Question

If $u={\cos }^{-1}\left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right),$ then
$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=$

• A. $\frac{1}{2}\cot u$
• B. $2\cot u$
• C. $\frac{-1}{2}\cot u$
• D. $3\cot u$

Answer 1

The correct option is C $\frac{-1}{2}\cot u$

$cosu=\frac{x+y}{\sqrt{x}+\sqrt{y}}$

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

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

$-sinu(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y})=\frac{(\sqrt{x}+\sqrt{y})(n+y)\frac{-1}{2}(\sqrt{x}+\sqrt{y})(x+y)}{(\sqrt{x}+\sqrt{y}{)}^2}$

$=\frac{1}{2}\frac{(x+y)}{\sqrt{x}+\sqrt{y}}$

$\frac{x\partial u}{\partial x}+\frac{y\partial u}{\partial y}=\frac{-1}{2}cotu$