# tangent line to $y=(1+x^{2/3})^{3/2}$ at $x = -1$

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.

-
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
show 1 more comment

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 $x=-1$.

-