In my calculus class, this theorem was presented :
Let $f :J\times I \rightarrow R$ where $I,J$ are two open set and $f$ a $C^1$ function. Let $a,b : I \rightarrow J $ two $C^1$ functions. Then : $$g(y)=\int_{a(y)}^{b(y)}f(x,y)dx$$ Is $C^1(I)$ and $$g'(y):=f(b(y),y)b'(y)-f(a(y),y)a'(y)+\int_{a(y)}^{b(y)}\frac{\partial f(x,y)}{\partial y}dx$$ But, in my series, I'm given this function : $$f(x,y)=\int_{-x^2}^{y^2}\cosh{(yt^2+x})dt$$ I have two compute the gradient of this function but i have no idea how to compute $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$. Can someone explain me how to compute this ?
Thank you,
In order to compute $\frac\partial{\partial y}f(x,y)$, you apply that theorem with $a(y)=-x^2$ (it will be a constant function then) and $b(y)=y^2$. And, in order to compute $\frac\partial{\partial x}f(x,y)$, you apply that theorem with $a^\star(x)=-x^2$ and $b^\star(x)=y^2$ (now, this one will be constant).