Find the derivative of an integral function 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...
 A: 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.
$$g'(x) = e^{\tan^2 x} \cdot \sec^2 x$$
A: 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 ;
$$g(x)=\int _{ 1 }^{ \tan(x) }{ e^ {t^ 2}dt }$$
$$\implies \frac{d}{dx}(g) =e^{\tan^2(x)}\cdot \sec^2{(x)}$$
A: 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:
$$F'(\tan(x))=e^{\tan(x)^2}$$
A: $$g(x)=\int _{ 1 }^{ \tan(x) }{ e^ {t^ 2}dt } $$
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 $$ g'(x) = e^ {\tan^ 2 x}sec^2 x $$
