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Find the derivative of the function: $$g(x)=\int _{ 1 }^{ \tan(x) }{ e^ {t^ 2}dt } $$

I have the answer as $\sec^2(x)\tan(x).$ Can anyone show the steps of how to solve this problem...

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    $\begingroup$ Use fundamental theorem of calculus $\endgroup$ – Martín Vacas Vignolo May 2 '18 at 17:15
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The Fundamental Theorem of Calculus states that if $$g(x) = \int_{a}^{f(x)} h(t)~{\rm d}t$$ where $a$ is any constant, then $$g'(x) = h(f(x)) \cdot f'(x)$$ Using this with the integral, $g(x) = g(x)$, $f(x) = \tan x$, and $h(x) = e^{x^2}$. The derivative of tangent is $\sec^2 x$. We may now plug in these values. $$g'(x) = e^{\tan^2 x} \cdot \sec^2 x$$

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You can use the Leibniz integral rule of differentiation under the integral sign

$\small\,\displaystyle\frac{d}{dx}\bigg(\large \int_{\small a(x)}^{ b(x)}\large f(x,t)\,dt\bigg) =\small f(x,b(x)).\frac{d(b(x))}{dx}-f(x,a(x)).\frac{d(a(x))}{dx}+\int_{\small{a(x)}}^{\small{b(x)}}\partial_xf(x,t)dt$

in your case ;

$$g(x)=\int _{ 1 }^{ \tan(x) }{ e^ {t^ 2}dt }$$

$$\implies \frac{d}{dx}(g) =e^{\tan^2(x)}\cdot \sec^2{(x)}$$

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Hint

Let: $$F(y)=\int_1^y e^{t^2} dt$$ thus $g(x)=F(\tan(x))$ using the chain rule: $$g'(x)=F'(\tan(x)) \tan'(x)$$ and by the fundamental theorem of calculus: $$F'(\tan(x))=e^{\tan(x)^2}$$

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$$g(x)=\int _{ 1 }^{ \tan(x) }{ e^ {t^ 2}dt } $$

This is a case of a composite function with the inner function being $\tan x$

According to the Fundamental theorem of calculus derivative of the integral is the integrand, and using the chain rule we get $$ g'(x) = e^ {\tan^ 2 x}sec^2 x $$

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