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

Question

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!

Answer 1

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

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

Answer 2

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

by the fundamental theorem of calculus and the chain rule.