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If $u$ and $v$ are functions of $x$ and $y$ defined by $ x=u+e^{-v}sinu, y=v+e^{-v}cosu$ , then prove that $$\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$$

My Attempt:

$\frac{\partial x}{\partial v}=0+(sinu)e^{-v}(-1)=-e^{-v}sinu$

and then I reciprocated it. Then I did similar with $\frac{\partial y}{\partial u}$.

Is my method correct?? Can we reciprocate in partial differentiation.?

Udit Chauhan
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    You can't take a reciprocal in partial differentiation, because the top and bottom actually have different meanings. A more explicit notation for the partial differentiation of $u$ with respect to $y$ would be something like $\frac{\partial_y u}{\mathrm{d}y}$. So, as you can see, the reciprocal would not match. – johnnyb Jan 17 '21 at 14:40

1 Answers1

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It is incorrect to take reciprocal of a dx/dv to get dv/dx. Take partial differentiation of first equation with respect to y and second equation with respect to x. Since x and y are not dependent on each other dx/dy and dy/dx is equal to 0. Eventually you will prove the result asked in the question i.e. du/dy = dv/dx.

programmer_pg
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