Derivatives of composite functions

Most functions in Class 12 calculus are not raw expressions but functions built by feeding one function into another — $\sin(x^2)$ , $(3x+1)^5$ , $e^{\cos x}$ . These are composite functions, and differentiating them needs a rule of its own: the chain rule. It follows directly after differentiability in the Continuity and Differentiability chapter, and it is the single most-used tool for the rest of the chapter — inverse trig derivatives, implicit differentiation, logarithmic differentiation and parametric differentiation all lean on it. The skill being tested is recognising the "outer" and "inner" layers of an expression and multiplying the right factors together in the right order.

The core insight

For $y = f(g(x))$ , differentiate the outer function $f$ at the inner value $g(x)$ , then multiply by the derivative of the inner function $g(x)$ : $\frac{dy}{dx} = f'(g(x))\cdot g'(x)$ .

The core idea

A composite function has the form $y = f(g(x))$ : the "outer" function $f$ is applied to the "inner" function $g(x)$ . If $y=f(u)$ where $u=g(x)$ , and both $f$ and $g$ are differentiable, the chain rule states:

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = f'(g(x))\cdot g'(x)

where: $y$ is the composite function, $u = g(x)$ is the inner function, $\frac{dy}{du}$ is the derivative of the outer function evaluated at $u$ , and $\frac{du}{dx}$ is the derivative of the inner function. In words — differentiate the outer function, keeping the inner function untouched, then multiply by the derivative of the inner function. The rule extends to any number of nested layers. For a three-function chain $y = f(g(h(x)))$ :

\frac{dy}{dx} = f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)

where: each factor is the derivative of one layer, evaluated with everything inside that layer left as it is — the outermost layer peeled first, working inward. Identifying the layers correctly is the whole battle. For $\sin(x^2)$ the outer function is $\sin(\cdot)$ and the inner function is $x^2$ . For $(3x+1)^5$ the outer function is $(\cdot)^5$ and the inner function is $3x+1$ . For $\sin^2 x = (\sin x)^2$ the outer function is "square" and the inner function is $\sin x$ — a genuinely different function from $\sin(x^2)$ , even though both "have a sine and a square" in them. Think of the chain rule as peeling off layers of an expression from the outside in. Each peel contributes one multiplicative factor — the derivative of that layer's function, with its own contents left alone — and the running answer is the product of all the factors collected along the way. This picture scales to any number of layers without needing a new rule for each case. The chain rule also combines with the product and quotient rules whenever a problem needs more than one tool at once. For instance, differentiating $y = x^2\sin(3x)$ requires the product rule for the two factors $x^2$ and $\sin(3x)$ , and inside that, the chain rule for $\sin(3x)$ itself, whose derivative is $3\cos(3x)$ . Recognising which rule governs the outermost structure of an expression — product, quotient, or composite — before diving into the algebra saves most of the errors students make on mixed problems.

Figure

Diagram of the chain rule as peeling layers from outside in for sine of x squared

Derivation AThe chain rule for a composite function

The goal is to show why the derivative of $y=f(g(x))$ must be the product $f'(g(x))\cdot g'(x)$ rather than, say, just $f'(g(x))$ alone, by building the derivative directly from the limit definition and the idea of a small change propagating through two linked functions. We use only the definition of the derivative as a limit of a difference quotient, applied twice — once to the outer function and once to the inner function.

Step 1. Let $u = g(x)$ and $y = f(u)$ , so $y = f(g(x))$ . Give $x$ a small increment $\Delta x$ . This produces a corresponding small change $\Delta u = g(x+\Delta x) - g(x)$ in $u$ , which in turn produces a small change $\Delta y = f(u+\Delta u) - f(u)$ in $y$ . A change in $x$ propagates through $u$ before it reaches $y$ .

Step 2. Write the difference quotient for $y$ with respect to $x$ and multiply and divide by $\Delta u$ (valid whenever $\Delta u \neq 0$ near the point of interest): $\dfrac{\Delta y}{\Delta x} = \dfrac{\Delta y}{\Delta u}\cdot\dfrac{\Delta u}{\Delta x}$ . This algebraic trick is the heart of the proof — it splits one ratio into two ratios chained together.

Step 3. Take the limit as $\Delta x \to 0$ . Because $g$ is differentiable (hence continuous), $\Delta u \to 0$ as well, so both factors tend to genuine derivatives: $\displaystyle\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta u} = \frac{dy}{du} = f'(u)$ and $\displaystyle\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} = \frac{du}{dx} = g'(x)$ .

Step 4. Multiply the two limits together (the limit of a product is the product of the limits, since both limits exist): $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx} = f'(g(x))\cdot g'(x)$ . This confirms the chain rule needs both factors — the outer derivative alone would ignore how fast the inner function itself is changing.

