implicit differentiation

**yoman360** · September 12th 2009, 12:50 PM

Find $\frac{d^2y}{dx^2}$ by implicit differentiation: $2x^3-3y^2=8$

So far I get
$\frac{dy}{dx}=\frac{x^2}{y}$
$\frac{d^2y}{dx^2}=\frac{2y}{x}$ <----- incorrect

correct answer= $\frac{2xy^2-x^4}{y^3}$

How do i find the correct answer?

**skeeter** · September 12th 2009, 01:25 PM

Originally Posted by yoman360

Find $\frac{d^2y}{dx^2}$ by implicit differentiation: $2x^3-3y^2=8$

show some effort ...

what did you get for $\frac{dy}{dx}$ ?

**yoman360** · September 12th 2009, 01:30 PM

I get $\frac{dy}{dx}=\frac{x^2}{y}$

i keep getting the same answer

this one is difficult

**skeeter** · September 12th 2009, 03:08 PM

$2x^3-3y^2=8$

$6x^2 - 6y \cdot y' = 0$

$y' = \frac{x^2}{y}$

$y'' = \frac{2xy - x^2y'}{y^2}$

$y'' = \frac{2xy - \frac{x^4}{y}}{y^2}$

$y'' = \frac{2xy^2 - x^4}{y^3}<br />$

**yoman360** · September 12th 2009, 03:15 PM

Originally Posted by skeeter

$2x^3-3y^2=8$

$6x^2 - 6y \cdot y' = 0$

$y' = \frac{x^2}{y}$

$y'' = \frac{2xy - x^2y'}{y^2}$

$y'' = \frac{2xy - \frac{x^4}{y}}{y^2}$

$y'' = \frac{2xy^2 - x^4}{y^3}<br />$

oh i see now I forgot to use the quotient rule

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