If $f(x,y)=\int^x_y \cos(t^2)\,dt$, find the first partial derivatives of the function

Question

Problem: If $f(x,y)=\int^x_y \cos(t^2) \, dt$, find the first partial derivatives of the function.

My thoughts: By the Fundamental Theorem of Calculus, I know that $f_x=\cos(x^2)$, since $y$ is just treated as a constant.

But what about $f_y$? Am I right in assuming that it would be $-\cos(x^2)$, since $\int^x_a f(t) \, dt=-\int^a_x f(t) \, dt$? Or are things more complicated than that?

@Micah Thanks, good catch- did you mean it should be $-cos(y^2)$ instead of $-cos(x^2)$? — Asker, Apr 30 '15 at 21:08

score 3 · Accepted Answer · answered Apr 30 '15 at 21:15

3

The way I would think of this (because it works better in more complicated situations where the limits are functions of $x$ and $y$) is to let $g$ be an antiderivative of $t \mapsto \cos t^2$. Then the Fundamental Theorem of Calculus tells us that $$ f(x,y)=g(x)-g(y) $$ which is easy to take partial derivatives of when you remember that $g'(t)=\cos t^2$.

But your way works too (except for the issue I mentioned in the comments).

answered Apr 30 '15 at 21:15

Micah

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When I try doing it this way for $f_x$, I get $cos(x^2)-g(y)$, but how do we know that $g(y)$ is 0? – Asker Apr 30 '15 at 21:24
$g(y)$ is independent of $x$, so $\frac{\partial}{\partial x} g(y)=0$. – Micah Apr 30 '15 at 21:29
Oops, I had a brain fart there- it all makes sense now. Thanks! – Asker Apr 30 '15 at 21:31

If $f(x,y)=\int^x_y \cos(t^2)\,dt$, find the first partial derivatives of the function

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