I know how you would do this:
$${d\over dx}\int_2^{x^4} \tan (t^2)\,dt,$$ but how would you do this: $${d\over dx}\int_2^{x^4} \tan (x^2)\,dx.$$ I am confused on the argument $x$ being in the integral and as the bound of the integral.
Edit: Since everyone is solving in terms of $t$ as the inside, I will assume that it is in terms of $t$, not $x$. I was confused because I was unsure if there were another method to solve the second integral, or if it was just a misprint. I will assume that this is a misprint and carry on. Thanks!
Best Answer
Using Leibniz Formula \begin{equation} \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. \end{equation} where \begin{equation} a(x) = 2 \end{equation} and \begin{equation} b(x) = x^4 \end{equation} and \begin{equation} f(x,t) = \tan t^2 \end{equation} You get \begin{equation} f(x,b(x)) = \tan x^8 \end{equation} \begin{equation} \frac{d}{dx}b(x) = 4x^3 \end{equation} Also \begin{equation} f(x,a(x)) = \tan 2^2 = \tan 4 \end{equation} but \begin{equation} \frac{d}{dx}a(x) = 0 \end{equation} and also notice that \begin{equation} \frac{\partial}{\partial x} f(x,t) = \frac{\partial}{\partial x} \tan t^2 = 0 \end{equation} So, we get that \begin{equation} {d\over dx}\int_{2}^{x^4} \tan (t^2)dt, = 4x^3 \tan x^8 \end{equation}