1. ## Implicit Differentiation

Given
$x^2 + 25y^2 = 100$
show that $\frac{d^2y}{dx^2} = - \frac{4}{25y^3}$

Thankies

2. Differentiate and solve for y':

$
x^2 + 25y^2 = 100 \Rightarrow 2x + 50yy' = 0 \Leftrightarrow y' = - \frac{x}{{25y}}
$

Differentiate again and substitute y' with the previous expression:

$
y'' = \frac{{xy' - y}}{{25y^2 }} = \frac{{x\left( { - \frac{x}{{25y}}} \right) - y}}{{25y^2 }} = \frac{{ - \frac{{x^2 + 25y^2 }}{{25y}}}}{{25y^2 }}
$

Now you can use the initial relation again in the numerator:

$
y'' = \frac{{ - \frac{{x^2 + 25y^2 }}{{25y}}}}{{25y^2 }} = \frac{{ - \frac{{100}}{{25y}}}}{{25y^2 }} = \frac{{ - \frac{4}{y}}}{{25y^2 }} = \boxed{- \frac{4}{{25y^3 }}}
$

Given: .[1] $x^2 + 25y^2 = 100,$ .show that: . $\frac{d^2y}{dx^2} = - \frac{4}{25y^3}$

Differentiate: . $2x + 50y\left(\frac{dy}{dx}\right) \:=\:0\quad\Rightarrow\quad\frac{dy}{dx}\:=\:-\frac{1}{25}\,\frac{x}{y}$ [2]

Differentiate again: . $\frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{y - x(\frac{dy}{dx})}{y^2}\right]$

Substitute [2]: . $\frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{y - x(-\frac{x}{25y})}{y^2}\right] \:=\:-\frac{1}{25}\left[\frac{y + \frac{x^2}{25y}}{y^2}\right]$

Multiply top and bottom by $25y\!:\;\;\frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{25y^2 + x^2}{25y^3}\right]$

From [1], the numerator equals 100: . $\frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{100}{25y^3}\right]$

Therefore: . $\boxed{\frac{d^2y}{dx^2} \:=\:-\frac{4}{25y^3}}$