Question

If $z=\frac{y}{x}[\sin \frac{x}{y}+\cos (1+\frac{y}{x}\left)\right]$, then $x\frac{\partial z}{\partial x}=$

• A. $y\frac{\partial z}{\partial y}$
• B. $-y\frac{\partial z}{\partial y}$
• C. $2y\frac{\partial z}{\partial y}$
• D. $2y\frac{\partial z}{\partial x}$

Answer 1

Answer: (B)

$-y\frac{\partial z}{\partial y}$

z is a homogeneous $e{q}^n$ of degree$=0$
$\sum \frac{x\partial z}{\partial x}=nz$
$\sum \frac{x\partial z}{\partial x}=0$
$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$

$x\frac{\partial z}{\partial x}=-y\frac{\partial z}{\partial y}$