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$F(x) = \int_0^x {\frac {e^t}{t+2} dt} $

$G(x) = \int_0^{x^2} {\frac {e^t}{t+2} dt} $

"Get $F'(x)$ and $G'(x) $ using the fundamental theorem of calculus and the chain rule"

Okay, I would assume that $F'(x)$ is equal to $\frac {e^x}{x+2}$ since that's basically what the fundamental theorem says. Am I correct? Feeling really unconfident about how the fundamental theorem actually works to be honest.

Regarding $G'(x)$ I'm pretty much blank. I would guess the use of the chain rule has something to do with that the upper limit of the integral is $x^2$ and not just $x$ but I'm not sure how to use it.

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    $\begingroup$ 1. You are right. $F'(x)=\frac{e^x}{x+2}$. 2. Notice that $G(x)=F(x^2)$ so that $G'(x)=F'(x^2)\cdot(x^2)'$. You may find the answer by yourself. $\endgroup$ – ChoF Mar 10 '18 at 11:26
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As $ChoF$ said, $$ G'\left(x\right)=\left(F\left(x^2\right)\right)'=2xF'\left(x^2\right)=\frac{2xe^{x^2}}{x^2+2} $$ You can also put $u=\sqrt{t}$ then $\displaystyle \text{d}u=\frac{\text{d}t}{2\sqrt{t}}=\frac{\text{d}t}{2u}$ Then $$ G\left(x\right)=\int_{0}^{x}\frac{e^{u^2}}{u^2+2}2u\text{d}u $$ And you can reapply the fondamental theorem of calculus to found again

$$G'\left(x\right)=\frac{2xe^{x^2}}{x^2+2}$$

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