Implicit Derivative and Square Root

[Jason76]

Solving for y'

\(\displaystyle f(x) = 2\sqrt{x} + \sqrt{y} = 5\)

\(\displaystyle f(x) = 2 x^{1/2} + y^{1/2} = 5\)

\(\displaystyle \dfrac{1}{2} 2x^{-1/2} + \dfrac{1}{2} y^{-1/2}y' = 0\)

\(\displaystyle x^{-1/2} + \dfrac{1}{2} y^{-1/2}y' = 0\)

\(\displaystyle x^{-1/2} / \dfrac{1}{2} y^{-1/2} + \dfrac{1}{2} y^{-1/2}y'/ \dfrac{1}{2} y^{-1/2} = 0/ \dfrac{1}{2} y^{-1/2}\)

\(\displaystyle x^{-1/2} / \dfrac{1}{2} y^{-1/2} + y' = 0\)

\(\displaystyle y' = - x^{-1/2} / \dfrac{1}{2} y^{-1/2}\)

 

[srmichael]

Solving for y' (sorry for unreadable writing style - problem with library computer)
\(\displaystyle f(x) = 2\sqrt{x} + \sqrt{y} = 5\)
\(\displaystyle f(x) = 2 x^{1/2} + y^{1/2} = 5\)
\(\displaystyle \dfrac{1}{2} 2x^{-1/2} + \dfrac{1}{2} y^{-1/2}y' = 0\)
\(\displaystyle x^{-1/2} + \dfrac{1}{2} y^{-1/2}y' = 0\)
\(\displaystyle x^{-1/2} / \dfrac{1}{2} y^{-1/2} + \dfrac{1}{2} y^{-1/2}y'/ \dfrac{1}{2} y^{-1/2} = 0/ \dfrac{1}{2} y^{-1/2}\) ===> Too complicated. Try easier approach below.
\(\displaystyle x^{-1/2} / \dfrac{1}{2} y^{-1/2} + y' = 0\)
\(\displaystyle y' = - x^{-1/2} / \dfrac{1}{2} y^{-1/2}\)

\(\displaystyle x^{-1/2} + \dfrac{1}{2} y^{-1/2}y' = 0\)

\(\displaystyle \dfrac{1}{2} y^{-1/2}y' = -x^{-1/2}\)

\(\displaystyle y^{-1/2}y' = -2x^{-1/2}\)

\(\displaystyle y' = \dfrac{-2x^{-1/2}}{y^{-1/2}}\)

\(\displaystyle y' = \dfrac{-2y^{1/2}}{x^{1/2}}\)

\(\displaystyle y' =-2\sqrt{\dfrac{y}{x}}\)