Find $\frac d{dx} \int_2^{x^4} \tan(t^2) dt$ I've been stuck on trying to figure out how to solve this for quite some time and I haven't found the solution or method I need to figure this out. Please help!
$$
\frac{d}{dx}\int_2^{x^4}\tan(t^2)dt
$$
 A: When $F(x)=\int^x_0\ \tan\ t^2\ dt$, then $$ \frac{d}{dx}\ F =_{{\rm Fundamental \ Theorem\ of\ Calculus}} \tan\
x^2$$ so that $$ \frac{d}{dx}\int_2^{x^4}\ \tan\ t^2\ dt
=\frac{d}{dx}\{F(x^4) +C\} =_{{\rm Chain\ Rule}} F'(x^4)(4x^3)=\tan\ x^8\cdot (4x^3)$$
A: I think you mean
$$\frac{\mathrm d}{\mathrm dx}\int_2^{x^4}\tan(t^2)\,\mathrm dt$$
as the integral is usually done with respect to a "dummy variable"; reusing variables like that is bad practice.
In this case, using Leibniz's integral rule, we get 
$$4x^3\tan(x^8)$$
A: 
Leibniz Integral Rule (Differentiation under the integral sign):
Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$,
$$\frac{d}{dx}(\int_{a(x)}^{b(x)} f(x,t) dt)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$


Here $f(x,t)=\tan(t^2)$, $b(x)=x^4$ and $a(x)=2$
So $\frac d{dx} \int_2^{x^4} \tan(t^2) dt=\int_2^{x^4}0 dt +\tan(x^8) 4x^3-\tan(4)0= 4x^3\tan(x^8) $
