Question

I'm having an issue simplifying and getting the "right" answer that's in my book on this one, and am not sure what I should do differently....
The question is:

(x^3 + y^3) = 1; Find y'' by Implicit Differentiation.

Any and all help is greatly appreciated!

Answer 1

How I went about solving this one:
\[x^3+y^3 = 1\]
\[3x^2+3y^2dy/dx = (0)\]
\[3x^2=-3y^2dy/dx\]
\[dy/dx=\frac{ 3x^2 }{-3y^2}\]
\[-x^2/y^2\]

Answer 2

Then, I did the following to find y'':
\[y' =-x^2/y^2\]
\[= -x^2y^{-2}\]
\[y'' = (-2x)y^{-2}+(-x^2)(-2ydy/dx)\]

Answer 3

Looks good!
maybe plugin the value of \(dy/dx\) and simplify

Answer 4

don't be scared of the quotient rule here either
might make it nicer to see what you really have

Answer 5

*Substituting in "(-x^2y^-2)" for dy/dx...
\[= -2xy^{-2}+(2x^2y)(-x^2y^{-2})\]
\[= -2xy^{-2}+(-2x^4y^{-1})\]
\[= -2xy^{-1}(y^3-x^3)\]

Answer 6

\[-\frac{y^2-2yx^2y'}{y^4}\]

Answer 7

oops that was wrong

Answer 8

And that's all the farther I know how to go, but my book has:
\[\frac{-2x}{y^5}\]

Answer 9

\[-\frac{2xy^2-2yx^2y'}{y^4}\]

Answer 10

where you see \(y'\) put \(-\frac{x^2}{y^2}\) then simplify the compound fraction

Answer 11

\[-\frac{2xy^2-2yx^2y'}{y^4}\\
-\frac{2xy^2+2yx^2\frac{x^2}{y^2}}{y^4}\]

Answer 12

Are you saying that that equals -2x/y^5 ?

Answer 13

I worked through it using the Quotient Rule, but where the heck does the exponent of 5 in the denominator come from?