if $u = \arccos \left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right)$ then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ Edit: this is a repost
if possible i need some sort of shorcut, 80 questions, 90 min to solve these type of questions i have only 1 min time solt for each question
Context: i've looked at PYQ's and this year mock tests and found out that this question was one of the most repeated ones and i'm having an exam next week and my teachers are not available
The answer i got
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{-x\left(\sqrt{x}+\sqrt{y}-(x+y)\frac{1}{2\sqrt{x}}\right)} {\sin^2(u)(\sqrt{x}+\sqrt{y})^2}- \frac{y\left(\sqrt{x}+\sqrt{y}-(x+y)\frac{1}{2\sqrt{y}}\right) }   {\sin^2(u)\left(\sqrt{x}+\sqrt{y}\right)^2}$


The answer im getting is nowhere near the options that have been given

Options
 1.$\frac{-1}{2}\sin(u)$
 2. $\frac{-1}{2}\cot(u)$
 3. $\frac{-1}{2}\tan(u)$
 4. $\frac{-1}{2}\cos(u)$

I got :
$$\tag{1} \boxed{   x \frac{\partial u}{\partial x} = -~ \frac{2x \sqrt{y} + x \sqrt{x} -y \sqrt{x}}{2~\sin(u) ~(\sqrt{x} + \sqrt{y})^2}} $$
$$\tag{2} \boxed{   y \frac{\partial u}{\partial y} = -~ \frac{2y \sqrt{x} + y\sqrt{y} -x \sqrt{y}}{ 2~\sin(u)~ (\sqrt{x} + \sqrt{y})^2}} $$
$$x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y} = - \frac{1}{2\sin(u)}~ \frac{\sqrt{y} (~ 2x+y-x)+\sqrt{x}(~x-y+2y)}{(\sqrt{x}+\sqrt{y})^2}$$
$$x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y} = - \frac{1}{2\sin(u)}~ \frac{ x+y}{(\sqrt{x}+\sqrt{y})} \tag{3}$$
From the equation : $$ u = \arccos( \frac{x+y}{(\sqrt{x}+\sqrt{y})})$$
$$ \cos(u) = \frac{x+y}{\sqrt{x}+\sqrt{y}}$$
replace in $(3)$ :
$$  \tag{4}\boxed{- \frac{\cos(u)}{2\sin(u)} = - \frac{\cot(u)}{2}}$$
Do confirm yourself the answer and partial derivatives
