Question

Calculate δf/δx and δf/δy for the function:
f^2 +cos(xy)=f

How do I do partial differentiation in this form?
(Normally it is f(x,y)=...)

Answer 1

woah, thats funky.

Answer 2

to me, f^2 -f + cos(xy) =0
2f df/dx -df/dx -y sin(xy)=0
df/dx(2f-1) =y sin(xy)
df/dx = \(\dfrac{y sin(xy)}{2f-1}\)

Answer 3

I see, so, df/dy= \[\frac{ xsinxy }{ 2f-1 }\]\

Answer 4

:)

Answer 5

Thanks for your help in the previous question btw. Although it was not the actual answer, but I figured it out.