find derivative of $\sqrt{x}+\sqrt{y}=\sqrt{a}$ I have an Implicit Function $\sqrt{x}+\sqrt{y}=\sqrt{a}$
the graph of the function is

I need to prove that $p+q=a$

and I need to find $\frac{d}{dx}$ to find the the slop to prove that.

result:
after rearranged the function I got 
$a=x+2\sqrt{xy}+y$
and the derivative of that function is 

$\frac{d}{dx}=\frac{-\sqrt{y}}{\sqrt{x}}$

how can I prove $p+q=a$ with $\frac{d}{dx}$?
 A: Let $y'$ denote $\displaystyle\frac{dy}{dx}.$
Given: $\sqrt{x} + \sqrt{y} = \sqrt{a}.$
Using implicit differentiation, you have that
$\displaystyle \frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0 \implies y' = \frac{-\sqrt{y}}{x} = 
\frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$

The distance $Q$ may be represented by the point $(q,0)$.
Then, the slope going from $(x,y)$ to $(q,0)$ is $y'$.
Therefore,
$\displaystyle \frac{-y}{q - x} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}
\implies
\frac{-\left(\sqrt{a} - \sqrt{x}\right)^2}{q - x} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$
Dividing both sides by
$\displaystyle -\left(\sqrt{a} - \sqrt{x}\right)$
gives
$\displaystyle \frac{\sqrt{a} - \sqrt{x}}{q - x} = \frac{1}{\sqrt{x}} \implies
q - x = \sqrt{ax} - x \implies q = \sqrt{ax}.$

The distance $P$ may be represented by the point $(0,p)$.
Then, the slope going from $(0,p)$ to $(x,y)$ is $y'$.
Therefore,
$\displaystyle \frac{y - p}{x - 0} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}
\implies
\frac{\left(\sqrt{a} - \sqrt{x}\right)^2 - p}{x} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$
This implies that
$\displaystyle \frac{\left(\sqrt{a} - \sqrt{x}\right)^2 - p}{\sqrt{x}} 
= \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{1}.$
This implies that
$\displaystyle a + x - 2\sqrt{ax} - p = -\sqrt{ax} + x \implies
a - \sqrt{ax} = p.$

Thus, $\displaystyle (p + q) = \left(a - \sqrt{ax}\right) + \sqrt{ax} = a.$
