Question

$I$f $u=\frac{x^2{y}^2}{x+y}$, then $\left(\frac{x\frac{{\partial }^{2}u}{\partial x^2}+y\frac{{\partial }^{2}u}{\partial x\partial y}}{\frac{\partial u}{\partial x}}\right)$

• A. $0$
• B. $u$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

$\frac{\partial u}{\partial x}=\frac{2x{y}^2(x+y)-x^2{y}^2}{(x+y{)}^2}=y^2\frac{(x^2+2xy)}{(x+y{)}^2}$

$\frac{d^{2}u}{\partial x^2}=\frac{(2x{y}^2+2y^3)(x+y{)}^2-2(x+y\left)\right(x^2{y}^2+2x{y}^3)}{(x+y{)}^4}$

$\frac{d^{2}u}{\partial y\partial x}=\frac{(2y{x}^2+4x{y}^2)(x+y{)}^2-2(x^2{y}^2+2x{y}^3\left)\right(x+y)}{(x+y{)}^4}$

$\frac{x\frac{d^{2}u}{\partial x^2}+y\frac{d^{2}y}{\partial y\partial x}}{\frac{\partial u}{\partial x}}=\frac{(2x^2{y}^2+2x{y}^3)(x+y{)}^2-2(x+y{)}^2(x^2{y}^2+2x{y}^3+(2y^2{x}^2+2x{y}^3\left)\right(x+y{)}^2)}{(x+y{)}^4}$

$=\frac{y^2(x^2+2xy)}{(x+y{)}^2}$

$=\frac{2(x^2{y}^2+2x{y}^3)}{x^2{y}^2+2x{y}^3}$

$=2$