Question

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

• A. $0$
• B. $e^u$
• C. $1$
• D. $2e^u$

Answer 1

The correct option is A $0$

If $e^u=\frac{x+\sqrt{x^2-y^2}}{x-\sqrt{x^2-y^2}}Ify=e^{x}x={\log }_{e}y$

SO the equation becomes $e^u=\frac{x+\sqrt{x^2-y^2}}{x-\sqrt{x^2-y^2}}u={\log }_e\left(\frac{x+\sqrt{x^2-y^2}}{x-\sqrt{x^2-y^2}}\right)\to \left(1\right)$

Differentiating ${}^{\prime }{u}^{\prime }$ w.r.t to ${}^{\prime }{x}^{\prime }$ partially

Apply chain rule $\frac{df}{dx}\left(u\right)=\frac{df}{du}.\frac{du}{dx}$

$f=\ell n\left(u\right)andu=\frac{x+\sqrt{x^2-y^2}}{x-\sqrt{x^2-y^2}}$

$\frac{\partial }{\partial u}f\left(u\right)=\frac{\partial }{\partial u}f\left(\ell n\left(u\right)\right)\frac{\partial }{\partial x}\left(\frac{x+\sqrt{x^2-y^2}}{x-\sqrt{x^2-y^2}}\right)$

$=\frac{1}{u}\left[\frac{\frac{\partial }{\partial x}(x+\sqrt{x^2+y^2})(x-\sqrt{x^2-y^2})-\frac{\partial }{\partial x}(x-\sqrt{x^2-y^2})(x+\sqrt{x^2+y^2})}{{(x-\sqrt{x^2-y^2})}^2}\right]$

Substitute ${}^{\prime }{u}^{\prime }$

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

Similarly $\frac{\partial }{\partial y}\left[\ell n\left(\frac{x+\sqrt{x^2+y^2}}{x-\sqrt{x^2-y^2}}\right)\right]=\frac{xy\sqrt{x^2-y^2}-2x{y}^2-xy\sqrt{x^2+y^2}}{(x+\sqrt{x^2+y^2})(x-\sqrt{x^2-y^2})\sqrt{x^2-y^2}\sqrt{x^2{+y}^2}}\to \left(2\right)$

To find $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$

SUbstituting equation $\left(1\right)$ and $\left(2\right)$

$x\left[ \frac { 2x{ y }^{ 2 }+{ y }^{ 2 }\sqrt { x^{ 2 }+{ y }^{ 2 } } -{ y }^{ 2 }\sqrt { -y^{ 2 }+{ x }^{ 2 } } }{ \left( x+\sqrt { x^{ 2 }+{ y }^{ 2 } } \right) \left( x-\sqrt { x^{ 2 }{ -y }^{ 2 } } \right) \left( \sqrt { x^{ 2 }-{ y }^{ 2 } } \right) \left( \sqrt { x^{ 2 }+{ y }^{ 2 } } \right) } \right] $

$+y\left[ \frac { y\sqrt { x^{ 2 }{ -y }^{ 2 } } -2x^{ 2 }y-xy\sqrt { x^{ 2 }+{ y }^{ 2 } } }{ \left( x+\sqrt { x^{ 2 }+{ y }^{ 2 } } \right) \left( x-\sqrt { x^{ 2 }{ -y }^{ 2 } } \right) \left( \sqrt { x^{ 2 }-{ y }^{ 2 } } \right) \left( \sqrt { x^{ 2 }+{ y }^{ 2 } } \right) } \right] $

$=\frac { 2x^{ 2 }y^{ 2 }+xy^{ 2 }\sqrt { x^{ 2 }+{ y }^{ 2 } } -xy^{ 2 }\sqrt { x^{ 2 }{ -y }^{ 2 } } +xy^{ 2 }\sqrt { x^{ 2 }{ -y }^{ 2 } } -2x^{ 2 }y^{ 2 }-xy^{ 2 }\sqrt { x^{ 2 }+{ y }^{ 2 } } }{ \left( x+\sqrt { x^{ 2 }+{ y }^{ 2 } } \right) \left( x-\sqrt { x^{ 2 }{ -y }^{ 2 } } \right) \left( \sqrt { x^{ 2 }-{ y }^{ 2 } } \right) \left( \sqrt { x^{ 2 }+{ y }^{ 2 } } \right) } $

$=0$

Therefor the correct answer is option $A$

$\\ \\ x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =0$