The Chain Rule: Differentiating sin²(x)

The chain rule is the engine that lets you differentiate composite functions. The slogan is:

Take the derivative of the outer function evaluated at the inner function, then multiply by the derivative of the inner.

In symbols, if $h(x) = f(g(x))$, then:

$h'(x) = f'(g(x)) \cdot g'(x)$

This visualization shows the composition $\sin^{2}(x)$ — simple enough to compute by hand, rich enough to feel the chain rule in action.

Setting Up

Define:

• Inner: $g(x) = \sin x$, so $g'(x) = \cos x$
• Outer: $f(u) = u^{2}$, so $f'(u) = 2u$

The composite is $h(x) = f(g(x)) = (\sin x)^{2} = \sin^{2}(x)$.

Step-by-Step Solution

Given: $h(x) = \sin^{2}(x)$. Find: $h'(x)$ using the chain rule, and verify by simplifying.

---

Step 1 — Identify the inner and outer functions.

$g(x) = \sin x \qquad f(u) = u^{2}$

So $h(x) = f(g(x))$.

Step 2 — Differentiate each separately.

$g'(x) = \cos x$

$f'(u) = 2u$

Step 3 — Apply the chain rule formula.

$h'(x) = f'(g(x)) \cdot g'(x)$

Plug in: $f'(g(x)) = 2 \cdot g(x) = 2\sin x$, and $g'(x) = \cos x$:

$h'(x) = 2\sin x \cdot \cos x$

Step 4 — Simplify using a double-angle identity.

Recall the identity $2\sin x \cos x = \sin(2x)$. So:

$h'(x) = \sin(2x)$

Step 5 — Verify at a specific point. Try $x = \pi/4$.

Direct: $h(\pi/4) = \sin^{2}(\pi/4) = (\sqrt{2}/2)^{2} = 1/2$.

Slope from chain rule: $h'(\pi/4) = 2\sin(\pi/4)\cos(\pi/4) = 2 \cdot \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{2}}{2} = 2 \cdot \dfrac{1}{2} = 1$.

Or simplified: $h'(\pi/4) = \sin(2 \cdot \pi/4) = \sin(\pi/2) = 1$ ✓.

Both forms agree. The slope of $\sin^{2}(x)$ at $x = \pi/4$ is exactly 1.

Step 6 — Interpret the answer.

$h'(x) = \sin(2x)$ has period $\pi$ (half the period of $\sin x$), which makes sense: $\sin^{2}(x)$ has two peaks per $2\pi$ cycle (one at $\pi/2$ and one at $3\pi/2$, both equal to 1), so its derivative should oscillate twice as fast.

---

Answer:

$\boxed{\;\dfrac{d}{dx}\bigl[\sin^{2}(x)\bigr] = 2\sin x \cos x = \sin(2x)\;}$

The chain rule gives us $2\sin x\cos x$ directly, and a double-angle identity simplifies it to $\sin(2x)$.

Try It

• Slide the point along the curve — watch how the slope of $\sin^{2}(x)$ matches $\sin(2x)$ at every point.
• Toggle show inner to see $\sin(x)$ separately. Notice that $h(x) = \sin^{2}(x)$ stays non-negative (it's a square) and doubles the frequency of zero-crossings.
• The HUD shows both forms of the answer: the raw chain-rule output $2\sin x\cos x$ and the simplified $\sin(2x)$.