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I have to find $\frac{\partial z}{\partial u}$ if $$z=\arctan\left(\frac{x}{y}\right),\quad x=4\sin u,\quad y=e^v.$$

I can find it directly by replacing $x$ and $y$ and get $$z=\arctan\left(\frac{4\sin u}{e^v}\right).$$

Now I find $$\frac{\partial z}{\partial u} = \frac{1}{1 + \frac{4\sin^2u}{e^{2v}}}\left(\frac{4\sin u}{e^v}\right)'.$$

When I do my calculations, the result is $$\frac{4\cos(u)e^v}{e^{2v} + 4\sin^2 u}$$ but the result in my book is $$\frac{e^v\sin v}{u^2\sin^2u +e^{2v}}.$$

How do I get this result?

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Are you sure about the formula in your original exercise? The only minor mistake I could see in your calculations was that you forgot to square the 4 when computing the derivative of the $\arctan()$ function. – Matt L. Apr 21 '13 at 8:30
yes Mat,I triple-checked it :) – eeee Apr 21 '13 at 8:33
In this case there's no way you can get the result in the book. The result is $\frac{4e^v\cos u}{16\sin^2u + e^{2v}}$. – Matt L. Apr 21 '13 at 8:54
Thank you Matt. – eeee Apr 21 '13 at 8:58
LOL,this is the first time that someone in mathstack doesnt answer to me haha! – eeee Apr 21 '13 at 8:58

Try instead to use the chain rule when you differentiate. It's much easier: $$ \frac{\partial z}{ \partial u } = \frac{\partial z}{ \partial x }\frac{\partial x}{ \partial u } + \frac{\partial z}{ \partial y }\frac{\partial y}{ \partial u } $$ $$ = \frac{1}{y}\frac{1}{1+\left(\frac{x}{y} \right)^2 }\frac{\partial x}{ \partial u } + \frac{x}{-y^2}\frac{1}{1+\left(\frac{x}{y} \right)^2 }\frac{\partial y}{ \partial u } . $$ Since there is no ralationship between $u$ and $v$ we have $ \frac{\partial y}{ \partial u } = 0$. Hence $$ \frac{\partial z}{ \partial u } = \frac{1}{y}\frac{1}{1+\left(\frac{x}{y} \right)^2 }\frac{\partial x}{ \partial u } = \frac{y}{y^2+x^2}\frac{\partial x}{ \partial u } = \frac{ 4 \cos{u} \cdot e^v}{e^{2v} + 16 \sin^2(u)}$$ as Matt said.

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