# derivative of the function $\int_{0}^{\sqrt{x}} e^{-t^{2}}dt$

by Naruto_007   Last Updated January 20, 2018 07:20 AM

The derivative of the function

$\int_{0}^{\sqrt{x}} e^{-t^{2}}dt$

at $x = 1$ is? I used libnith rule and got $e^{-1}/2$ am I correct!

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Yes if you are differentiating with respect to $x$.

$$\frac{d}{dx}\int_0^{\sqrt{x}}e^{-t^2}\, dt=e^{-x}\frac{d}{dx}{\sqrt{x}}=e^{-x}\frac{1}{2\sqrt{x}}$$

Siong Thye Goh
January 20, 2018 06:49 AM

$$\frac{d}{dx}\int_{0}^{\sqrt{x}} e^{-t^{2}}dt = e^{-(\sqrt{x})^2} \cdot \frac{d}{dx}\sqrt{x} = e^{-x}\frac{1}{2\sqrt{x}}$$

by the fundamental theorem of calculus and the chain rule.

Henno Brandsma
January 20, 2018 06:51 AM

## How to demonstrate this?

Updated April 16, 2015 01:08 AM