I have a small problem which I just have no idea on how to calculate, I am trying to find the derivative of $p$ of an integral that is with respect to $y$ (and ultimately solve the equation).
$$\frac{d}{dp}[2\int_0^1(p-y)^2{y^{x-1}(1-y)^{-x}}dy]=0$$
$\endgroup$1 Answer
$\begingroup$Your tool is differentiation under the integral. Essentially: $$\frac{d}{dp}\int_a^bf(y,p)\,dy = \int_a^b\frac{\partial}{\partial p}f(y,p)\,dy$$
So: $$\begin{align} \frac{d}{dp}[2\int_0^1(p-y)^2{y^{x-1}(1-y)^{-x}}dy]&=2\int_0^1\frac{\partial}{\partial p}(p-y)^2{y^{x-1}(1-y)^{-x}}dy\\ &=2\int_0^1 2(p-y){y^{x-1}(1-y)^{-x}}dy\\ \end{align}$$
To go further, we'd need to know for what variable you are trying to solve, and what in the world the $x$'s are doing there. :)
