Question

If \(z=sin^{-1}\frac{x^3+y^3+z^3}{x+y+z}\) then, \(x\frac{∂z}{∂x}+y\frac{∂z}{∂y}\).

(a) 2 tan(z)

(b) 2 cot(z)

(c) tan(z)

(d) cot(z)

The question was asked in an interview.

The above asked question is from Euler’s Theorem in division Partial Differentiation of Engineering Mathematics

Answer 1

Correct answer is (a) 2 tan(z)

To explain: Given \(z=sin^{-1}⁡\frac{x^3+y^3+z^3}{x+y+z}\), put u=sin⁡(z)=\(\frac{x^3+y^3+z^3}{x+y+z}=x^2 f(\frac{y}{x},\frac{z}{x})\)

Hence, \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=2u\)

Putting u = sin(z), we get

\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=\frac{2Sin(z)}{Cos(z)}=2Tan(z)\)