Let $\displaystyle u= {e}^{ -2x }\cos 4y$ and $\displaystyle v = { e }^{ -2x }\sin 4y$
Use implicit differentiation to evaluate $\displaystyle \frac { \partial x }{ \partial u }$ at $(x,y)= (1,2).$
I'm not really sure how to do this one.
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For example
$$u=e^{-2x}\cos 4y\implies x=-\frac{1}{2}\log\frac{u}{\cos 4y}$$
$$\frac{dx}{du}=-\frac12\frac1u\;,\;\;\text{and now just substitute the point you want.}$$
To differentiate partially, say according to x, everything that is not $\,x\,$ is taken as a constant, and the other way around if you want to differentiate wrt $\,y\,$
DonAntonio
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Would the point be $, u\ =\ { e }^{ -2 }\cos(8)$ ? To differentiate partially, do I do it to u or v? – Dexter Apr 28 '13 at 19:58
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Yes, that's the point. The above is already the partial derivative of $,x,$ wrt $,u,$ . – DonAntonio Apr 28 '13 at 21:49