Find partial derivative of $\displaystyle u= {e}^{ -2x }\cos 4y$ and $\displaystyle v = { e }^{ -2x }\sin 4y$

Question

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.

Answer 1

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 $\,y\,$