if $f(x,y) = \int_{x}^{y}p(p-1)(p-2)dp$ then calculate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ How do i take derivative of this function : $f(x,y) = \int_{x}^{y}p(p-1)(p-2)dp$.
For single variable I can evaluate but this involves two variables. 
Any hint please I am stuck here.
I want to calculate the stationary points, so I would like to evaluate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

$$\int_x^yp(p-1)(p-2)\,dp=\int_x^y (p^3-3p^2+2p)\,dp=-\frac{x^4}{4}+x^3-x^2+\frac{y^4}{4}-y^3+y^2$$
Now differentiate with respect to whatever variable you want.
In general, if we are given a continuous function $f(p)$ and we want to differentiate the integral 
$$G(x,y)=\int_x^yf(p)\,dp$$
as a function of $y$, say, then we can use the fundamental theorem of calculus which allows us to deduce that 
$$\frac{\partial G}{\partial y}(x,y)=f(y)$$
similarly,
$$\frac{\partial G}{\partial x}(x,y)=-f(x)$$
the negative sign is because the $x$ appears as a lower limit.
