Question

Find the derivative: 2ysin^2(xy)

Thanks,
Find the derivative: 2ysin^2(xy)

Thanks,
@Mathematics

Answer 1

i assume this is set equal to something, or are you looking for a partial?

Answer 2

Its part of a bigger function, but I'm just confused on how to find the derivative of this,

Answer 3

you can't unless it is equal something.

Answer 4

oh my bad, so the function is basically 2ysin^2(xy) + 6x = 3 + (pi)

Answer 5

ok

Answer 6

\[\[2(y'\sin^2(xy)+2y\cos(xy)y'+6=0\]

Answer 7

solve for y', and the 2 is in front of the first part, not the 6

Answer 8

So would i use the chain rule for ysin^2 and not the (xy)?

Answer 9

\[2(y'\sin^2(xy)+2y\cos(xy)y')+6=0\] without an equation you cannot solve. yes i used the chain rule for both

Answer 10

oh damn
good catch

Answer 11

\[2(y'\sin^2(xy)+2y\sin(xy)\cos(xy)y')+6=0\] is more like it. my mistake, good call

Answer 12

Oh i see, thanks a lot, I appreciate it.

Answer 13

yw, and hope i didn't confuse with the wrong solution!

Answer 14

not at all, haha thanks.

Answer 15

haha?

Answer 16

Whats wrong with typing haha? I was just trying to be friendly.