Chain Rule Formula

Chain rule is the derivative of a composite function f(g(x)) equals f'(g(x)) x g'(x): the derivative of the outer function evaluated at the inner, times.

The Formula

(f∘g)′(x)=f′(g(x))⋅g′(x)(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

When to use: Derivative of outside times derivative of inside. Unpack layers.

Quick Example

ddx(sin⁡(x2))=cos⁡(x2)⋅2x\frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot 2x Outside: sin⁡\sin. Inside: x2x^2.

Notation

In Leibniz notation: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} where y=f(u)y = f(u) and u=g(x)u = g(x).

What This Formula Means

The derivative of a composite function f(g(x))f(g(x)) equals f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x): the derivative of the outer function evaluated at the inner, times the derivative of the inner.

Derivative of outside times derivative of inside. Unpack layers.

Formal View

If gg is differentiable at xx and ff is differentiable at g(x)g(x), then (f∘g)′(x)=f′(g(x))⋅g′(x)(f \circ g)'(x) = f'(g(x)) \cdot g'(x).

Worked Examples

Example 1

easy
Find the derivative of f(x)=(3x+1)4f(x) = (3x + 1)^4.

Answer

f′(x)=12(3x+1)3f'(x) = 12(3x + 1)^3

First step

1
Identify the outer function u4u^4 and the inner function u=3x+1u = 3x + 1.

Full solution

  1. 2
    Apply the chain rule: ddx[u4]=4u3⋅dudx\frac{d}{dx}[u^4] = 4u^3 \cdot \frac{du}{dx}.
  2. 3
    The derivative of the inner function: dudx=3\frac{du}{dx} = 3.
  3. 4
    Combine: f′(x)=4(3x+1)3⋅3=12(3x+1)3f'(x) = 4(3x+1)^3 \cdot 3 = 12(3x+1)^3.
The chain rule says: differentiate the outer function, keep the inner function, then multiply by the derivative of the inner function. Think of it as peeling layers.

Example 2

medium
Find the derivative of f(x)=(x2+1)5f(x) = (x^2 + 1)^5.

Example 3

hard
Find the derivative of f(x)=sin⁡(x3)f(x) = \sin(x^3).

Common Mistakes

  • Forgetting the inner-derivative factor g′(x)g'(x) — every composite contributes a multiply-by-the-inside step.
  • Evaluating the outer derivative at xx instead of at g(x)g(x) — f′f' must be applied to the inner function, e.g. cos⁡(x2)\cos(x^2) not cos⁡x\cos x.
  • Confusing nesting with multiplication — sin⁡(x2)\sin(x^2) is composite (chain rule) while x2sin⁡xx^2\sin x is a product (product rule).

Why This Formula Matters

Most real derivatives in physics and modeling are composites — cos⁡(2t)\cos(2t), ekxe^{kx}, x2+1\sqrt{x^2+1} — so without the chain rule the differentiation rules cover almost nothing. The extra g′(x)g'(x) factor is exactly what students forget, and it is the difference between a correct answer and one that's off by the inner rate. Recognizing it by "Is one function plugged into another, so I must multiply the outer derivative by the inner derivative?" — rather than by familiar numbers — is what lets a student tell it apart from product rule and power rule alone and u-substitution in a mixed problem set.

Frequently Asked Questions

What is the Chain Rule formula?

The derivative of a composite function f(g(x))f(g(x)) equals f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x): the derivative of the outer function evaluated at the inner, times the derivative of the inner.

How do you use the Chain Rule formula?

Derivative of outside times derivative of inside. Unpack layers.

What do the symbols mean in the Chain Rule formula?

In Leibniz notation: dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} where y=f(u)y = f(u) and u=g(x)u = g(x).

Why is the Chain Rule formula important in Math?

Most real derivatives in physics and modeling are composites — cos⁡(2t)\cos(2t), ekxe^{kx}, x2+1\sqrt{x^2+1} — so without the chain rule the differentiation rules cover almost nothing. The extra g′(x)g'(x) factor is exactly what students forget, and it is the difference between a correct answer and one that's off by the inner rate. Recognizing it by "Is one function plugged into another, so I must multiply the outer derivative by the inner derivative?" — rather than by familiar numbers — is what lets a student tell it apart from product rule and power rule alone and u-substitution in a mixed problem set.

What do students get wrong about Chain Rule?

The procedure for chain rule is the easy part; the trap is forgetting the inner-derivative factor g′(x)g'(x). Asking "Is one function plugged into another, so I must multiply the outer derivative by the inner derivative?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

What should I learn before the Chain Rule formula?

Before studying the Chain Rule formula, you should understand: derivative, composition.

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Derivatives Explained: Rules, Interpretation, and Applications →
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