Question

If u=sin^-1(x2+y2/x+y)prove that xdu/dx)+ydu/dy)=tan u

Answer 1

Given:

$u=sin^{-1}(\frac{x^2+y^2}{x+y})$

$\implies\;sinu=\frac{x^2+y^2}{x+y}=f(x,y)$

$f(tx,ty)=\frac{t^2x^2+t^2y^2}{tx+ty}$

$f(tx,ty)=\frac{t^2(x^2+y^2)}{t(x+y)}$

$f(tx,ty)=t^1(\frac{x^2+y^2}{x+y})$

$\therefore\text{f is a homogeneous function of degree 1}$

$\text{By Euler's theorem}$

$x\frac{{\partial}f}{{\partial}x}+y\frac{{\partial}f}{{\partial}y}=1.f$

$x\frac{{\partial}(sinu)}{{\partial}x}+y\frac{{\partial}(sinu)}{{\partial}y}=1.sinu$

$x\;cosu\;\frac{{\partial}u}{{\partial}x}+y\;cosu\;\frac{{\partial}u}{{\partial}y}=sinu$

$\text{Divide both sides by cosu}$

$x\frac{{\partial}u}{{\partial}x}+y\frac{{\partial}u}{{\partial}y}=\frac{sinu}{cosu}$

$\implies\boxed{\bf\;x\frac{{\partial}u}{{\partial}x}+y\frac{{\partial}u}{{\partial}y}=tanu}$