How to compute the partial derivatives of this function? $$f(x,y) = \int_{-x}^{y}\sinh(xyt^2)dt$$
I tried replacing $\sinh(xyt^2)$ by its taylor series $$\sum_{n=1}^{\infty} \frac{(xyt^2)^{2n+1}}{(2n+1)!} = \sum_{n=1}^{\infty} \frac{x^{2n+1}y^{2n+1}t^{4n+2}}{(2n+1)!} $$
so I think that an anti derivative of this function with respect to $t$ would be $$\sum_{n=1}^{\infty} \frac{x^{2n+1}y^{2n+1}t^{4n+3}}{(2n+1)!\cdot(4n+3)}$$
so $$ \int_{-x}^{y}\sinh(xyt^2)dt = \sum_{n=1}^{\infty} \frac{x^{2n+1}y^{6n+4}}{(2n+1)!\cdot(4n+3)} +\sum_{n=1}^{\infty} \frac{x^{6n+4}y^{2n+1}}{(2n+1)!\cdot(4n+3)} $$
computing the partial derivatives of the above thing is a pain in the neck and  recognizing the function that matches the results is even painful.
maybe I'm doing it wrong... I don't know but anyway I don't like the way I'm solving it show me an easier/softer method.
thank you ! 

If $x,y,t$ are independent variables, you  can consider $y$ as a constant when deriving with respect to $x$ and use the Leibintz Integral Rule. So we have:
$$
\frac{\partial}{\partial x}\int_{-x}^y\sinh(xyt^2)dt =\sinh(x^3y)+\int_{-x}^y\left[\frac{\partial}{\partial x}\sinh(xyt^2)\right]dt
$$
and do the same for the derivative with respect to $y$.
