Mathematics · Ch 5 — Continuity and Differentiability

Derivatives of Composite Functions

5.3.1

Derivatives of Composite Functions

5.3.1 Derivatives of Composite Functions

The Core Idea: Why We Need a Special Rule

For a function like $f(x) = (2x+1)^3$ you could expand and differentiate term by term, but that is hopeless for $(2x+1)^{100}$ . The better view is that $f(x) = (2x+1)^3$ is two functions nested: with $g(x) = 2x+1$ and $h(t) = t^3$ , we have $f(x) = h(g(x))$ , a composite function ( $g$ applied first, then $h$ ).

Setting $t = g(x) = 2x+1$ , so $f(x) = h(t) = t^3$ :

\frac{df}{dx} = 6(2x+1)^2 = 3(2x+1)^2 \cdot 2 = 3t^2 \cdot 2 = \frac{dh}{dt} \cdot \frac{dt}{dx}

Multiplying the derivative of the outer function (w.r.t. the inner variable) by the derivative of the inner function is the essence of the chain rule.

Theorem 4: The Chain Rule

Statement: Let $f$ be a composite of two functions $u$ and $v$ ; i.e., $f = v \circ u$ . Suppose $t = u(x)$ and both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist. Then:

\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}

The proof is omitted: differentiate the outer function $v$ with respect to its argument $t$ , then multiply by the derivative of the inner function $u$ with respect to $x$ .

Watch out

A common mistake is to forget that $\frac{dv}{dt}$ must be evaluated at $t = u(x)$ , not at $x$ . Always substitute back after differentiating the outer function.

Extending the Chain Rule to Three Functions

For a composite of three functions, $f = (w \circ u) \circ v$ , with $t = v(x)$ and $s = u(t)$ : …

Theorem 4

The Chain Rule (Theorem 4)

Statement. Let $f$ be a real-valued function which is a composite of two functions $u$ and $v$ ; i.e., $f = v \circ u$ . Suppose $t = u(x)$ and if both $\frac{dt}{dx}$ and $\frac{dv}{dt}$ exist, then

\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}

The theorem requires three conditions: (i) $f$ must be expressible as $v(u(x))$ , (ii) $u$ must be differentiable at $x$ , and (iii) $v$ must be differentiable at $t = u(x)$ . When these hold, the derivative of the composite function is the product of the derivative of the outer function (evaluated at the inner function) and the derivative of the inner function.

›Proof

Proof. Let $f = v \circ u$ , so that $f(x) = v(u(x))$ for all $x$ in the domain. Define $t = u(x)$ . Then $f(x) = v(t)$ .

By definition of the derivative,

$\frac{df}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Since $f(x) = v(u(x))$ and $f(x+h) = v(u(x+h))$ , we have

$\frac{df}{dx} = \lim_{h \to 0} \frac{v(u(x+h)) - v(u(x))}{h}$

Let $\Delta t = u(x+h) - u(x)$ . Then $u(x+h) = u(x) + \Delta t = t + \Delta t$ . The numerator becomes $v(t + \Delta t) - v(t)$ . So

$\frac{df}{dx} = \lim_{h \to 0} \frac{v(t + \Delta t) - v(t)}{h}$

Multiply and divide by $\Delta t$ (provided $\Delta t \neq 0$ for $h$ sufficiently small — the case $\Delta t = 0$ is handled separately by a standard argument):

$\frac{df}{dx} = \lim_{h \to 0} \frac{v(t + \Delta t) - v(t)}{\Delta t} \cdot \frac{\Delta t}{h}$

Now $\frac{\Delta t}{h} = \frac{u(x+h) - u(x)}{h}$ , so as $h \to 0$ , $\frac{\Delta t}{h} \to \frac{du}{dx} = \frac{dt}{dx}$ (since $t = u(x)$ ). Also, as $h \to 0$ , $\Delta t \to 0$ because $u$ is continuous (differentiability implies continuity). Hence

$\lim_{h \to 0} \frac{v(t + \Delta t) - v(t)}{\Delta t} = \frac{dv}{dt}$

Therefore,

$\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$

This completes the proof. …

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