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While trying to find the tangent line to

$y=(1+x^\frac{2}{3})^\frac{3}{2}$

at $x=-1$, I determined the derivative,

$\frac{dy}{dx}=\frac{\sqrt{x^\frac{2}{3}+1}}{\sqrt[3]{x}},$

to get the slope of the tangent. Now that doesn't really help much, since $\sqrt{(-1)^{2/3}+1} = 0$.

I know the result is supposed to be $y=2^{3/2}-\sqrt{2}(x+1)$, though I have no idea why.

Thanks for your help!

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    people edit x^{2/3} to x^(2/3) and bac$k$... although it doesn't seem to make any difference at all. wtf is wrong with that editing thing !? do they get points for it or why keep ppl editing unimportant things in my posts?2012-10-04

1 Answers 1

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Obviously, $\frac{dy}{dx}=\frac{\sqrt{x^\frac{2}{3}+1}}{\sqrt[3]{x}} = \frac{-\sqrt{2}}{1}=-\sqrt{2}$., which is the slope at $x=-1$.