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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.

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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.

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