# implicit differentiation 2nd derivative

• Oct 28th 2009, 10:11 AM
haebinpark
implicit differentiation 2nd derivative
i have to find y' and y" using implicit differentiation
2x^3 - y^2 = 8

what i did was

d2x^3/dx - dy^2/dx = d8/dx

d2x^3/dx = 6x^2
dy^2/dx = 2y*y'
d8/dx = 0

6x^2 - 2yy' = 0
-6x^2=-2yy'
y'=6x^2/2y
= 3x^2 / y

but i don't know how to find 2nd derivative from here

how do i do it?
• Oct 28th 2009, 12:12 PM
Quote:

Originally Posted by haebinpark
i have to find y' and y" using implicit differentiation
2x^3 - y^2 = 8

what i did was

d2x^3/dx - dy^2/dx = d8/dx

d2x^3/dx = 6x^2
dy^2/dx = 2y*y'
d8/dx = 0

6x^2 - 2yy' = 0
-6x^2=-2yy'
y'=6x^2/2y
= 3x^2 / y

but i don't know how to find 2nd derivative from here

how do i do it?

This is correct so far, all you have to do is use the quotient rule and implicit differentiation.

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

$\displaystyle y''=\frac{6yx-3y'x^2}{y^2}$
• Oct 28th 2009, 03:55 PM
haebinpark
would you show me a little detail please?
like from y' to y"
how the implicit differentiation is applied
• Oct 28th 2009, 04:03 PM
Quote:

Originally Posted by haebinpark
would you show me a little detail please?
like from y' to y"
how the implicit differentiation is applied

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

Apply the quotient rule:

$\displaystyle y''=\frac{y(3x^2)'-(3x^2)y'}{y^2}$

$\displaystyle =\frac{6yx-3y'x^2}{y^2}$