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$$ z = f(x,y) $$ and $z$ is differentiable.

Let $$ x = s^2 - t, y = t^3 \ln(1+s) $$

Then $$ \frac{\partial z}{\partial s}$$ at $s = 0$ and $t = 0$ is?

What assumption do I have to make? $$ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}(2s) + \frac{\partial z}{\partial y}\left(\frac{1}{1+s}\right) $$

Because I need to know the partial of $z$ with respect to $x$ and the partial of $z$ with respect to $y$.

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I think you should have $$\frac{\partial z}{\partial s} = (2s)\frac{\partial z}{\partial x} + \left(\frac{t^3}{1+s}\right)\frac{\partial z}{\partial y}$$ in which case when you evaluate at $(s,t) = (0,0)$ you get ... –  John Martin Dec 11 '12 at 3:56
    
Indeed. Thank you. –  40Plot Dec 11 '12 at 4:15

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