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Second derivative via the chain rule

Let f(x)=cos⁡(x2)f(x) = \cos(x^2).

Find f′′(x)f''(x) for f(x)=cos⁡(x2)f(x) = \cos(x^2). Show your work, applying the chain rule to compute f′(x)f'(x) first, then differentiate again (using the product rule where needed) to obtain f′′(x)f''(x).

Hint

First find f′(x)f'(x) using the chain rule with outer function cos⁡(u)\cos(u) and inner function u=x2u = x^2. Then differentiate f′(x)f'(x) again — since it's a product of two functions of xx, you'll need the product rule along with the chain rule.

Step-by-step solution

Using the chain rule on f(x)=cos⁡(x2)f(x) = \cos(x^2) with outer function cos⁡(u)\cos(u) and inner function u=x2u = x^2, we get f′(x)=−sin⁡(x2)⋅2x=−2xsin⁡(x2)f'(x) = -\sin(x^2)\cdot 2x = -2x\sin(x^2).

Now differentiate f′(x)=−2xsin⁡(x2)f'(x) = -2x\sin(x^2) using the product rule: f′′(x)=−2sin⁡(x2)+(−2x)⋅cos⁡(x2)⋅2xf''(x) = -2\sin(x^2) + (-2x)\cdot\cos(x^2)\cdot 2x. Simplifying gives f′′(x)=−2sin⁡(x2)−4x2cos⁡(x2)f''(x) = -2\sin(x^2) - 4x^2\cos(x^2).

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