Partial differentiation of a integral

Question

If I have: $$f(x,y)=\int_{x}^{y}e^{-t^2}dt$$ To calculate that, I change the t to y and x to get: $$e^{-y^2}(e^{-y^2})'-e^{-x^2}(e^{-x^2})$$ With the differentials being in respect to x and y for each partial derivative?

Answer 1

We have: $\displaystyle \int_{x}^y = \displaystyle \int_{0}^y - \displaystyle \int_{0}^x \Rightarrow f_x = -e^{-x^2}, f_y = e^{-y^2}$.

Answer 2

Let $F$ be antiderivative of $g(t)=e^{-t^2}$(F'(t)=g(t)), then by Fundamental Theorem of Calculus:

$$f(x,y)=\int_{y}^{x}e^{-t^2}dt=F(x)-F(y)$$

So

$$\frac{\partial f}{\partial x}(x,y)=F'(x)=e^{-x^2}$$

The same with second partial derivative.