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I am having troubles finding partial derivatives.

If $f(x,y)=2x^2+y^2$ then, $$f_x=4x$$ $$f_y=2y$$

That's simple enough. But when I see a $z$ in the equation, I get stumped. I know $z=f(x,y)$. I don't really see the process.

For example, if $z=2x^2+y^2$ then do we differentiate both sides with respect to $x$ like this?

$$f_x:\frac{dz}{dx}=4x$$ $$f_y:\frac{dz}{dy}=2y$$

Even worse, what am I supposed to do for something like this? Same thing?

$$x+y^2+z^3=3$$ $$z^3=-x-y^2+3$$

$$f_x:3z^2\frac{dz}{dx}=-1$$ $$f_x:\frac{dz}{dx}=\frac{-1}{3z^2}$$

Likewise,

$$f_y:\frac{dz}{dy}=\frac{-2y}{3z^2}$$

I'm pretty sure that's wrong but I don't know why. Can someone please help me understand? Thanks.

Edit Some context for the last example (it's from a homework problem):

Find the equation of the tangent plane to the surface with equation $x+y^2+z^3=3$ at the point (2,1,0).

I know that the equation for the tangent plane is $$z=f(x_0,y_0)+[f_x(x_0,y_0)](x-x_0)+[f_y(x_0,y_0)](y-y_0)$$

Since the $f_x$ and $f_y$ found above contain z, do I plug in $z_0$?

So I'm trying to find $f_x$ and $f_y$.

Edit 2

I get it now

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    $\begingroup$ That looks correct to me! $\endgroup$ – William Stagner May 16 '13 at 3:44
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    $\begingroup$ well are you trying to find the partial of $z$? in the equation $z = 2x^2 + y^2$ there is no "$f$" so how can you have $f_x$? I may be totally offbase and misunderstanding the question... $\endgroup$ – DanZimm May 16 '13 at 3:46
  • $\begingroup$ @DanZimm I believe he set $z=f(x, y)$. $\endgroup$ – William Stagner May 16 '13 at 3:47
  • $\begingroup$ @WilliamStagner then this all makes sense :D $\endgroup$ – DanZimm May 16 '13 at 3:49
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    $\begingroup$ @Jey I believe what you're looking for is solving an "implicit equation" of $z$. This appears to be exactly what you've done so all seems well! $\endgroup$ – DanZimm May 16 '13 at 3:55
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What you are looking for is called the chain rule. If you have: $$g(z(x,y))=0$$ Then: $$\frac{\partial g}{\partial x}=\frac{dg}{dz}\cdot\frac{\partial z}{\partial x}$$

So if $x + y^2 + z^3 = 3$, you take the derivative according to $x$: $$1+0+3z^2 \cdot \frac{dz}{dx}=0 \ \ \to \ \ \frac{dz}{dx}=\frac{-1}{3z^2}$$ What does mean? this is a differential equation that may or may nor be easy to solve for $z$, but if you do know what $z$ is, you could just plug n' play: $$\frac{dz}{dx}=\frac{-1}{3z^2}=\frac{-1}{3(3-x-y^2)^{2/3}}$$

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