# Urgent second differential problem

• Jun 17th 2012, 04:57 PM
righteous818
Urgent second differential problem

(x3) + (y3) = 1
• Jun 17th 2012, 05:03 PM
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Re: Urgent second differential problem
Quote:

Originally Posted by righteous818

(x3) + (y3) = 1

Try using implicit differentiation.
• Jun 17th 2012, 05:06 PM
righteous818
Re: Urgent second differential problem
i did but i am not the gettin the answer the book said the answer is -2x/(y5)
• Jun 17th 2012, 05:07 PM
Prove It
Re: Urgent second differential problem
Quote:

Originally Posted by righteous818
i did but i am not the gettin the answer the book said the answer is -2x/(y5)

• Jun 17th 2012, 05:07 PM
righteous818
Re: Urgent second differential problem
how do i get fractions
• Jun 17th 2012, 05:11 PM
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Re: Urgent second differential problem
If you are unsure how to use LaTeX, just use a / for the vinculum and use brackets where they are necessary. Otherwise inside tex tags, use the command \frac{}{}, with the numerator in the first set of brackets and the denominator in the second.
• Jun 17th 2012, 05:15 PM
righteous818
Re: Urgent second differential problem
well for te first differential i got

3x2 + 3y2(dy/dx)=0

dy/dx = (-x2 )/(y2)

Then seoncd differntial using quotient rule

[(y2)(-2x)-(-x2 )(2y(dy/dx))]/(y4)

then finally the second differntial came down to for me

[-2xy3 + 2x4 ]/y5

• Jun 17th 2012, 05:27 PM
Prove It
Re: Urgent second differential problem
Quote:

Originally Posted by righteous818
well for te first differential i got

3x2 + 3y2(dy/dx)=0

dy/dx = (-x2 )/(y2)

Then seoncd differntial using quotient rule

[(y2)(-2x)-(-x2 )(2y(dy/dx))]/(y4)

then finally the second differntial came down to for me

[-2xy3 + 2x4 ]/y5

\displaystyle \displaystyle \begin{align*} \frac{d^2y}{dx^2} &= \frac{-2xy^2 + 2x^2y\,\frac{dy}{dx}}{y^4} \\ &= \frac{-2xy + 2x^2\frac{dy}{dx}}{y^3} \\ &= \frac{-2xy + 2x^2\left(-\frac{x^2}{y^2}\right)}{y^3} \\ &= \frac{-2xy - \frac{2x^4}{y^2}}{y^3} \\ &= \frac{-2xy^3 - 2x^4}{y^5} \end{align*}