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. $\frac{-1}{2}\cos(u)$
 A: 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
A: Write $\cos(u)=\frac{x+y}{\sqrt{x}+\sqrt{y}}$ and differentiate implicitly wrt $x$ and $y$. The RHS is symmetric in $x$ and $y$ so you only need to actually compute the derivative once. This gives
$$-\sin(u)\frac{\partial u}{\partial x}=\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{1}{2\sqrt{x}}\frac{x+y}{(\sqrt{x}+\sqrt{y})^2},$$
$$-\sin(u)\frac{\partial u}{\partial y}=\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{1}{2\sqrt{y}}\frac{x+y}{(\sqrt{x}+\sqrt{y})^2}.$$
Multiply the first equation by $x$ and the second by $y$ and add the equations to find
\begin{align}
-\sin(u)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)&=\frac{x+y}{\sqrt{x}+\sqrt{y}}-\frac{x+y}{\sqrt{x}+\sqrt{y}}\left(\frac{1}{2}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{1}{2}\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right)\\
&=\frac{1}{2}\cos(u)
\end{align}
where the last equality follows from the original equation. Divide by $-\sin(u)$ and you're done.
A: There's a shortcut if you know that $x \, \frac{\partial u}{\partial x} + y \, \frac{\partial u}{\partial y} = r \, \frac{\partial u}{\partial r}$ in terms of polar coordinates $(r,\varphi)$. Then you can write
$$
\cos(u)
= \frac{x+y}{\sqrt{x}+\sqrt{y}}
= \frac{r \, (\cos\varphi+\sin\varphi)}{\sqrt{r} \, (\sqrt{\cos\varphi}+\sqrt{\sin\varphi})}
= \sqrt{r} \, g(\varphi)
,
$$
differentiate with respect to $r$ to find
$$
-\sin(u) \, \frac{\partial u}{\partial r} = \frac{1}{2\sqrt{r}} \, g(\varphi)
,
$$
and deduce that
$$
x \, \frac{\partial u}{\partial x} + y \, \frac{\partial u}{\partial y} = r \, \frac{\partial u}{\partial r}
= r \, \frac{-1}{\sin(u)} \, \frac{1}{2\sqrt{r}} \, g(\varphi)
= \frac{-1}{2 \sin(u)} \, \sqrt{r} \, g(\varphi)
= \frac{-1}{2 \sin(u)} \, \cos(u)
.
$$
(As a bonus, you see that this works for any function of the form $u=\arccos\bigl( \sqrt{r} \, g(\varphi) \bigr)$ in polar coordinates, so there's nothing special about the particular function $g(\varphi) = \frac{\cos\varphi+\sin\varphi}{\sqrt{\cos\varphi}+\sqrt{\sin\varphi}}$ that we have in this case. See also Euler's theorem about homogeneous functions.)
