implicit differentiation 2nd derivative

**haebinpark** · October 28th 2009, 10:11 AM

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?

**adkinsjr** · October 28th 2009, 12:12 PM

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.

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

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

**haebinpark** · October 28th 2009, 03:55 PM

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

**adkinsjr** · October 28th 2009, 04:03 PM

Originally Posted by haebinpark

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

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

Apply the quotient rule:

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

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

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