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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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    To get multicharacter exponents (or any time you want multiple characters in one space) enclose them in braces. So x^{(2/3)} gives $x^{(2/3)}$2012-10-03
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    It's a cube root in the denominator.2012-10-03
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    ups... changing2012-10-03
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    Look very, very carefully at $\sqrt{(-1)^\frac{2}{3}+1}$. Work it out step by step. It is not zero.2012-10-03
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    weird... i could have sworn my calculater said -1 ! now it doesn't anymore. maybe it's just too late >.< sorry !2012-10-03
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    people edit x^{2/3} to x^(2/3) and back... 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 $x=-1$.