What is the derivative of y = (4x - 1)^5 with respect to x?

20(4x - 1)^4. Outer derivative is 5(4x-1)^4, inner derivative is 4. Multiplying gives 5 4(4x-1)^4 = 20(4x-1)^4.

-e^ (x) (x). Derivative of e^u is e^u u'. Here u = (x), so u' = - (x), giving -e^ (x) (x).

Differentiate y = x^3 + 7.

3x^22 x^3 + 7. Rewrite as (x^3+7)^1/2. Derivative is 12(x^3+7)^-1/2 3x^2 = 3x^22 x^3+7.

If y = (3x), what is dydx?

3 ^2(3x). Outer derivative of (u) is ^2(u), inner derivative of 3x is 3. Thus 3 ^2(3x).

If h(x) = f(g(x)), g(2) = 3, g'(2) = -4, and f'(3) = 5, what is the value of h'(2)?

-20 1. Apply the chain rule: h'(x) = f'(g(x)) g'(x). 2. Substitute x = 2: h'(2) = f'(g(2)) g'(2). 3. We know g(2) = 3, so f'(g(2)) = f'(3) = 5. 4. Therefore, h'(2) = 5 (-4) = -20.

Mathematics · AP Calculus · College Level

Derivatives and the Chain Rule

TheChain Ruleis the fundamental differentiation technique in calculus used to compute the derivative of acomposite function $f(g(x))$ , multiplying the derivative of the outer function by the derivative of the inner function.

1What is the Chain Rule? Visual Overview

In calculus, simple formulas like the power rule $\frac{d}{dx}[x^n] = n x^{n-1}$ only work when the base is simply $x$ . When an expression is "nested" inside another function—such as $(3x^2 + 1)^5$ or $\sin(4x)$ —we must apply theChain Rule. Think of it as coupled gears transmitting rates of change.

The Chain Rule: Rate of Change Transmission: Input x, Inner Function g(x), Inner Rate du/dx, Outer Function f(u), Outer Rate dy/du, Chain Multiplier — The Chain Rule: Rate of Change Transmission
Think of composite functions as coupled gears: the rate of change of y with respect to x equals the rate of change of y relative to u multiplied by the rate of change of u relative to x.
1Input x — The base independent variable driving the system
2Inner Function g(x) — Transforms input x into intermediate variable u = g(x)
3Inner Rate du/dx — How fast the intermediate value u changes as x moves
4Outer Function f(u) — Operates on u to produce the final output y = f(u)
5Outer Rate dy/du — Differentiates the outer structure while holding the inside frozen
6Chain Multiplier — Multiplying the two rates yields dy/dx = (dy/du) * (du/dx)

Definition

Composite Function — A function $y = f(g(x))$ where the output of an inner function $u = g(x)$ serves as the input to an outer function $f(u)$ .

The Chain Rule (Function Notation)

\frac{d}{dx}\Big[f(g(x))\Big] = f'(g(x)) \cdot g'(x)

f(g(x)) = The composite function
f'(g(x)) = Derivative of outer function evaluated at unchanged inner function
g'(x) = Derivative of the inner function

The Chain Rule (Leibniz Notation)

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

y = Outer dependent variable ($y = f(u)$)
u = Intermediate inner variable ($u = g(x)$)
x = Independent input variable

24-Step Algorithm & Worked Examples

4-Step Algorithm to Differentiate Any Composite Function

1
1. Identify Layers
Isolate the outer shell f(u) and the inner expression u = g(x)
2
2. Differentiate Outer
Take derivative of f, leaving the inner argument completely untouched
3
3. Differentiate Inner
Compute g'(x) separately on scrap paper
4
4. Multiply & Simplify
Multiply outer result by g'(x) and simplify algebraic factors

Real-life example · Example 1: General Power Rule

Differentiate $y = (2x^3 - 5x)^4$ .
• Step 1: Let inner $u = 2x^3 - 5x$ , outer $y = u^4$ .
• Step 2: $\frac{dy}{du} = 4u^3 = 4(2x^3 - 5x)^3$ .
• Step 3: $\frac{du}{dx} = 6x^2 - 5$ .
• Step 4: $\frac{dy}{dx} = 4(2x^3 - 5x)^3 \cdot (6x^2 - 5)$ .

