# Thread: Urgent second differential problem

1. ## Urgent second differential problem

(x3) + (y3) = 1

2. ## Re: Urgent second differential problem

Originally Posted by righteous818

(x3) + (y3) = 1
Try using implicit differentiation.

3. ## Re: Urgent second differential problem

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

4. ## Re: Urgent second differential problem

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

5. ## Re: Urgent second differential problem

how do i get fractions

6. ## 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.

7. ## 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

8. ## Re: Urgent second differential problem

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 \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*}