Question

If $u=\log \left(\frac{x^3+y^3}{x-y}\right)$

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

• A. $1$
• B. $0$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

$e^u=\frac{x^3+y^3}{x-y}$

$e^u\frac{du}{dx}=\frac{3x^2(x-y)-(x^3+y^3)}{(x-y{)}^2}=\frac{2x^3-3y{x}^2-y^3}{(x-y{)}^2}$

$e^u\frac{du}{dy}=\frac{3y^2(x-y)-(x^3+y^3)}{(x-y{)}^2}=\frac{3y^{2}x-2y^3-x^3}{(x-y{)}^2}$

$e^u(x\frac{du}{dx}+y\frac{du}{dy})=\frac{2x^4-3y{x}^3-x{y}^3}{(x-y{)}^2}+\frac{3y^{3}x-2y^4+x^{3}y}{(x-y{)}^2}$

$e^u(x\frac{du}{dx}+y\frac{du}{dy})=\frac{2x^4-2y{x}^3+2y^{3}x-2y^4}{(x-y{)}^2}$

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

$e^u(x\frac{du}{dx}+y\frac{du}{dy})=2e^u$

Thus, $(x\frac{du}{dx}+y\frac{du}{dy})=2$