Real-life example · Example 2: Trigonometric Composite

Differentiate $y = \cos(5x^2 + 1)$ .
• Outer is $\cos(u)$ , so derivative is $-\sin(u) = -\sin(5x^2 + 1)$ .
• Inner is $5x^2 + 1$ , derivative is $10x$ .
• Result: $\frac{dy}{dx} = -10x \sin(5x^2 + 1)$ .

3Common Derivative Patterns & Traps

Common Standard Chain Rule Patterns

Function Type	Composite Formula y	Derivative dy/dx	Quick Example
Power Chain	$[g(x)]^n$	$n[g(x)]^{n-1} \cdot g'(x)$	$\frac{d}{dx}[(x^2+1)^3] = 3(x^2+1)^2(2x)$
Exponential	$e^{g(x)}$	$e^{g(x)} \cdot g'(x)$	$\frac{d}{dx}[e^{3x}] = 3e^{3x}$
Natural Log	$\ln(g(x))$	$\frac{g'(x)}{g(x)}$	$\frac{d}{dx}[\ln(x^2+4)] = \frac{2x}{x^2+4}$
Trig (Sine)	$\sin(g(x))$	$\cos(g(x)) \cdot g'(x)$	$\frac{d}{dx}[\sin(4x)] = 4\cos(4x)$

Power Chain
Composite Formula y
$[g(x)]^n$
Derivative dy/dx
$n[g(x)]^{n-1} \cdot g'(x)$
Quick Example
$\frac{d}{dx}[(x^2+1)^3] = 3(x^2+1)^2(2x)$
Exponential
Composite Formula y
$e^{g(x)}$
Derivative dy/dx
$e^{g(x)} \cdot g'(x)$
Quick Example
$\frac{d}{dx}[e^{3x}] = 3e^{3x}$
Natural Log
Composite Formula y
$\ln(g(x))$
Derivative dy/dx
$\frac{g'(x)}{g(x)}$
Quick Example
$\frac{d}{dx}[\ln(x^2+4)] = \frac{2x}{x^2+4}$
Trig (Sine)
Composite Formula y
$\sin(g(x))$
Derivative dy/dx
$\cos(g(x)) \cdot g'(x)$
Quick Example
$\frac{d}{dx}[\sin(4x)] = 4\cos(4x)$

Common mistake: The Most Common Student Mistake: Differentiating the inside function INSIDE the outer function! For example, writing

\frac{d}{dx}[\sin(x^2)] = \cos(2x)

. This is WRONG! The correct answer is

\cos(x^2) \cdot 2x

. The inner function must remain intact during outer differentiation!

4Exam-Style AP Question

Exam question · AP Calculus AB / BC

h(x) = f(g(x))

g(2) = 3

g'(2) = -4

, and

f'(3) = 5

, what is the value of

h'(2)

A.-20
B.15
C.-12
D.1

Remember

The chain rule is required whenever a function has another function nested inside it.
Core formula: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$ .
Never differentiate the inner function inside the outer function — differentiate the outer shell first, keeping the inner argument intact, then multiply by the inner derivative.
In Leibniz notation: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ .
For nested compositions $f(g(h(x)))$ , apply the chain rule iteratively from outside to innermost.

Check yourself

Question 1
What is the derivative of $y = (4x - 1)^5$ with respect to $x$ ?
Question 2
Find $\frac{d}{dx}[e^{\cos(x)}]$ .
Question 3
Differentiate $y = \sqrt{x^3 + 7}$ .
Question 4
If $y = \tan(3x)$ , what is $\frac{dy}{dx}$ ?