# Thread: Need help with a partial derivative of a multivariate function

1. ## Need help with a partial derivative of a multivariate function

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

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

So, what I am getting is:

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

But the answer is supposed to be almost identical, except that it is $-((x^3+y)2x)$ instead of $-((x^3+y)2y)$. 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!!!

2. ## Re: Need help with a partial derivative of a multivariate function

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

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

So, what I am getting is:

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

But the answer is supposed to be almost identical, except that it is $-((x^3+y)2x)$ instead of $-((x^3+y)2y)$. 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!!!
Are you evaluating \displaystyle \begin{align*} \frac{\partial f}{\partial x} \end{align*} or \displaystyle \begin{align*} \frac{\partial f}{\partial y} \end{align*}?

3. ## Re: Need help with a partial derivative of a multivariate function

Sorry, I understand it now.... I mixed up the formulae and was somehow trying to evaluate both :/