S8.2.6.3 differenciate by separation and power rule

[karush]

$\tiny{s8.2.6.3}$
Find y' $\sqrt{x}+\sqrt{y}=1$
\begin{array}{lll}
\textit{isolate }y
&\sqrt{y}=1-\sqrt{x}
&(1)\\ \\
\textit{square both sides}
&y=(1-\sqrt{x})^2
&(2)\\ \\
\textit{differentiate both sides}
&y'=2\left(1-\sqrt{x}\right)\left(-\dfrac{1}{2\sqrt{x}}\right)&
(3)\\ \\
\textit{simplify}
&y'=-\dfrac{1-\sqrt{x}}{\sqrt{x}}
&(4)
\end{array}

well we could rationalize the denominator but why?
hopefully correct

 

[skeeter]

implicitly ...

$\dfrac{1}{2\sqrt{x}} + \dfrac{y’}{2\sqrt{y}} = 0$

$y’ = -\dfrac{\sqrt{y}}{\sqrt{x}} = \dfrac{\sqrt{x}-1}{\sqrt{x}}$

 

[karush]

implicitly ...

$\dfrac{1}{2\sqrt{x}} + \dfrac{y’}{2\sqrt{y}} = 0$

$y’ = -\dfrac{\sqrt{y}}{\sqrt{x}} = \dfrac{\sqrt{x}-1}{\sqrt{x}}$

that looks like magic wand stuff

mahalo tho,,,