Differentiating an accumulation function with a composite upper limit

Let $F(x) = \int_5^{\sin(x)} e^t \, dt.$ Find $F'(x)$ .

Hint

This is the Fundamental Theorem of Calculus, Part 1, but the upper limit is $\sin(x)$ rather than $x$ itself. Treat the integrand as a function of the upper limit, apply FTC1, and then multiply by the derivative of the upper limit (chain rule).

Step-by-step solution

By FTC Part 1, if $G(u) = \int_5^{u} e^t \, dt$ , then $G'(u) = e^u$ . Here the upper limit is $u = \sin(x)$ , so $F(x) = G(\sin(x))$ is a composition of functions.

Applying the chain rule, $F'(x) = G'(\sin(x)) \cdot \frac{d}{dx}[\sin(x)] = e^{\sin(x)} \cdot \cos(x)$ . Thus $F'(x) = \cos(x) e^{\sin(x)}$ .

Differentiating an accumulation function with a composite upper limit

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