Partial Derivative of Function

[triden]

Homework Statement

I am doing some gradient questions and having a little trouble understanding the partial derivatives to obtain the gradient. Most particularly in this question:

f(x,y) = $$\frac{1}{3}$$$$(x^{2}+y^{2})^{2}$$

Homework Equations

So to find the gradient we take the partial derivative with respect to x and then y.

The Attempt at a Solution

I know the answer is

$$\nu$$(x,y) = $$\frac{4}{3}x$$$$(x^{2}+y^{2})$$

but my brain says it should be

$$\nu$$(x,y) = $$\frac{4}{3}x$$

why does the (x^2 + y^2) term still stick around? When I do the partial derivative of the x term, do I not treat the Y term as constant? If so, shouldt it disappear?

Thanks

 

[lanedance]

treat the y as a constant when you differntiate only. You must still retain the y, in general the partial derivative w.r.t. x is still a function of both x & y

using chain rule below
$$f_x<br /> = \frac{\partial}{\partial x}(x^{2}+y^{2})^{2}<br /> = 2(x^{2}+y^{2})\frac{\partial}{\partial x}(x^{2}+y^{2})<br /> = 2(x^{2}+y^{2})2x$$

 

[triden]

Ah yes, I got it. It was just a stupid logic error in my head. Thanks for the help!