Given h(u) = \int _u^5 \cos (x) dx, find \frac {dh}{du}.


Given {eq}h(u) = \int _u^5 \cos (x) dx, {/eq} find {eq}\frac {dh}{du}. {/eq}

The First Fundamental Theorem of Calculus:

Suppose that {eq}f {/eq} and {eq}F {/eq} are continuous functions on the closed interval {eq}[a,b] {/eq}, and {eq}F {/eq} is differentiable on the open interval {eq}(a,b) {/eq}. If {eq}F'(x)=f(x) {/eq} for all {eq}x {/eq} in the open interval {eq}(a,b) {/eq}, then we say that {eq}F {/eq} is an antiderivative of {eq}f {/eq} on the closed interval {eq}[a,b] {/eq}.

If {eq}F {/eq} is an antiderivative of {eq}f {/eq} on {eq}[a,b] {/eq}, then the fundamental theorem of calculus states that

{eq}\displaystyle \int_a^b f(x) \, dx = F(b)-F(a) \, . {/eq}

Answer and Explanation:

Let {eq}F(x) {/eq} be an antiderivative of {eq}\cos x {/eq}. By the first fundamental theorem of calculus, we have

{eq}\begin{align*} h(u)&=\int_u^5 \cos x \, dx\\ &=F(5)-F(u) \, . \end{align*} {/eq}

So differentiating {eq}h {/eq} gives

{eq}\begin{align*} \frac{dh}{du}&=\frac{d}{du}\left[F(5)-F(u)\right]\\ &=0-F'(u)\\ &=-F'(u)\\ &=-\cos u&&\text{(since }F\text{ is an antiderivative of }\cos\text{).} \end{align*} {/eq}

That is, we've computed that {eq}\boxed{\frac{dh}{du}=-\cos u}\, {/eq}.

Learn more about this topic:

Antiderivative: Rules, Formula & Examples

from Calculus: Help and Review

Chapter 8 / Lesson 12

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