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Quick question on how to handle negative fraction exponents when differentiating:

I have this problem to differentiate.

$$x^{2/3} + y^{2/3} = 1$$

So my textbook and I both did the first thing the same way:

$$\frac23x^{-1/3} + \frac23y^{-1/3} \frac{dy}{dx} = 0$$

But here is where I got a bit lost.. I simplified the negative exponent and came up with:

$$\frac{dy}{dx} = -\frac{2}{3x^3} \bigg/ \frac{3}{3y^3} \implies -\frac{y^3}{x^3}$$

My textbook gives this instead:

$$-\left(\frac{y}{x}\right)^{1/3}$$

Can anyone explain how that happened? Unfortunately I'm taking this college subject online and the college forum where we communicate with our teachers isn't very active.

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    $\begingroup$ $\displaystyle y^{-1/3} = {1\over y^{1/3}}$, not $\displaystyle{1\over y^3}$. $\endgroup$ – Christopher Carl Heckman Apr 26 '16 at 7:05
  • $\begingroup$ Your mistake occurs when you claim $a^{-1/3} = 1/a^3$ instead of $1/a^{1/3} $. $\endgroup$ – fleablood Apr 26 '16 at 7:31
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$x^{-1/3} = \frac{1}{x^{1/3}}$, not $\frac{1}{x^3}$.

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    $\begingroup$ Is there an echo in here? 8-) $\endgroup$ – Christopher Carl Heckman Apr 26 '16 at 7:07
  • $\begingroup$ @ChristopherCarlHeckman No, you beat me by 30s but I was scrolled too far down to see the comment. $\endgroup$ – lastresort Apr 26 '16 at 7:08
  • $\begingroup$ ... Ahh... So careless! Thanks :) $\endgroup$ – momo Apr 26 '16 at 7:10

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