# Partial differentiation dz/ds when function of z is differentiable function of x and y but not given explicitly

$$z = f(x,y)$$ 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