Question

If $u={\sec }^{-1}\left(\frac{x^3-y^3}{x+y}\right)$

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

• A. $\cot u$
• B. $\tan u$
• C. $2\cot u$
• D. $2\tan u$

Answer 1

Answer: (C)

$2\cot u$

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

$\sec u\tan u\frac{\partial u}{\partial x}=\frac{3x^2(x+y)-(x^3-y^3)}{(x+y{)}^2}=\frac{2x^3+3x^{2}y+y^3}{(x+y{)}^2}$

$\sec u\tan u(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y})=\frac{2x^4+3x^{3}y+x{y}^3-2y^4-3x{y}^3-x^{3}y}{(x+y{)}^2}$

$=\frac{2(x^4+x^{3}y-x{y}^3-y^4)}{(x+y{)}^2}$

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

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

$\sec u\tan u(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y})=2\sec u$

$x\frac{du}{dx}+y\frac{du}{dy}=2\cot u$