Find the derivative of y = sin^2(x).

Step-by-step solution

1. 1

   Decode the notation first. $\sin^2 x$ means $\left(\sin x\right)^2$, not $\sin\left(x^2\right)$. The outer function is the squaring and the inner function is the sine; getting this backwards produces $2x\cos(x^2)$, an entirely different answer.

2. 2

   Set up the chain rule with an explicit substitution. Let

   $u=\sin x\ \Longrightarrow\ y=u^2$

   The chain rule then says

   $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2u\cdot\frac{du}{dx}$

3. 3

   Differentiate the inner function.

   $\frac{du}{dx}=\frac{d}{dx}\sin x=\cos x$

   (This is where radians matter: $\frac{d}{dx}\sin x=\cos x$ holds only for radian measure.)

4. 4

   Substitute back.

   $\frac{dy}{dx}=2\sin x\cos x$

   Note that $u$ must be replaced by $\sin x$ — leaving the answer as $2u\cos x$ would be incomplete.

5. 5

   Rewrite with the double-angle identity. Since $\sin 2x=2\sin x\cos x$,

   $\frac{dy}{dx}=\sin 2x$

   Both forms are correct; the compact one makes the zeros obvious, at $x=\tfrac{k\pi}{2}$.

6. 6

   Check numerically. At $x=0.7$ a central difference gives $0.98544973$, and $\sin(1.4)=0.98544973$ ✓. As a structural check, $y=\sin^2x$ has minima at $x=0,\pi$ and a maximum at $x=\tfrac\pi2$, and $\sin 2x$ vanishes at exactly those points.