if the derivative of y=1+3x^2/3 at x=-8 is y'= 2x^ -1/3
How do we get the x-value appropriate rule
2(-8)^-1/3
how do i simply from there please provide steps and rules thanks.

Question

if the derivative of y=1+3x^2/3 at x=-8  is   y'= 2x^ -1/3
How do we get the x-value appropriate rule 
2(-8)^-1/3
how do i simply from there please provide steps and rules thanks.

anonymous · Answer

are you asking how you would evaluate 2(-8)^-1/3

anonymous · Answer

Yes on my book it shows answer to be -1. I am lost as to how was that simplied.

anonymous · Answer

the 3rd root of 8 is 2

aum · Answer

$$ \large
2(-8)^{-1/3} = \frac{2}{\sqrt[3]{-8}} = ?
$$
What is the cube root of -8?

aum · Answer

-8 = (-2)^3
So cube root of -8 is -2.
2 / (-2) = -1.

anonymous · Answer

aum thanks your so awesome. All easy steps and i just cant figure this algebra out. THANKS AGAIN!

aum · Answer

You are welcome.

aum · Answer

$$
\large  2(-8)^{-1/3} = \frac{2}{(-8)^{1/3}} = \frac{2}{\{(-2)^3\}^{1/3}}  = 
\frac{2}{(-2)^{3*1/3}}  = \ \large  \frac{2}{(-2)^{1}}  = \frac{2}{-2}  = -1
$$