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

Question

I want to compute the partial derivative of the following function:

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

Since $f$ is a continous function it has an antiderivative. Let $F$ be the antiderivative of f.

Then:

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

Now computing the partial derivatives of $F(y) -F(x)$ results in:

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

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

Is that valid?

Answer 1

Recall from the Fundamental Theorem of Calculus, if we define ${f(x)=\int_{a}^{x}g(t)dx}$ then we have ${f'(x)=g(x)}$. In your case, a multivariate function is defined to be

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

Remember when computing the partial derivatives of ${f(x,y)}$, you always treat the other variable as a constant. So in actuality we get

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

Furthermore, we see $${\int_{x}^{y}e^{-t^2}dt=-\int_{y}^{x}e^{-t^2}dt}$$

And so ${\frac{\partial f}{\partial x}=-e^{-x^2}}$

Answer 2

Your calculation is not compleatly correct, see https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

For $F(x):=\int_a^x f(t) dt$ it holds $F'(x)=f(x)$ (note: you put $x$ in $f$ instead of $t$)

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

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

Additionally, $F$ in your example in not the antiderivative for $f(x,y)$ but the antiderivative of $\exp(t-2)$