Mathematics · Calculus

Chain Rule of Differentiation

The chain rule differentiates composite functions f(g(x)): outer derivative times inner derivative.

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Formula

LaTeX: (f \circ g)'(x) = f'(g(x)) \cdot g'(x)
Dimensionless (calculus)

Variables & units – Chain Rule of Differentiation

SymbolMeaningUnit
fOuter functiondimensionless
gInner functiondimensionless
f', g'Derivativesdimensionless

Derivation & background – Chain Rule of Differentiation

Mnemonic: "differentiate the outer function, leave the inner one in place, then multiply by the inner derivative." Leibniz notation: dy/dx = (dy/du)·(du/dx). Foundation of the backpropagation algorithm in neural networks.

Exam blueprint

Validity range

Applies to composite functions when inner and outer functions are differentiable at the relevant points.

Derivation steps

A change in x first changes the inner variable and then the outer function value.

  1. 1Set u = g(x).
  2. 2Then dy/dx = dy/du · du/dx.

Rearrangements

Isolate inner derivative

Only meaningful when f'(g(x)) is not zero.

Task variant

Differentiate e^{3x²}.

Outer derivative e^{3x²}, inner derivative 6x. Result: 6x e^{3x²}.

Common mistakes

Forgetting the inner derivative.

After differentiating the outside, always multiply by g'(x).

Exam context

  • Core rule for exponential, logarithmic, trigonometric and substitution tasks.

These mistakes cost points in real exams. The set drills them until they stick.

Worked example

sin(x²): outer function sin, inner function x². → cos(x²) · 2x. Check at x = 0: cos(0)·0 = 0 (local minimum) ✓

Applications

Machine learning (backpropagation), physics (relativity, chained coordinate transformations)

Quanta exam set

Curated exam set for "Chain Rule of Differentiation":

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Which formula describes Chain Rule of Differentiation?

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How do you rearrange (f∘g)' = f'(g(x))·g'(x) for Isolate inner derivative?

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Which common mistake happens with Chain Rule of Differentiation?

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Scientific sources

Common notations & search queries

(f(g(x)))'=f'(g(x))*g'(x)außere mal innere Ableitungd/dx f(g(x))Kettenregel Ableitungverkettete Funktionen ableitenchain rule derivativeinnere äußere Ableitung Formel

Related formulas

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Frequently asked questions about Chain Rule of Differentiation

How do you differentiate a composite function with the chain rule?+

By the chain rule (f∘g)′(x) = f′(g(x))·g′(x). So you differentiate the outer function and insert the unchanged inner function into it, then multiply by the derivative of the inner function. Remember it as outer derivative times inner derivative. Example: for e^(3x²) the outer function is e to the power of something with derivative e^(3x²), the inner function is 3x² with derivative 6x. The result is therefore 6x·e^(3x²). The most important step is not to forget the inner derivative; it is the most common reason for wrong results with composite functions.

What is the inner and what is the outer function?+

In a composite function f(g(x)), g is the inner function, because it acts on x first, and f is the outer function, because it is applied to the result of g. You recognize the inner function because it sits in the argument of the outer one, that is in the bracket or under the function symbol. In sin(x²), x² is inside and the sine outside. In √(2x + 1), 2x + 1 is inside and the root outside. A helpful question is: what would I enter first into the calculator? This quantity computed first is the inner function. Its derivative provides the decisive factor of the chain rule.

Why must you not forget the inner derivative?+

Because a change in x first changes the inner quantity and this change is then passed on through the outer function. The chain rule dy/dx = dy/du · du/dx shows that the two rates of change multiply. If you omit the inner derivative, you only account for how strongly the outer function reacts to its argument, but not how fast the argument itself changes with x. Example: for sin(3x) the derivative is 3·cos(3x), not just cos(3x); the factor 3 comes from the inner derivative. Without it the result would be too small by exactly this factor. That is why the inner derivative is indispensable.

How do you differentiate multiply nested functions?+

For multiply nested functions you apply the chain rule repeatedly, so to speak from outside to inside. For f(g(h(x))) the derivative is f′(g(h(x)))·g′(h(x))·h′(x). So you differentiate each layer individually and multiply all inner derivatives one after another. Example: sin(√(x² + 1)). The outermost is the sine, then the root, then x² + 1. The derivative is cos(√(x² + 1))·(1/(2√(x² + 1)))·2x. Work systematically from the outermost function inward and multiply all factors. If you forget a layer, a factor is missing and the result is wrong.

How do you differentiate e^(kx) with the chain rule?+

For e^(kx) the outer function is the exponential function, whose derivative is again e^(kx), and the inner function is kx with derivative k. By the chain rule you multiply both and get k·e^(kx). The constant factor k from the exponent therefore comes to the front as a prefactor. For e^(3x) the derivative is 3·e^(3x), for e^(−0.5x) you get −0.5·e^(−0.5x). This result is the basis of all growth and decay processes: the derivative is proportional to the function itself, with the rate k as the proportionality factor. That is exactly why the exponential function describes exponential growth and radioactive decay so naturally.

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Create a curated FSRS exam set for (f∘g)' = f'(g(x))·g'(x): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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How do you calculate with Chain Rule of Differentiation?

Here is how to work through a typical Chain Rule of Differentiation ((f∘g)' = f'(g(x))·g'(x)) task step by step:

  1. 1

    Task

    Differentiate e^{3x²}.

    Solution path

    Outer derivative e^{3x²}, inner derivative 6x. Result: 6x e^{3x²}.