Question

If \(u=\frac{e^{x+y}}{e^x-e^y}\), what is \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\)?

(a) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2} \)

(b) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x+e^y)^2} \)

(c) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x-e^y)}{(e^x-e^y)^2} \)

(d) u

I have been asked this question in an online quiz.

This question is from First Order PDE topic in section Partial Differential Equations of Engineering Mathematics

Answer 1

Correct choice is (a) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2} \)

Easiest explanation: Here we use partial differentiation.

\(\frac{\partial u}{\partial x}=\frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^x)}{(e^x-e^y)^2}\)

\(\frac{\partial u}{\partial y}=\frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^y)}{(e^x-e^y)^2}\)

\(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^x)}{(e^x-e^y)^2} + \frac{(e^x-e^y)×e^{x+y}-(e^(x+y))(e^y)}{(e^x-e^y)^2} \)

\(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2}\).