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Calculate $\frac{d}{dx}f(x)=\frac{d}{dx}(\int_0^xe^{-t^2}dt)^2$

So I thought of applying the leibniz integral rule:

$\frac{d}{dx}\int_{u(x)}^{o(x)}h(t,x)dt=o'(x)h(o(x),x)-u'(x)h(u(x),x)+\int_{u(x)}^{o(x)}\frac{\partial}{\partial x}h(t,x)dt$

So we got :

$u(x)=0 \Rightarrow u'(x)=0$

$o(x)=x \Rightarrow o'(x)=1$

$\frac{\partial}{\partial x}h(t,x)=\frac{\partial}{\partial x}(e^{-t^2})=0$

$h(o(x),x)=e^{-x^2}$

$h(u(x),x)=e^0=1$

Finally:

$\frac{d}{dx}\int_0^xe^{-t^2}dt=1\cdot e^{-x^2}- 0 \cdot 1 + \int_0^x0 dt=e^{-x^2}+const$

Now I'm not sure , whether just to quare the whole or to use product and chain rule,like:

$\frac{d}{dx}f(x)=\frac{d}{dx}(\int_0^xe^{-t^2}dt)^2=\frac{d}{dx}(\int_0^xe^{-t^2}dt\int_0^xe^{-t^2}dt)$

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    i got $$\sqrt {\pi}{\rm erf} \left(x\right){{\rm e}^{-{x}^{2}}}$$ here – 2017-01-21
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    @Dr.SonnhardGraubner: WolframAlpha says $\frac{1}{2}\sqrt(\pi) erf(x)e^{-x^2}$ – 2017-01-21
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    are you sure? what gave you in Wolfram alpha? – 2017-01-21
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    i'll try it again – 2017-01-21
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    i think my result is right – 2017-01-21
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    @Dr.SonnhardGraubner: Alright , sorry, I just used $e^{-t^2}$ – 2017-01-21
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    you are welcome! – 2017-01-21
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    @Dr.SonnhardGraubner: But how would I calculate this without an calculator ? There should be a trick or no ? – 2017-01-21
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    you must study something about the error function – 2017-01-21
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52205/discussion-between-pinkpanther5-and-dr-sonnhard-graubner). – 2017-01-21

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To close the question:

Let $erf(x)=\frac{2}{\sqrt\pi}\int_0^xe^{-t^2}dt$

So $\frac{d}{dx}f(x)=\frac{d}{dx}(\int_0^xe^{-t^2}dt)^2=\frac{d}{dx}(\frac{\sqrt\pi}{2}\cdot erf(x))^2=\frac{d}{dx}\frac{\pi}{4} \cdot erf^2(x)=\frac{\pi}{2} \cdot erf(x)\cdot erf'(x)=\frac{2}{\sqrt\pi}\frac{\pi}{2}\cdot e^{-x^2}\cdot erf(x)=\sqrt\pi \cdot e^{-x^2}\cdot erf(x)$