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The derivative of the function

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

at $x = 1$ is $e^{-1}$ .
in Calculus by Boss (30.8k points) | 167 views

1 Answer

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No the Answer is $\frac{1}{2} e^{-1}$... Apply Leibniz Rule {\displaystyle {\frac {d}{dx}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)=f{\big (}x,b(x){\big )}\cdot {\frac {d}{dx}}b(x)-f{\big (}x,a(x){\big )}\cdot {\frac {d}{dx}}a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt,}

 

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