Question

If $z=a{x}^2+2hxy+b{y}^2$,

then $x^2\frac{{\partial }^{2}z}{\partial x^2}+2xy\frac{{\partial }^{2}z}{\partial x\partial y}+y^2\frac{{\partial }^{2}z}{\partial y^2}=?$

• A. $0$
• B. $z$
• C. $2z$
• D. $4z$

Answer 1

Answer: (C)

$2z$

$z=a{x}^2+2hxy+b{y}^2$
$\frac{\partial z}{\partial x}=2ax+2hy$

$\frac{{\partial }^{2}z}{\partial x^2}=2a$

$\frac{\partial }{\partial y}\frac{\partial z}{\partial x}=2h$$\frac{\partial y}{\partial y}=2h$

$\frac{{\partial }^{2}z}{\partial y^2}=2b$
So, $x^2\frac{{\partial }^{2}z}{\partial x^2}+2xy\frac{{\partial }^{2}z}{\partial x\partial y}+y^2\frac{{\partial }^{2}z}{\partial y^2}=x^2\left(2a\right)+2xy\left(2h\right)+y^2\left(2b\right)=2z$