Question

How do you differentiate `g(t)=(t-sqrtt)/(t^(1/3))`?

Answer 1

I would rewrite to avoid the quotient rule.

Explanation:

`g(t) = (t-sqrtt)/t^(1/3) = (t-t^(1/2))/t^(1/3)`

`= t/t^(1/3) - t^(1/2)/t^(1/3)`

`= t^(1-1/3) - t^(1/2-1/3)`

`= t^(2/3) - t^(1/6)`

`g'(t) = 2/3t^(-1/3) - 1/6 t^(-5/6)`

`= 2/(3root(3)x) - 1/(6 root(6)(x^5))`

If you want a single ratio for the derivative, use `root(3)x = root(3)(x^2)` to get a common denominator.

`= (4x-1)/(6root(6)(x^5))`