Worked example

Differentiate $y = \sin(x^2+1)$ and then $y=(3x+1)^5$ with respect to $x$ , and compare the two chain-rule applications.

Step 1. For $y=\sin(x^2+1)$ , identify the layers: outer function $f(u)=\sin u$ , inner function $u=g(x)=x^2+1$ .

Step 2. Differentiate the outer function with respect to $u$ and substitute back $u=x^2+1$ : $\dfrac{dy}{du}=\cos u=\cos(x^2+1)$ . Differentiate the inner function: $\dfrac{du}{dx}=2x$ .

Step 3. Apply the chain rule: $\dfrac{dy}{dx}=\cos(x^2+1)\cdot 2x = 2x\cos(x^2+1)$ .

Step 4. For $y=(3x+1)^5$ , identify the layers: outer function $f(u)=u^5$ , inner function $u=g(x)=3x+1$ . The outer derivative is $\dfrac{dy}{du}=5u^4=5(3x+1)^4$ and the inner derivative is $\dfrac{du}{dx}=3$ , a constant since the inner function is linear.

Step 5. Apply the chain rule: $\dfrac{dy}{dx}=5(3x+1)^4\cdot 3 = 15(3x+1)^4$ .

Answer: $\dfrac{dy}{dx} = 2x\cos(x^2+1)$ for the first function, and $\dfrac{dy}{dx} = 15(3x+1)^4$ for the second. The second result also follows instantly from the shortcut $\dfrac{d}{dx}(ax+b)^n = an(ax+b)^{n-1}$ , since a linear inner function always contributes a constant factor.

Competitive stretch

For a three-layer chain like $y=\sin(\cos(x^2))$ , peel one layer at a time: outer $\sin(\cdot)$ gives $\cos(\cos(x^2))$ ; the middle layer $\cos(\cdot)$ contributes $-\sin(x^2)$ ; the innermost layer $x^2$ contributes $2x$ . Multiplying all three factors, $\dfrac{dy}{dx} = \cos(\cos(x^2))\cdot(-\sin(x^2))\cdot 2x = -2x\sin(x^2)\cos(\cos(x^2))$ .

Common traps

Forgetting the inner-function factor. Writing $\frac{d}{dx}\sin(x^2)=\cos(x^2)$ and stopping, omitting the essential $2x$ .
Misidentifying outer and inner functions. Confusing $\sin^2 x = (\sin x)^2$ , whose outer function is "square" of $\sin x$ , with $\sin(x^2)$ , which is entirely different.
Using the chain rule where it does not belong. Treating a product like $\sin x\cdot\cos x$ as a composite instead of applying the product rule.
Dropping a layer in longer chains. For $\sin(\cos(x^2))$ , differentiating only the outermost layer and skipping one or more inner factors.
Overthinking a simple constant inner-derivative. When the inner function is linear, such as $ax+b$ , the chain rule contributes just one constant factor $a$ , not a whole new function — do not expect extra complexity where there is none.

Ask the tutor: If a multi-layer chain like $\sin(\cos(x^2))$ still feels confusing, ask the AI tutor to peel it one layer at a time with you, checking the inner function at each step.

How do I quickly tell if a function is composite or a product, so I know which rule to use?

Ask whether one function is being applied to the output of another, or whether two separate functions are simply multiplied together. In $\sin(x^2)$ , the sine acts on the whole expression $x^2$ — that is composite, chain rule. In $\sin x \cdot \cos x$ , sine of $x$ and cosine of $x$ are two independent functions of $x$ multiplied side by side — that is a product, product rule. If you can rewrite the expression as $f(g(x))$ with a single innermost variable, it is composite.

Why do I multiply by the inner derivative instead of just differentiating the outer function?

Because the inner function is also changing as $x$ changes, and that changing rate feeds into how fast the outer function changes. The derivation shows this precisely: $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ , where $\frac{dy}{du}$ measures how $y$ responds to a change in $u$ , and $\frac{du}{dx}$ measures how much $u$ itself changes per unit change in $x$ . Skipping the second factor silently assumes the inner function is $x$ itself, changing at rate $1$ — true only when there is no real inner layer.

Does the chain rule still apply if the inner function is something simple like a constant times x plus a constant?

Yes, and it becomes especially quick. If the inner function is linear, $u=ax+b$ , its derivative $\frac{du}{dx}=a$ is just a constant, so the chain rule contributes a single multiplicative factor rather than a whole new function of $x$ . This is exactly the shortcut $\frac{d}{dx}(ax+b)^n = an(ax+b)^{n-1}$ used for $(3x+1)^5$ above — worth memorising since linear-inner composites appear constantly in problems.

Derivatives of composite functions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads