.
Where $f$ is a straight line, $f'$ is a constant (a flat segment).

Build f′ point by point

Lab 2 · drag on the top graph

function x

x–

slope of tangent = f′(x)–

differentiable here?–

When is f not differentiable?

Corner

Left slope ≠ right slope. Example: $|x|$ at 0 (slopes −1 and +1).

Discontinuity

A jump or hole. Theorem 4: differentiable ⇒ continuous, so not continuous ⇒ not differentiable.

Vertical tangent

The slope blows up: $|f'(x)|\to\infty$ . Example: $\sqrt[3]{x}$ at 0.

Careful

Continuous does not imply differentiable. $|x|$ is continuous at 0 but has a corner.

Notation and higher derivatives

Same thing, many names

$f'(x) = y' = \dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{d}{dx}f(x) = Df(x)$

Second derivative

$f''=(f')'$ , $\dfrac{d^2y}{dx^2}$ . If $s(t)$ is position: $v=s'$ , $a=v'=s''$ , jerk $j=s'''$ .

Domain of f′

$f'$ can have a smaller domain than $f$ . $\sqrt{x}$ : domain $[0,\infty)$ , but $f'(x)=\frac1{2\sqrt x}$ only on $(0,\infty)$ .

Worked examples

Example · definition with a square root (Stewart Ex. 3 style)

Find $f'(x)$ for $f(x)=\sqrt{x}$ using the definition. State the domain of $f'$ .

Write the difference quotient: $\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$ .
Plugging in $h=0$ gives $\frac00$ . For square roots, multiply by the conjugate.
$\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} = \dfrac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$
The $x$ 's cancel, leaving $\dfrac{h}{h(\sqrt{x+h}+\sqrt x)} = \dfrac{1}{\sqrt{x+h}+\sqrt x}$ . The $h$ cancels.
Now let $h\to0$ : $f'(x)=\dfrac{1}{2\sqrt x}$ .
Domain of $f'$ : $(0,\infty)$ . At $x=0$ we would divide by zero (vertical tangent).

Example · definition with a fraction (Stewart Ex. 4 style)

Find $f'(x)$ for $f(x)=\dfrac{1-x}{2+x}$ .

$\dfrac{f(x+h)-f(x)}{h} = \dfrac{1}{h}\left[\dfrac{1-x-h}{2+x+h}-\dfrac{1-x}{2+x}\right]$
Fractions: combine over a common denominator. Numerator: $(1-x-h)(2+x)-(1-x)(2+x+h)$ .
Expand: $(2 - x - 2h - x^2 - xh) - (2 - x + h - x^2 - xh) = -3h$ .
So the quotient is $\dfrac{-3h}{h(2+x+h)(2+x)} = \dfrac{-3}{(2+x+h)(2+x)}$ .
Let $h\to0$ : $f'(x) = \dfrac{-3}{(2+x)^2}$ . Domain: $x\neq-2$ .

Practice

§2.8 #21

Use the definition to find

f'(x)

for

f(x)=3x-8

. State the domains of

f

and

f'

§2.8 #23

Use the definition to find

f'(t)

for

f(t)=2.5t^2+6t

§2.8 #28

Use the definition to find

F'(v)

for

F(v)=\dfrac{v}{v+2}

. State domains.

§2.8 #29

Use the definition to find

g'(u)

for

g(u)=\dfrac{u+1}{4u-1}

§2.8 #31

Use the definition to find

f'(x)

for

f(x)=\dfrac{1}{\sqrt{1+x}}

. State domains.

§2.8 #33(c)

For

f(x)=1+\sqrt{x+3}

, use the definition to find

f'(x)

. What are the domains of

f

and

f'

§2.8 #30

Use the definition to show

\dfrac{d}{dx}x^4=4x^3

Concept · §2.8 #41–46 style

Where is

f(x)=|x-2|+\begin{cases}0&x\lt 5\\ 1 & x\ge5\end{cases}

not differentiable, and why?

§2.8 · higher derivatives

f(x)=x^3-x

, find

f'

f''

f'''

f^{(4)}

Traps

Plugging in h = 0 too early

You always get 0/0 at first. Simplify until the $h$ in the denominator cancels, then set $h=0$ .

Forgetting to distribute the minus

$f(x+h)-f(x)$ : put $f(x)$ in brackets. Most definition errors are sign errors here.

Domain of f′ = domain of f

Not always. Check for division by zero in $f'$ (e.g. $\sqrt{x}$ at 0).