Question

If $u=x^3-3x^{2}y+y^3$ then $x\frac{{\partial }^{2}u}{\partial x\partial y}+y\frac{{\partial }^{2}u}{\partial y^2}=$

• A. $6(y^2-x^2)$
• B. $6(x^2-y^2)$
• C. $6(x^2+y^2)$
• D. $6(x-2y)$

Answer 1

Answer: (A)

$6(y^2-x^2)$

$u=x^3-3x^{2}y+y^3$
$\frac{\partial u}{\partial x}=3x^2-6xy$
$\frac{{\partial }^{2}u}{\partial x\partial y}=-6x$

Hence $x\frac{{\partial }^{2}u}{\partial x\partial y}=x(-6x)=-6x^2$ ....(i)

Similarly$\frac{\partial u}{\partial y}=-3x^2+3y^2$
$\frac{{\partial }^{2}u}{\partial y^2}=6y$
$y\frac{{\partial }^{2}u}{\partial y^2}=y\left(6y\right)=6y^2$ ...(ii)
Hence $x\frac{{\partial }^{2}u}{\partial x\partial y}+y\frac{{\partial }^{2}u}{\partial y^2}=-6x^2+6y^2$
$=6(y^2-x^2)$