$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$ I'm having trouble understanding how to apply the $\frac{d}{dx}$when taking the anti-derivative.
$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$$
In class it was mentioned we'll end up taking the derivative of $e^{x^{2}}$ which is $2xe^{x^{2}}$
My guess of solving this is
$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt = [2t^{\frac{1}{2}}]_0^{e^{x^2}}$$
Then
$$[2(e^{x^{2}})^{\frac{1}{2}}]-[2(0)^{\frac{1}{2}}]=2(e^{x^{2}})^{\frac{1}{2}}$$
Am i missing something?
 A: Let $$f(x)=e^{x^2}$$ $$g(x)=\int_0^x \frac{dt}{\sqrt t}$$
Then $$h(x)=\int_0^{e^{x^2}}\frac{dt}{\sqrt t}=g\circ f(x)$$
Use the chain rule and FTC to find $h'(x)$, which is what you want.
A: You just got a little "sloppy" with notation, and forgot that your aim is to take the derivative of the integral you found:
First, when integrating, after evaluating the indefinite integral, drop the integral sign:
$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt = \frac d{dx}\left[ 2t^{\frac{1}{2}}\Big|_0^{e^{x^2}}\right]$$
Then
$$\dfrac d{dx}\left[[2(e^{x^{2}})^{\frac{1}{2}}]-[2(0)^{\frac{1}{2}}]\right]= \frac{d}{dx} \left[2(e^{x^{2}})^{\frac{1}{2}}\right]$$
Precisely...Now, use what you know about $\frac d{dx}\left(e^{x^2}\right) = \frac d{dx}\left(x^2\right)\cdot \left(e^{x^2}\right)$ and apply the same strategy (chain rule) to your result of evaluating the indefinite integral, as you did well.
$$\frac{d}{dx} \left[2(e^{x^{2}})^{\frac{1}{2}}\right] = \frac{d}{dx} \left[2\left(e^{\large\frac{x^{2}}{2}}\right)\right] = 2\frac{d}{dx}\left(\frac{x^2}{2}\right)\cdot e^{\large\frac{x^2}{2}}= 2\cdot \dfrac 12\cdot 2xe^{\large\frac{x^{2}}{2}} = 2xe^{\frac{x^{2}}{2}}$$
A: No you aren't. Just continue to get
$\Large2(e^{x^{2}})^{\frac{1}{2}}=2e^{\frac{x^{2}}{2}}$
and then differentiate with respect to $x$ to get
$\Large 2xe^{\frac{x^{2}}{2}}$
A: $$f(t)=t^{-\frac{1}{2}} \quad ,\int_{t=0}^{t=e^{x^2}} f(t)\, \mathrm dt =F(e^{x^2})-F(0)$$
$$\frac {\mathrm d} {\mathrm dx}\left[F(e^{x^2})-F(0) \right]=\frac {\mathrm d} {\mathrm dx}\left[F(e^{x^2}) \right]=\frac {\mathrm d F(e^{x^2})} {\mathrm dx}=\frac {\mathrm d \, F(e^{x^2})} {\mathrm d \, e^{x^2}}\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x}=f(e^{x^2})\,\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x}$$
$$f(e^{x^2})\,\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x}=f(e^{x^2})\,\frac {\mathrm d \, e^{x^2} } {\mathrm d \,x^2}\,\frac {\mathrm d \, x^2 } {\mathrm d \,x}=(e^{x^2})^{-\frac{1}{2}} \, e^{x^2} \, 2x= e^{-\frac{x^2}{2}+{x^2}} \, 2x = e^{\frac{x^2}{2}} \, 2x $$
