Question

If  z = Sin^-1 (^x⁄y) + Tan^-1 (^y⁄x) then x ^∂z⁄∂x + y ^∂z⁄∂y is?

(a) 0

(b) y

(c) 1 + ^x⁄y Sin^-1 (^x⁄y)

(d) 1 + ^y⁄x Tan^-1 (^y⁄x)

I had been asked this question in homework.

The question is from Euler’s Theorem topic in portion Partial Differentiation of Engineering Mathematics

Answer 1

Correct option is (a) 0

The best explanation: Given z = Sin^-1 (^x⁄y) + Tan^-1 (^y⁄x)

Let, u = Sin^-1 (^x⁄y) and v = Tan^-1 (^y⁄x) hence z = u + v

Now, let u’ = Sin(u) = ^x⁄y = f(^x⁄y) hence u’ satisfies euler’s theorem,

Hence,

\(x \frac{∂u’}{∂x}+y \frac{∂u’}{∂y}=0\)

Hence, by putting u’=e^u, we get

\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=0/e^u = 0\) ,……(1)

Now, let v’= Tan(v)=y/x=f(y/x) hence v’ satisfies euler’s theorem,

Hence,

\(x \frac{∂v’}{∂x}+y \frac{∂v’}{∂y}=0\)

Hence, by putting v’=ln(v), we get

\(x \frac{∂v}{∂x}+y \frac{∂v}{∂y}=0 \),……(2)

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

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