# implicit differentiation

• Sep 12th 2009, 12:50 PM
yoman360
implicit differentiation
Find $\displaystyle \frac{d^2y}{dx^2}$ by implicit differentiation: $\displaystyle 2x^3-3y^2=8$

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

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

How do i find the correct answer?
• Sep 12th 2009, 01:25 PM
skeeter
Quote:

Originally Posted by yoman360
Find $\displaystyle \frac{d^2y}{dx^2}$ by implicit differentiation: $\displaystyle 2x^3-3y^2=8$

show some effort ...

what did you get for $\displaystyle \frac{dy}{dx}$ ?
• Sep 12th 2009, 01:30 PM
yoman360
I get $\displaystyle \frac{dy}{dx}=\frac{x^2}{y}$

i keep getting the same answer

this one is difficult
• Sep 12th 2009, 03:08 PM
skeeter
$\displaystyle 2x^3-3y^2=8$

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

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

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

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

$\displaystyle y'' = \frac{2xy^2 - x^4}{y^3}$
• Sep 12th 2009, 03:15 PM
yoman360
Quote:

Originally Posted by skeeter
$\displaystyle 2x^3-3y^2=8$

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

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

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

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

$\displaystyle y'' = \frac{2xy^2 - x^4}{y^3}$

oh i see now I forgot to use the quotient rule