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**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 $\displaystyle \textcolor{red}{y^2 - x^2}$ ?