Derivative of a Composite Function

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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 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:

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, 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}

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