# It is know that G(x)= \int _0^{h(x)} f(t) dt. We have \frac{d}{dx} G(x)=F'(h(x))h'(x) where...

## Question:

It is know that {eq}G(x)= \int _0^{h(x)} f(t) dt. {/eq} We have {eq}\frac{d}{dx} G(x)=F'(h(x))h'(x) {/eq} where {eq}F(x)=\int _0^x f(t) dt. {/eq}

Find a formula for {eq}\frac {d}{dx} G(x) {/eq}, where {eq}g(x) = \int _{g(x)}^o f(t) dt {/eq}

## The First Fundamental Theorem of Calculus:

The first fundamental theorem of calculus provides a connection between derivatives and integrals.

Specifically, suppose that {eq}F {/eq} is an antiderivative of {eq}f {/eq} on the interval {eq}[a,b] {/eq}: that is, {eq}F {/eq} is continuous on {eq}[a,b] {/eq} and differentiable, and {eq}F'(x)=f(x) {/eq} for all {eq}x {/eq} in {eq}(a,b) {/eq}. Then the first fundamental theorem of calculus states that

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

Suppose that {eq}\displaystyle G(x)=\int_{g(x)}^0 f(t) \, dt {/eq}, and let {eq}F(t) {/eq} be an antiderivative of the function {eq}f(t) {/eq}. Then by the first fundamental theorem of calculus, we have

{eq}\begin{align*} G(x)&=\int_{g(x)}^0 f(t) \, dt\\ &=F(0)-F(g(x)) \, . \end{align*} {/eq}

Differentiating {eq}G {/eq}, we therefore have

{eq}\begin{align*} \frac{d}{dx}G(x)&=\frac{d}{dx}\left[F(0)-F(g(x))\right]\\ &=0-\frac{d}{dx}F(g(x))\\ &=-F'(g(x))g'(x)&&\text{(by the chain rule)}\\ &=-f(g(x))g'(x)&&\text{(since }F\text{ is an antiderivative of }f\text{).} \end{align*} {/eq}

That is, we've shown that {eq}\boxed{\frac{d}{dx}G(x)=-f(g(x))g'(x)}\, {/eq}. 