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AP Calculus AB/BC

Differentiation

AP CALCULUS · UNIT 2 — DIFFERENTIATION: DEFINITION & FUNDAMENTAL PROPERTIES

The theory, at a glance

The limit definition, differentiability, and the power, product and quotient rules. ✎

no calculator

① The Limit Definition

f′(x) = lim h→0 f(x+h) − f(x) / h

The alternate form, useful at a specific point x = a:

f′(a) = lim x→a f(x) − f(a) / x − a

② Differentiability

differentiable ⟹ continuous (not the reverse!)

Four ways a derivative fails to exist:

vertical tangent

discontinuity

|x| at x = 0

one-sided derivatives

③ The Basic Rules

sec²x

1⁄x

powers first

④ Product & Quotient Rules

(uv)′ = u′v + uv′

u / v ′ = u′v − uv′ / v²

"low d-high minus high d-low, over low squared"

order matters

★ Worked example ✎

Let f(x) = x²·sin x. Find f′(x), then find the equation of the tangent line at x = 0.

This is a product, so I'll name the two factors and their derivatives before combining.

u = x², u′ = 2x · v = sin x, v′ = cos x

Apply the product rule u′v + uv′.

f′(x) = 2x·sin x + x²·cos x

For the tangent I need the slope at x = 0.

f′(0) = 2(0)(0) + 0·(1) = 0

And the point itself, from the original function.

f(0) = 0²·sin 0 = 0, so the point is (0, 0)

Point-slope form with m = 0 through the origin.

y = 0

AP tip

Never differentiate a product term-by-term.

AP CALCULUS · UNIT 3 — COMPOSITE, IMPLICIT & INVERSE FUNCTIONS

Chain, Implicit & Inverse

The chain rule, implicit differentiation, inverse functions and higher derivatives. ✎

① The Chain Rule

[ f(g(x)) ]′ = f′(g(x)) · g′(x)

Outside first, then multiply by the inside's derivative.

2x

x²

5x

② Implicit Differentiation

When y isn't isolated, differentiate both sides with respect to x — every y term picks up a dy/dx by the chain rule:

Differentiate every term; y² → 2y·dy/dx.

Collect all dy/dx terms on one side.

Factor it out and divide.

Watch outProducts of x and y need the product rule too: (xy)′ = y + x·dy/dx. The answer usually contains both x and y — that's expected, not a mistake.

③ Inverse Functions & Inverse Trig

(f −1 )′(a) = 1 / f′(f −1 (a))

1 ⁄ √(1 − x²)

−1 ⁄ √(1 − x²)

1 ⁄ (1 + x²)

To use the inverse formula you often must find f −1 (a) first — ask "what input gives output a?" Note arcsin and arccos differ only by a sign.

④ Higher Derivatives

Just differentiate again. Notation: f″(x) or d²y/dx².

For implicit problems, differentiate dy/dx implicitly again — and substitute the first derivative back in to simplify.

Find dy/dx for x² + xy + y² = 7, then the slope of the tangent at (1, 2).

Differentiate every term with respect to x. The middle term xy needs the product rule.

2x + (y + x·y′) + 2y·y′ = 0

Gather the y′ terms on one side, everything else on the other.

x·y′ + 2y·y′ = −2x − y

Factor out y′ and divide.

y′(x + 2y) = −(2x + y) → y′ = −(2x + y) / x + 2y

Substitute the point. Check it's on the curve first: 1 + 2 + 4 = 7 ✓

y′ = −(2 + 2) / 1 + 4

slope = −4⁄5

Don't forget dy/dx on every y term

Download the PDF cheatsheet

One page, free — along with nine others.

↓ PDF

Practice, with the working

Three questions from the bank, each with its full solution.

1Easier
The limit lim⁡h→0(2+h)5−32h\displaystyle\lim_{h \to 0} \dfrac{(2 + h)^5 - 32}{h} represents f′(c)f'(c) for some function ff and number cc. Identify ff and cc, and evaluate the limit.
›Show solution
**Solution**
[M] The expression matches lim⁡h→0f(c+h)−f(c)h\displaystyle\lim_{h \to 0} \dfrac{f(c + h) - f(c)}{h} with f(x)=x5f(x) = x^5, c=2c = 2 (note 32=2532 = 2^5).
[M] f′(x)=5x4f'(x) = 5x^4.
[A] Limit =5⋅24=80= 5 \cdot 2^4 = 80.
**Answer:** f(x)=x5f(x) = x^5, c=2c = 2; limit =80= 80
2Medium
Let f(x)=∣x−2∣f(x) = |x - 2|. Show that ff is continuous at x=2x = 2 but not differentiable there.
›Show solution
**Solution**
[M] lim⁡x→2∣x−2∣=0=f(2)\displaystyle\lim_{x \to 2} |x - 2| = 0 = f(2), so ff is continuous at 2.
[M] Left difference quotient: ∣2+h−2∣−0h=−hh=−1\dfrac{|2 + h - 2| - 0}{h} = \dfrac{-h}{h} = -1 for h<0h < 0.
[M] Right: hh=1\dfrac{h}{h} = 1 for h>0h > 0.
[A] One-sided derivatives differ (−1≠1-1 \ne 1), so f′(2)f'(2) does not exist — a corner. ■\blacksquare
**Answer:** Continuous; one-sided derivatives −1≠1-1 \ne 1
3Harder
Let f(x)=x2ln⁡xf(x) = x^2 \ln x (for x>0x > 0) and g(x)=exxg(x) = \dfrac{e^x}{x} (for x≠0x \ne 0).
›Show solution
**Part (a)**
_Find f′(x)f'(x) and evaluate f′(e)f'(e)._
[M] Product rule: f′(x)=2xln⁡x+x2⋅1x=2xln⁡x+xf'(x) = 2x\ln x + x^2 \cdot \dfrac{1}{x} = 2x\ln x + x.
[A] f′(e)=2e⋅1+e=3ef'(e) = 2e \cdot 1 + e = 3e.
**Answer:** f′(x)=2xln⁡x+xf'(x) = 2x\ln x + x; f′(e)=3ef'(e) = 3e
**Part (b)**
_Find g′(x)g'(x) and the xx-value where the graph of gg has a horizontal tangent._
[M] Quotient rule: g′(x)=xex−exx2=ex(x−1)x2g'(x) = \dfrac{x e^x - e^x}{x^2} = \dfrac{e^x(x - 1)}{x^2}.
[M] ex≠0e^x \ne 0, so g′(x)=0g'(x) = 0 only when x=1x = 1.
[A] Horizontal tangent at x=1x = 1 (point (1,e)(1, e)).
**Answer:** g′(x)=ex(x−1)x2g'(x) = \dfrac{e^x(x - 1)}{x^2}; x=1x = 1

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