Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

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 $x=-1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.