Implicit Derivative

**VitaX** · October 16th 2009, 11:57 PM

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

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

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

There something wrong with my second derivative?

**skeeter** · October 17th 2009, 12:13 AM

Originally Posted by VitaX

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

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

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

There something wrong with my second derivative?

nothing wrong ...

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

...

**VitaX** · October 17th 2009, 12:16 AM

Eh I did not see that. Thanks

**harbottle** · October 17th 2009, 12:20 AM

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

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

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

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