1. ## Implicit Derivative

a. By differentiating $\displaystyle x^2 - y^2 = 1$ implicitly, show that $\displaystyle \frac{dy}{dx} = \frac{x}{y}$
b. Then show that $\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{y^3}$

a. $\displaystyle x^2 - y^2 = 1$
$\displaystyle 2x - 2yy' = 0$
$\displaystyle 2yy' = 2x$
$\displaystyle y' = \frac{2x}{2y}$
$\displaystyle y' = \frac{x}{y}$

b. $\displaystyle y'' = \frac{(y)(1) - (x)(y')}{y^2}$
$\displaystyle y'' = \frac{y - xy'}{y^2}$
$\displaystyle y'' = \frac{y - x \left(\frac{x}{y}\right)}{y^2}$
$\displaystyle y'' = \frac{\frac{y^2 - x^2}{y}}{y^2}$
$\displaystyle y'' = \frac{y^2 -x^2}{y^3}$

There something wrong with my second derivative?

2. Originally Posted by VitaX
a. By differentiating $\displaystyle x^2 - y^2 = 1$ implicitly, show that $\displaystyle \frac{dy}{dx} = \frac{x}{y}$
b. Then show that $\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{y^3}$

a. $\displaystyle x^2 - y^2 = 1$
$\displaystyle 2x - 2yy' = 0$
$\displaystyle 2yy' = 2x$
$\displaystyle y' = \frac{2x}{2y}$
$\displaystyle y' = \frac{x}{y}$

b. $\displaystyle y'' = \frac{(y)(1) - (x)(y')}{y^2}$
$\displaystyle y'' = \frac{y - xy'}{y^2}$
$\displaystyle y'' = \frac{y - x \left(\frac{x}{y}\right)}{y^2}$
$\displaystyle y'' = \frac{\frac{y^2 - x^2}{y}}{y^2}$
$\displaystyle y'' = \frac{\textcolor{red}{y^2 -x^2}}{y^3}$

There something wrong with my second derivative?

nothing wrong ...

if $\displaystyle \textcolor{red}{x^2 - y^2 = 1}$ , then what is the value of $\displaystyle \textcolor{red}{y^2 - x^2}$ ?
...

3. Eh I did not see that. Thanks

4. I wrote y' = x(1/y) and used the product rule. I get

$\displaystyle y^{\prime\prime} = \frac{1}{y}-\frac{x}{y^2}\cdot y^\prime = \frac{1}{y}-\frac{x^2}{y^3}$

This is $\displaystyle \frac{y^2-x^2}{y^3} = \frac{-(y^2-x^2)}{y^3} = -\frac{1}{y^3}$