Derivative and 2nd derivative of x^3+y^3=1?

[pr1nny]

For the first derivative I got -x/y, is that correct? According to the book the second derivative is -2x/y^5. How does this happen? When I take the derivative of -x/y, I get (y[-1]-[y${\displaystyle ]-x\frac{)}{y^2}\left(quotientr\underline{e}\right).To\ge tthederivativeofthe\text{or}ig\in alequationIdha\in r\underline{e}\text{and}got3x^2+(y}$)3y^2 =>y`= -3x^2/-3y^2 => -x/y

 

[pr1nny]

Ok I think I just figured out that you can't reduce x^2/y^2 to x/y. I am still confused though, I tried it from there.

 

[nikkor180]

Differentiating the given function implicitly gives 3x^2 + 3y^2y' = 0 and, therefore, y' = -x^2 / y^2. Differentiating y' gives
y'' = [-2xy^2 + 2x^2yy'] / y^3 which simplifies to -2x/y^5 [x^3 + y^3]. But we are given x^3 + y^3 = 1 and hence y'' = -2x/y^5.

Regards,

Rich B.

 

[BigGlenntheHeavy]

$\displaystyle First \ one, \ assuming \ we \ want \ y' \ = \ \frac{dy}{dx} \ of \ x^3+y^3 \ = \ 1.$

$\displaystyle Implicit \ differentation: \ 3x^2+3y^2y' \ = \ 0, \ \implies \ y' \ = \ \frac{-x^2}{y^2}$

$\displaystyle Explicit \ differentation: \ y \ = \ (1-x^3)^{1/3}, \ y' \ = \ \frac{-x^2}{(1-x^3)^{2/3}}$

$\displaystyle I'll \ leave \ it \ up \ to \ you \ to \ show \ that \ y^2 \ = \ (1-x^3)^{2/3}.$

 

[lookagain]

y' = -x^2 / y^2. Differentiating y' gives

$\displaystyle >>$ y'' = [-2xy^2 + 2x^2yy'] / y^3 $\displaystyle <<$

which simplifies to -2x/y^5 [x^3 + y^3].

But we are given x^3 + y^3 = 1 and hence y'' = -2x/y^5.

nikkor180,

the highlighted expression needs a different exponent on the y variable in the denominator:

y'' = [-2xy^2 + 2(x^2)yy']/(y^2)^2 =

[-2xy^2 + 2(x^2)yy']/y^4 =

$\displaystyle \frac{-2xy^2 + 2x^2yy'}{y^4} =$

$\displaystyle \frac{y(-2xy^2 + 2x^2yy')}{y(y^4)} =$

$\displaystyle \frac{-2xy^3 + 2x^2y^2y'}{y^5} =$

Substitute in the y' expression:

$\displaystyle \frac{-2xy^3 + 2x^2y^2(\frac{-x^2}{y^2})}{y^5}=$

$\displaystyle \frac{-2xy^3 - 2x^4}{y^5} =$

$\displaystyle \frac{-2x}{y^5}(y^3 + x^3) =$

$\displaystyle Note: \ \ x^3 + y^3 = 1 \ (or \ y^3 + x^3 = 1) \ from \ the \ given.$

$\displaystyle \frac{-2x}{y^5}(1) =$

$\displaystyle \frac{-2x}{y^5}$