AP Calculus AB · Lesson review · September 18, 2026

Product and Quotient Rules

Prove the rules from the derivative definition, apply them to polynomial and trigonometric products, and derive the tangent and secant rules.

38-minute edited lesson10 chaptersCaptions + corrected transcript

01 · Watch

Lesson video

02 · Understand

Learning targets

03 · Rules

The derivative rules

These rules combine functions without pretending the derivative distributes over multiplication or division.

01Product rule
ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)Differentiate the first factor, then the second factor, and add the two products.
02Quotient rule
ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}Use low d-high minus high d-low, over low squared.
03Tangent
ddx(tanx)=sec2x\frac{d}{dx}(\tan x)=\sec^2xWrite tangent as sine over cosine, then simplify with the Pythagorean identity.
04Secant
ddx(secx)=secxtanx\frac{d}{dx}(\sec x)=\sec x\tan xWrite secant as one over cosine and apply the quotient rule.

04 · Derive and apply

Worked examples

Name the factors first, apply the correct rule, and simplify afterward.

Why it works

Product rule from the definition

h(x)=limh0f(x+h)g(x+h)f(x)g(x)hh'(x)=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}

Add and subtract the same middle expression so the numerator separates into two recognizable difference quotients.

h(x)=f(x)g(x)+f(x)g(x)h'(x)=f'(x)g(x)+f(x)g'(x)
Product rule

Polynomial times sine

f(x)=x4sinxf(x)=x^4\sin x

Keep each original factor once while differentiating the other.

f(x)=4x3sinx+x4cosxf'(x)=4x^3\sin x+x^4\cos x
Three factors

Apply the product rule repeatedly

f(x)=x5cosxexf(x)=x^5\cos x\,e^x

Differentiate one factor at a time. The final derivative has three terms.

f(x)=5x4cosxexx5sinxex+x5cosxexf'(x)=5x^4\cos x\,e^x-x^5\sin x\,e^x+x^5\cos x\,e^x
Quotient rule

Polynomial over sine

f(x)=x3sinxf(x)=\frac{x^3}{\sin x}

Differentiate the numerator and denominator, preserve the subtraction order, and square the full denominator.

f(x)=3x2sinxx3cosxsin2xf'(x)=\frac{3x^2\sin x-x^3\cos x}{\sin^2x}
Trig proof

Derivative of tangent

ddx(sinxcosx)=cos2x+sin2xcos2x\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)=\frac{\cos^2x+\sin^2x}{\cos^2x}

The numerator becomes cos2x+sin2x=1\cos^2x+\sin^2x=1.

ddx(tanx)=sec2x\frac{d}{dx}(\tan x)=\sec^2x
Trig proof

Derivative of secant

ddx(1cosx)=sinxcos2x=1cosxsinxcosx\frac{d}{dx}\left(\frac1{\cos x}\right)=\frac{\sin x}{\cos^2x}=\frac1{\cos x}\frac{\sin x}{\cos x}

Rewrite the result as a product of secant and tangent.

ddx(secx)=secxtanx\frac{d}{dx}(\sec x)=\sec x\tan x

05 · Avoid

Common mistakes

Multiplying the derivatives.

(fg)(fg)' is not fgf'g'. The product rule has two added terms.

Dropping an unchanged factor.

Each product-rule term contains one derivative and one original factor.

Reversing the quotient numerator.

Keep the order fgfgf'g-fg'.

Forgetting the square.

The denominator of the quotient-rule result is [g(x)]2[g(x)]^2.

Stopping before simplifying.

Use identities such as sin2x+cos2x=1\sin^2x+\cos^2x=1 to reveal standard trig derivatives.

Using only two terms for three factors.

A three-factor product produces three derivative terms.

06 · Check yourself

Quick check

Try each before revealing the answer.

1Differentiate x4sinxx^4\sin x.

4x3sinx+x4cosx4x^3\sin x+x^4\cos x.

2Which factor is squared in the quotient-rule denominator?

The original denominator function: [g(x)]2[g(x)]^2.

3Differentiate tanx\tan x.

sec2x\sec^2x.

4Differentiate secx\sec x.

secxtanx\sec x\tan x.

5Why is (fg)=fg(fg)'=f'g' incorrect?

Changing either factor changes the product. The correct derivative adds both contributions: fg+fgf'g+fg'.

07 · Revisit

Transcript and practice

Search the corrected transcript for a rule or example. Captions follow the edited video timeline.

Download corrected transcript

Return to the Friday 9/18 eKadence activity for the assigned practice.