Question

find
d/dx (integral[(x+y),(x-y)] sin(t^3) dt)

Answer 1

\[\frac{ d }{ dx } \int\limits_{x+y}^{x-y} \sin(t ^{3}) dt\]

Answer 2

This is a partial derivative again?

Answer 3

yes...

Answer 4

do i have to use chain rule for (x-y) and (x+y) or just leave it as sin (x-y)^3 -sin (x+y)^3?

Answer 5

You `do` have to apply the chain rule in each case.
But it looks the chain rule is just giving us 1. Sooooo, yah.

Answer 6

If the limits on our integral had been something like (x-c) and (x+c),
we would worry about applying the chain rule since the derivative of those limits is just 1.

Same idea here since we're applying partials :)

Answer 7

We wouldn't*

Answer 8

thanks!