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Derivative of an accumulation function with a composite upper limit

Let F(x)=∫0x2cos⁡(t) dt.F(x) = \int_0^{x^2} \cos(t)\, dt. Use the Fundamental Theorem of Calculus (Part 1) together with the chain rule to find F′(x)F'(x).

Hint

Think of F(x)F(x) as a composition: first define G(u)=∫0ucos⁡(t) dtG(u) = \int_0^u \cos(t)\,dt so that G′(u)=cos⁡(u)G'(u) = \cos(u) by FTC Part 1, then set u=x2u = x^2 and apply the chain rule F′(x)=G′(x2)⋅ddx[x2]F'(x) = G'(x^2)\cdot \frac{d}{dx}[x^2].

Step-by-step solution

By FTC Part 1, if G(u)=∫0ucos⁡(t) dtG(u) = \int_0^u \cos(t)\,dt, then G′(u)=cos⁡(u)G'(u) = \cos(u). Since F(x)=G(x2)F(x) = G(x^2), this is a composition of functions, so we must apply the chain rule rather than simply substituting x2x^2 into the integrand alone.

The chain rule gives F′(x)=G′(x2)⋅ddx[x2]=cos⁡(x2)⋅2xF'(x) = G'(x^2)\cdot \frac{d}{dx}[x^2] = \cos(x^2)\cdot 2x. Therefore F′(x)=2xcos⁡(x2)F'(x) = 2x\cos(x^2).

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