## anonymous 4 years ago Find y'' by implicit differentiation: x^3 + y^3 = 1

1. precal

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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. anonymous

can you help me precal please

7. precal
8. precal

Here is the final solution from wolfram alpha.

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

got to that but i'm stuck anyway

13. anonymous

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

14. anonymous

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

17. anonymous

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