# implicit differentiation

• Sep 12th 2009, 01:50 PM
yoman360
implicit differentiation
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?
• Sep 12th 2009, 02:25 PM
skeeter
Quote:

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 $\frac{dy}{dx}$ ?
• Sep 12th 2009, 02:30 PM
yoman360
I get $\frac{dy}{dx}=\frac{x^2}{y}$

i keep getting the same answer

this one is difficult
• Sep 12th 2009, 04:08 PM
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}
$
• Sep 12th 2009, 04:15 PM
yoman360
Quote:

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}
$

oh i see now I forgot to use the quotient rule