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# Derivative of a Composite Function

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/the-derivative-of-a-composite-function/
Fetched from algebrica.org post 6132; source modified 2026-03-06T21:00:41.

## The chain rule

Let \\( g \\) be differentiable at \\( x \\), and let \\( f \\) be differentiable at \\( z = g(x) \\). Then the composite function \\( y = f(g(x)) \\) is differentiable at \\( x \\), and its [derivative](https://algebrica.org/derivative/) is the product of the derivative of \\( f \\) evaluated at \\( g(x) \\) and the derivative of \\( g \\) at \\( x \\):

\\[
D\[f(g(x))\] = f’(g(x)) \\cdot g’(x)
\\]

This result is known as the chain rule. It states that to differentiate a composite function, one multiplies the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function.

In Leibniz notation, if \\( y = f(u) \\) and \\( u = g(x) \\), the chain rule takes the form:

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

## Proof

To prove that \\(D\[f(g(x))\] = f’(g(x)) \\cdot g’(x)\\) we calculate the following [limit](../limits.md):

\\[
D\[f(g(x))\] = \\lim\_{h \\to 0} \\frac{f(g(x + h))-f(g(x))}{h}
\\]

* * *

Let \\( z = g(x) \\), then \\( g(x+h)-g(x) = \\Delta z. \\) This implies that \\( g(x+h) = g(x) + \\Delta z. \\) The limit becomes:

\\[
D\[f(g(x))\] = \\lim\_{h \\to 0} \\frac{f(z+\\Delta z)-f(z)}{h}
\\]

* * *

Multiplying both the numerator and the denominator by \\( \\Delta z \\), we get:

\\[
\\begin{aligned} D\[f(g(x))\] &= \\lim\_{h \\to 0} \\frac{f(z + \\Delta z)-f(z)}{\\Delta z} \\cdot \\frac{\\Delta z}{h} \\[0.5em] &= \\lim\_{h \\to 0} \\frac{f(z + \\Delta z)-f(z)}{\\Delta z} \\cdot \\frac{g(x + h)-g(x)}{h} \\[0.8em] &= f’(z) \\cdot g’(x)\\[1em] &= f’(g(x)) \\cdot g’(x) \\end{aligned}
\\]

This argument assumes \\( \\Delta z \\neq 0 \\) for \\( h \\) sufficiently small. A complete proof handles the case \\( \\Delta z = 0 \\) separately via an auxiliary function; the conclusion is the same.

* * *

In the case of powers of a function, the rule generalizes as follows:

\\[
D\[f(x)^a\] = a\[f(x)\]^{a-1}f’(x)
\\]

## Example 1

Let’s compute the derivative of the following composite function:

\\[
y = f(g(x)) = \\sin(3x^2 + 2x)
\\]

In this case, we have:

-   The inner function \\( g(x) = 3x^2 + 2x \\)
-   The outer function \\( f(t) = \\sin(t) \\), where \\( t = g(x) = 3x^2 + 2x \\)

* * *

The outer function is \\( f(t) = \\sin(t) \\). Its derivative is:

\\[
f’(t) = \\cos(t)
\\]

Substituting \\( t = g(x) \\):

\\[
f’(g(x)) = \\cos(3x^2 + 2x)
\\]

* * *

The inner function is \\( g(x) = 3x^2 + 2x \\). Its derivative is:

\\[
g’(x) = 6x + 2
\\]

Applying the chain rule we obtain:

\\[
D\[f(g(x))\] = f’(g(x)) \\cdot g’(x)
\\]

The result is:

\\[
(6x + 2)\\cos(3x^2 + 2x)
\\]

###### Explore the case of [composite power functions](../derivative-of-composite-power-functions.md), specifically the calculation of the derivative of functions of the type:

\\[
D\[f(x)^{g(x)}\]
\\]

## Extension to multiple compositions

The chain rule can be extended to compositions involving three or more functions. For example, given \\( y = f(g(h(x))) \\), the derivative is:

\\[
D\[f(g(h(x)))\] = f’(g(h(x))) \\cdot g’(h(x)) \\cdot h’(x)
\\]

Each factor represents the derivative of a function in the composition, evaluated at the composition of all subsequent functions. This pattern generalises to any finite number of nested functions. For \\( y = f\_1(f\_2(\\cdots f\_n(x)\\cdots)) \\), the derivative is given by the product:

\\[
f\_1’(f\_2(\\cdots f\_n(x)\\cdots)) \\cdot f\_2’(f\_3(\\cdots f\_n(x)\\cdots)) \\cdots f\_{n-1}'(f\_n(x)) \\cdot f\_n’(x)
\\]

In practical applications, differentiation proceeds from the outermost function inward, with each derivative computed in sequence and the results multiplied together.

* * *

As an example, consider \\( y = \\sin(e^{3x}) \\). The composition involves three functions:

\\[
\\begin{align} h(x) &= 3x \\[6pt] g(t) &= e^t \\[6pt] f(s) &= \\sin(s) \\end{align}
\\]

Applying the chain rule from the outside inward we obtain:

\\[
\\begin{align} D\[\\sin(e^{3x})\] &= \\cos(e^{3x}) \\cdot e^{3x} \\cdot 3 \\[6pt] &= 3e^{3x}\\cos(e^{3x}) \\end{align}
\\]

## Example 2

Consider the following function:

\\[
y = \\ln\\left(e^{x^2} + 1\\right)
\\]

The composition involves three functions:

\\[
\\begin{align} h(x) &= x^2 \\[6pt] g(t) &= e^t + 1 \\[6pt] f(s) &= \\ln(s) \\end{align}
\\]

* * *

The derivative of the outer function \\( f(s) = \\ln(s) \\) is \\( f’(s) = \\frac{1}{s} \\), evaluated at \\( s = g(h(x)) = e^{x^2} + 1 \\):

\\[
f’(g(h(x))) = \\frac{1}{e^{x^2} + 1}
\\]

The derivative of the middle function \\( g(t) = e^t + 1 \\) is \\( g’(t) = e^t \\), evaluated at \\( t = h(x) = x^2 \\):

\\[
g’(h(x)) = e^{x^2}
\\]

The derivative of the inner function \\( h(x) = x^2 \\) is:

\\[
h’(x) = 2x
\\]

* * *

Applying the chain rule from the outside inward:

\\[
\\begin{align} D\\left\[\\ln\\left(e^{x^2} + 1\\right)\\right\] &= \\frac{1}{e^{x^2} + 1} \\cdot e^{x^2} \\cdot 2x \\[6pt] &= \\frac{2x e^{x^2}}{e^{x^2} + 1} \\end{align}
\\]

The result is:

\\[
\\frac{2x e^{x^2}}{e^{x^2} + 1}
\\]

## Selected references

-   **Harvard University, O. Knill**. [Chain Rule](https://people.math.harvard.edu/~knill/teaching/math1a2020/handouts/lecture10.pdf)
-   **MIT OpenCourseWare, G. Strang**. [Derivatives by the Chain Rule](https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/mitres_18_001_f17_ch04.pdf)
-   **University of Toronto, J. Campesato**. [Differentiability and the Chain Rule](https://www.math.toronto.edu/campesat/ens/1920/1114.pdf)