Question

If \(z=ln⁡(\frac{x^2+y^2}{x+y})-e^{\frac{x^2+y^2}{x+y}}\) then find \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\).

(a) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

(b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

(c) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1+\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

(d) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

I have been asked this question in an online interview.

Question is from Euler’s Theorem topic in section Partial Differentiation of Engineering Mathematics

Answer 1

Correct answer is (b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)

The best explanation: Given \(z=ln⁡(\frac{x^2+y^2}{x+y})-e^\frac{x^2+y^2}{x+y}\)

Let, \(u = ln⁡(\frac{x^2+y^2}{x+y})\) and \(v=e^(\frac{x^2+y^2}{x+y})\) hence z=u-v

Now, let \(u’ = e^u = \frac{x^2+y^2}{x+y}=xf(\frac{y}{x})\) hence u’ satisfies euler’s theorem,

Hence,

\(x \frac{∂u’}{∂x}+y \frac{∂u’}{∂y}=u’\)

Hence, by putting u’=e^u, we get

\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=\frac{e^u}{e^u} = 1\), ……(1)

Now, let v’ = ln(v)= \(\frac{x^2+y^2}{x+y}=xf(\frac{y}{x})\) hence v’ satisfies euler’s theorem,

Hence,

\(x \frac{∂v’}{∂x}+y \frac{∂v’}{∂y}=v’\)

Hence, by putting v’=ln(v),we get

\(x \frac{∂v}{∂x}+y \frac{∂v}{∂y}=vln(v)\), …… (2)

By subtracting eq(1) and eq(2), we get

\(x \frac{∂z}{∂x}-y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)