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

Dec 2, 2016

I would rewrite to avoid the quotient rule.

#### Explanation:

$g \left(t\right) = \frac{t - \sqrt{t}}{t} ^ \left(\frac{1}{3}\right) = \frac{t - {t}^{\frac{1}{2}}}{t} ^ \left(\frac{1}{3}\right)$

$= \frac{t}{t} ^ \left(\frac{1}{3}\right) - {t}^{\frac{1}{2}} / {t}^{\frac{1}{3}}$

$= {t}^{1 - \frac{1}{3}} - {t}^{\frac{1}{2} - \frac{1}{3}}$

$= {t}^{\frac{2}{3}} - {t}^{\frac{1}{6}}$

$g ' \left(t\right) = \frac{2}{3} {t}^{- \frac{1}{3}} - \frac{1}{6} {t}^{- \frac{5}{6}}$

$= \frac{2}{3 \sqrt[3]{x}} - \frac{1}{6 \sqrt[6]{{x}^{5}}}$

If you want a single ratio for the derivative, use $\sqrt[3]{x} = \sqrt[3]{{x}^{2}}$ to get a common denominator.

$= \frac{4 x - 1}{6 \sqrt[6]{{x}^{5}}}$