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Chain rule with a triple composition

Let h(x)=cos⁡3(2x)h(x) = \cos^3(2x), which means h(x)=(cos⁡(2x))3h(x) = \big(\cos(2x)\big)^3. Find h′(x)h'(x).

Hint

Think of h(x)h(x) as built from three layers: cubing (outermost), then cosine, then 2x2x (innermost). Apply the chain rule from the outside in, differentiating one layer at a time.

Step-by-step solution

View h(x)=[cos⁡(2x)]3h(x) = [\cos(2x)]^3 as an outer power function applied to the inner function cos⁡(2x)\cos(2x). By the chain rule, ddx[cos⁡(2x)]3=3[cos⁡(2x)]2⋅ddx[cos⁡(2x)]\frac{d}{dx}[\cos(2x)]^3 = 3[\cos(2x)]^2 \cdot \frac{d}{dx}[\cos(2x)].

Now differentiate cos⁡(2x)\cos(2x), which is itself a composite function with inner function 2x2x. Using the chain rule again, ddx[cos⁡(2x)]=−sin⁡(2x)⋅2=−2sin⁡(2x)\frac{d}{dx}[\cos(2x)] = -\sin(2x)\cdot 2 = -2\sin(2x).

Combining both steps gives h′(x)=3cos⁡2(2x)⋅(−2sin⁡(2x))=−6cos⁡2(2x)sin⁡(2x)h'(x) = 3\cos^2(2x)\cdot(-2\sin(2x)) = -6\cos^2(2x)\sin(2x).

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