Question

If \(z=e^{\frac{x^2+y^2}{x+y}}\) then, \(x \frac{∂z}{∂x} + y \frac{∂z}{∂y}\) is?

(a) 0

(b) zln(z)

(c) z^2 ln⁡(z)

(d) z

I had been asked this question during a job interview.

This interesting question is from Euler’s Theorem topic in section Partial Differentiation of Engineering Mathematics

Answer 1

Correct answer is (b) zln(z)

To elaborate: Given \(z=e^{\frac{x^2+y^2}{x+y}}\),let u=ln⁡(z)=\(\frac{x^2+y^2}{x+y}=\frac{x(1+(\frac{y}{x})^2)}{(1+\frac{y}{x})}\) = x f(y/x)

Hence u is homogeneous of order 1,

Hence,

\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}\)=u

Putting, u = ln(z)  we get,

\(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\) = zln(z)