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The Chain Rule Made Simple: A Step-by-Step Visual Guide

The Chain Rule is used when you need to find the derivative of a "function inside another function."

Core Theorem
If y=f(g(x))y = f(g(x)), then:
y′=f′(g(x))⋅g′(x)y' = f'(g(x)) \cdot g'(x)
Step-by-Step SOP
  1. 1

    Identify

    Use a 'U' to substitute the inner function g(x)g(x).
  2. 2

    Peel the Outer

    Differentiate f(u)f(u) normally.
  3. 3

    Chain the Inner

    Multiply by du/dxdu/dx.

Practice Exercises

Example 01Easy
Find f′(x)f'(x) for f(x)=(3x2+1)4f(x) = (3x^2 + 1)^4.
NEED A HINT?
Identify the 'outer' function as u4u^4 and the 'inner' function as 3x2+13x^2 + 1.
SHOW DETAILED EXPLANATION

Step 1: Outer Derivative

Differentiate the power: 4(3x2+1)34(3x^2 + 1)^3.

Step 2: Inner Derivative

Differentiate 3x2+13x^2 + 1 to get 6x6x.

Step 3: Combine

Multiply them: 4(3x2+1)3⋅6x=24x(3x2+1)34(3x^2 + 1)^3 \cdot 6x = 24x(3x^2 + 1)^3.
Example 02AB/BC Standard
Find h′(2)h'(2) if h(x)=f(g(x))h(x) = f(g(x)), given g(2)=3,g′(2)=5,f′(3)=7g(2)=3, g'(2)=5, f'(3)=7.
NEED A HINT?
Apply the formula h′(x)=f′(g(x))g′(x)h'(x) = f'(g(x))g'(x) directly with the given values.
SHOW DETAILED EXPLANATION

Step 1: Write Formula

h′(2)=f′(g(2))⋅g′(2)h'(2) = f'(g(2)) \cdot g'(2).

Step 2: Substitute $g(2)$

h′(2)=f′(3)⋅g′(2)h'(2) = f'(3) \cdot g'(2).

Step 3: Final Calculation

h′(2)=7⋅5=35h'(2) = 7 \cdot 5 = 35.
Common Pitfalls
  • ⚠
    The 'Inside' remains the sameDon't differentiate the inner part insideinside the outer derivative. d/dx[sin(x2)]d/dx [sin(x^2)] is cos⁡(x2)⋅2x\cos(x^2) \cdot 2x, NOT cos⁡(2x)\cos(2x).
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