Question

If $z{e}^{x^2+y^2}=1$

then $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=?$

• A. $2z\log z$
• B. $\frac{2}{z}$
• C. $\frac{2\log z}{z}$
• D. $2\log z$

Answer 1

The correct option is A $2z\log z$

$z=e^{-(x^2+y^2)}=e^{-x^2-y^2}$
$\log z=-(x^2+y^2)$

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=x\left(e^{-x^2-y^2}\right)(-2x)+y\left(e^{-x^2-y^2}\right)(-2y)$

$=-2\left(e^{-x^2-y^2}\right)(x^2+y^2)$

$=2z\log z$