Originally Posted by

**juanma101285** Hi, I have to work out the first partial derivative of the following function:

$\displaystyle f(x,y)=(x^3+y)sin(1/(x^2+y^2))$

So, what I am getting is:

$\displaystyle f'(x, y)=(3x^2)sin(1/(x^2+y^2))-((x^3+y)2y) / (x^2+y^2)^2 * cos(1/(x^2 + y^2))$

But the answer is supposed to be almost identical, except that it is $\displaystyle -((x^3+y)2x)$ instead of $\displaystyle -((x^3+y)2y)$. I have done it several times with the chain rule and I keep on getting the same thing... I would really appreciate it if someone could show me how to get the second half :/. Thanks a lot in advance!!!