Topic 3.1 The Chain Rule: differentiate composite functions using the chain rule.

What this topic is asking

The College Board (Topic 3.1) introduces the chain rule, the tool for differentiating a composite function - a function inside another function, such as $(3x^2 + 1)^4$, $\sin(e^x)$, or $\ln(5x + 2)$. Almost every interesting derivative beyond the basic rules needs it, so this is the single most important rule in Unit 3.

The rule

The phrase to internalise is "derivative of the outer (with the inner left untouched), times the derivative of the inner". The inner function is copied unchanged into the outer derivative, and then a separate factor $g'(x)$ is multiplied on.

Identifying outer and inner

A reliable procedure

Differentiating a composite cleanly is a matter of decomposition: name the inner function $u$, differentiate the outer treating $u$ as a single variable, then multiply by $u'$. Writing $u$ and $u'$ off to the side before assembling keeps the two factors straight.

Why the chain rule multiplies the rates

The Leibniz form $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ makes the meaning vivid: if $y$ changes $\frac{dy}{du}$ times as fast as $u$, and $u$ changes $\frac{du}{dx}$ times as fast as $x$, then $y$ changes the product of those rates as fast as $x$. A concrete picture helps. Suppose a car's distance depends on the engine's revolutions, and the revolutions depend on time. If the car moves $0.01$ km per revolution and the engine turns $3000$ revolutions per minute, the car moves $0.01 \times 3000 = 30$ km per minute - the two rates multiply. The chain rule is exactly this composition of rates, and it is why a small change in $x$ propagates through the inner function and then through the outer function, picking up a factor at each stage. This also explains why the inner derivative can never be dropped: it is the rate at which the inside responds to $x$, without which the outer rate has nothing to act on.

Nesting and combining with other rules

The chain rule extends to deeper nesting by applying it repeatedly, peeling one layer at a time from the outside in. For $\sin(e^{2x})$ you differentiate sine to get $\cos(e^{2x})$, multiply by the derivative of $e^{2x}$, which itself needs the chain rule to give $e^{2x}\cdot 2$, ending at $2e^{2x}\cos(e^{2x})$. The chain rule also combines with the product and quotient rules: in $x^2\sin(3x)$ you use the product rule, and the $\sin(3x)$ factor is differentiated by the chain rule to $3\cos(3x)$. On the exam, most no-calculator derivatives are really chain-rule problems with one or two extra layers, so fluency here pays off everywhere in Units 3 to 8. A good final check is to confirm that every inner function contributed its derivative as a factor; a missing factor is almost always a forgotten inner derivative.