## calyne 3 years ago Find y'' by implicit differentiation: x^3 + y^3 = 1

1. precal

|dw:1333397126713:dw|

2. precal

|dw:1333397156412:dw|

3. precal

|dw:1333397188197:dw|

4. precal

|dw:1333397226975:dw|

5. precal

ok now take the derivative again. Don't forget to use this statement in again for y'

6. chance1

can you help me precal please

7. precal
8. precal

Here is the final solution from wolfram alpha.

9. satellite73

$x^3 + y^3 = 1$ $3x^2+3y^2y'=0$ $y'=-\frac{x^2}{y^2}$

10. satellite73

$y''=\frac{2xy^2-2x^2yy'}{y^4}$ by the quotient rule

11. satellite73

replace $$y'$$ by $$-\frac{x^2}{y^2}$$ in the above, do some algebra and get the solution

12. calyne

got to that but i'm stuck anyway

13. satellite73

$y''=\frac{2xy^2-2x^2y\times \frac{x^2}{y^2}}{y^4}$

14. satellite73

multiply top and bottom by $$y^2)\] so clear the fraction then cancel some y's 15. satellite73 typo there i lost the minus sign, should be $y''=\frac{2xy^2+2x^2y\times \frac{x^2}{y^2}}{y^4}$ 16. satellite73 $y''=\frac{2xy^4+2x^4y}{y^6}$ $y''=\frac{2xy^3+2x^4}{y^5}$ $y''=\frac{2x(y^3+x^3)}{y^5}$ $y''=\frac{2x}{y^5}$ since \(x^36y^3=1$$

17. satellite73

rather since $$x^3+y^3=1$$