Q : Find $\frac{dz}{du}$ and $\frac{dz}{dv}$ if $z=\mathrm{arctan}(\frac{x}{y})$, $x=4\sin(u)$, $y=e^v$.

So,finding $\frac{dz}{du}$ means finding the partial derivative of $z=\mathrm{arctan}(\frac{x}{y})$ and $x=4\sin(u)$. but there is a problem, I don't understand, when I find this, is y constant? How do I relate this to $e^v$?

Does the same hold for $\frac{dz}{dv}$?

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In general, you have $\frac {dz}{du}=\frac {\partial z}{\partial x}\cdot \frac {dx}{du}+\frac {\partial z}{\partial y}\cdot \frac {dy}{du}$ In this case, $\frac {dy}{du}=0$ so you can ignore the last term. An alternate approach would be to invert the equations for $x,y$ and substitute them into the equation for $z$, which would let you find $\frac {dz}{du}$ directly. That is probably more work, though.
twinkletwinklelittlestar in case you are confused as to how partial derivative is mixed with normal derivative as given in the edited answer of Ross, here is what happen: the differential equation he wrote there is a chain rule and it should have involved partial derivatives throughout, but since $x$ is a function of single variable, then there is no need to do that. And to answer your question ..."is $y$ constant?", yes, you should regard it as constant. Also the same method hold for $\dfrac{dz}{dv}$ and I advice you to use this method and ignore the inversion method if you find it somewhat confusing.