Derivatives of Polynomials and Exponential Functions

The shortcuts that replace the limit definition.

0 / 10 problems mastered

Constant

\frac{d}{dx}(c)=0

A horizontal line has slope 0.

Power Rule (any real n)

\frac{d}{dx}x^n = n\,x^{n-1}

Bring the power down, subtract one.

Constant multiple

\frac{d}{dx}[c\,f]=c\,f'

Sum / Difference

(f\pm g)'=f'\pm g'

Natural exponential

\frac{d}{dx}e^x=e^x

e is the one base whose slope at 0 is exactly 1.

Rewrite before you differentiate. The Power Rule needs the form $x^n$ . Convert roots and fractions first:

\sqrt{x}=x^{1/2},\quad \sqrt[3]{x^2}=x^{2/3},\quad \frac{1}{x^3}=x^{-3},\quad \frac{4}{\sqrt[3]{x}}=4x^{-1/3}

If a fraction has a single term on the bottom, split it: $\dfrac{3x^2+x^3}{x}=3x+x^2$ . No quotient rule needed.

Why e is special

Lab 3 · slide the base

base a

f(x) = aˣa = 2

slope at x = 0, lim (aʰ − 1)/h–

so f′(x) = (that slope) · aˣ–

For any base, the derivative of aˣ is aˣ times a constant (its slope at 0). When a = e that constant is exactly 1, and the pink derivative curve lands right on top of the blue curve.

Tangent and normal lines

The tangent line at $(a,f(a))$ is $y-f(a)=f'(a)(x-a)$ . The normal line is perpendicular to the tangent, so its slope is $-\dfrac{1}{f'(a)}$ .

A horizontal tangent happens where $f'(x)=0$ . Set the derivative to zero and solve.

Slope of the tangent vs. the tangent line

These sound alike but ask for different things. The slope is one number. The tangent line is a whole equation, and the slope is just one of its two ingredients.

“Slope of the tangent line” → a number

One step: plug $a$ into the derivative.

m = f'(a)

Same answer for “rate of change at $a$ ” or “find $f'(a)$ ”.

“Tangent line” → an equation

Three steps: a point, a slope, then point-slope form.

y - f(a) = f'(a)\,(x - a)

The point comes from $f$ . The slope comes from $f'$ .

Slope only · practice quiz Q2

Find the slope of the tangent line to $f(x)=x^2-4x$ at $x=3$ .

Differentiate: $f'(x)=2x-4$ .
Plug in 3: $f'(3)=6-4=2$ .
Answer: 2. It's a number, so you're done.

Full line · practice quiz Q6

Find the tangent line to $y=x^3-2x$ at $x=1$ .

Point (use $y$ ): $y(1)=1-2=-1$ , so $(1,-1)$ .
Slope (use $y'$ ): $y'=3x^2-2$ , so $y'(1)=1$ .
Equation: $y-(-1)=1(x-1)$ , so $y=x-2$ .

Same function, both questions

Turn Q2 into a tangent-line question: find the tangent line to $f(x)=x^2-4x$ at $x=3$ .

Point: $f(3)=9-12=-3$ , so $(3,-3)$ .
Slope: $f'(3)=2$ (the Q2 answer).
Line: $y-(-3)=2(x-3)$ , so $y=2x-9$ .
Q2's answer (2) is the slope of this line.

	Slope of the tangent	Tangent line
Answer is	a number	an equation $y=\dots$
Uses $f(a)$ ?	No	Yes, for the point
Uses $f'(a)$ ?	Yes	Yes, for the slope
Q2 / Q6 answer	$2$	$y=x-2$

Read the wording: “slope”, “rate of change” or “ $f'(a)$ ” → give a number. “Tangent line” or “equation of the tangent line” → give $y=\dots$ . “Normal line” → same point, slope $-\dfrac{1}{f'(a)}$ .

Getting the point from f′

The y-coordinate of the point is $f(a)$ , never $f'(a)$ . In Q6, $y(1)=-1$ is the point and $y'(1)=1$ is the slope.

Stopping at the slope

If the question says “tangent line”, the number $f'(a)$ alone gets partial credit at best. Finish with $y=\dots$ .

Writing a line for a slope question

If it only asks for the slope, give the number. An equation answers a different question.

Worked examples

Example · rewrite, then differentiate

Differentiate $y=\dfrac{\sqrt{x}+x}{x^2}$ .

Split: $y=\dfrac{x^{1/2}}{x^2}+\dfrac{x}{x^2}=x^{-3/2}+x^{-1}$ .
Power rule on each: $y'=-\tfrac32x^{-5/2}-x^{-2}$ .
Optional cleanup: $y'=-\dfrac{3}{2x^{5/2}}-\dfrac1{x^2}$ .

Example · horizontal tangents (Stewart Ex. 7)

Find the points on $y=x^4-6x^2+4$ where the tangent line is horizontal.

$y'=4x^3-12x=4x(x^2-3)$ .
Set $y'=0$ : $x=0$ or $x=\pm\sqrt3$ .
Plug back into $y$ (not $y'$ ): $y(0)=4$ , $y(\pm\sqrt3)=9-18+4=-5$ .
Points: $(0,4)$ , $(\sqrt3,-5)$ , $(-\sqrt3,-5)$ .

Practice

§3.1 #18

Differentiate

W(t)=\sqrt t-2e^t

§3.1 #21

Differentiate

y=3e^x+\dfrac{4}{\sqrt[3]{x}}

§3.1 #23

Differentiate

f(x)=\dfrac{3x^2+x^3}{x}

§3.1 #25

Differentiate

G(r)=\dfrac{3r^{3/2}+r^{5/2}}{r}

§3.1 #28

Differentiate

k(r)=e^r+r^e

§3.1 #30

Differentiate

G(q)=(1+q^{-1})^2

§3.1 #37

Find the tangent line to

y=2x^3-x^2+2

(1,3)

§3.1 #41

Find the tangent and normal lines to

y=x^4+2e^x

(0,2)

§3.1 #49

Find

f'

and

f''

for

f(x)=0.001x^5-0.02x^3

§3.1 · motion

A particle moves by

s=t^3-6t^2+9t

(meters, seconds). When is it at rest? What is its acceleration then?

Traps

Power rule on eˣ

$\frac{d}{dx}e^x\neq xe^{x-1}$ . The power rule is for variable base, constant exponent.

Derivative of a constant like e² or π³

These are numbers. Their derivative is 0.

Negative exponents

$\frac{d}{dx}x^{-3}=-3x^{-4}$ , not $-3x^{-2}$ . Subtracting 1 makes it more negative.