Yes, so, letting and , we have so that .
Fortunately there is a similar, though a bit more complicated rule for third (and one-third) powers: so, letting and ,
so that .
Let f(x)= x1/3
If a does not = 0, use the eqaution f`(a)= [f(x)-f(a)]/(x-a) to find f`(a).
f`(a) = [x1/3-a1/3]/(x-a)
If the variables in the numerator were square roots rather than cube roots this would be much easier to solve. I'm not sure how to simplify this to find the derivative function. The answer in the back of the book is (1/3)a-2/3, but I can't get anywhere near this. Not sure how to even begin to break down the eqaution.