Solution

Let $y = [\cos(x^2)]^2$. Using the chain rule: 

 \[ \frac{dy}{dx} = 2\cos(x^2)\cdot \frac{d}{dx}\big(\cos(x^2)\big) \]  \[      = 2\cos(x^2)\cdot\big(-\sin(x^2)\cdot 2x\big) \] \[= -4x\,\cos(x^2)\sin(x^2). \] Using $\;2\sin u\cos u=\sin(2u)$ with $u=x^2$: \[ \frac{dy}{dx} = -2x\,\sin\!\big(2x^2\big). \]

Answer: ${\dfrac{dy}{dx}=-4x\,\sin\!\big(x^2\big) \cos (x^2)}$