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While trying to find the tangent line to


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


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 $x^{(2/3)}$ – Ross Millikan Oct 3 '12 at 17:02
It's a cube root in the denominator. – EuYu Oct 3 '12 at 17:04
ups... changing – foaly Oct 3 '12 at 17:05
Look very, very carefully at $\sqrt{(-1)^\frac{2}{3}+1}$. Work it out step by step. It is not zero. – EuYu Oct 3 '12 at 17:07
weird... i could have sworn my calculater said -1 ! now it doesn't anymore. maybe it's just too late >.< sorry ! – foaly Oct 3 '12 at 17:11
up vote 0 down vote accepted

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 $x=-1$.

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