# What is the second derivative of xy+y^3=1 using implicit differentiation?

• Aug 27th 2012, 09:57 AM
citcat
What is the second derivative of xy+y^3=1 using implicit differentiation?
y'' of xy+y^3=1
Thanks!
• Aug 27th 2012, 10:20 AM
skeeter
Re: What is the second derivative of xy+y^3=1 using implicit differentiation?
Quote:

Originally Posted by citcat
y'' of xy+y^3=1

what did you get for y' ?
• Aug 27th 2012, 10:29 AM
citcat
Re: What is the second derivative of xy+y^3=1 using implicit differentiation?
-(y/x-y)
• Aug 27th 2012, 02:44 PM
Soroban
Re: What is the second derivative of xy+y^3=1 using implicit differentiation?
Hello, citcat!

You're off to a bad start . . .

Do you have the correct answer?

Quote:

$\displaystyle \text{Find }y''\text{ for: }\:xy+y^3\:=\:1$

We have: .$\displaystyle 1\!\cdot\!y + x\!\cdot\!y' + 3y^2\!\cdot\!y' \:=\:0$

. . . . . . . . . . . . $\displaystyle x\!\cdot\!y' + 3y^2\!\cdot\!y' \:=\:-y$

. . . . . . . . . . . . . $\displaystyle (x + 3y^2)y' \:=\:-y$

. . . . . . . . . . . . . . . . . . . $\displaystyle y' \:=\:\frac{\text{-}y}{x+3y^2}$ .[1]

Then: .$\displaystyle y" \;=\;\frac{(x+3y^2)(\text{-}y') - (\text{-}y)(1 + 6yy')}{(x+3y^2)^2}$

. . . . . $\displaystyle y'' \;=\;\frac{-xy' - 3y^2y' + y + 6y^2y'}{(x+3y^2)^2}$

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

Substitute [1]:
. . $\displaystyle y'' \;=\;\frac{y - xy + 3y^2\left(\frac{\text{-}y}{x+3y^2}\right)}{(x+3y^2)^2}$

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

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

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