Question

1. - / 5.55 points differentiate. $g(t) = \frac{t - \sqrt{t}}{t^{1/3}}$ $g(t) = \square$

Explanation:

Step1: Simplify the function

Rewrite terms with exponents:
$g(t) = \frac{t - t^{1/2}}{t^{1/3}} = t^{1 - 1/3} - t^{1/2 - 1/3} = t^{2/3} - t^{1/6}$

Step2: Differentiate term by term

Use power rule $\frac{d}{dt}(t^n)=nt^{n-1}$:
$g'(t) = \frac{2}{3}t^{2/3 - 1} - \frac{1}{6}t^{1/6 - 1} = \frac{2}{3}t^{-1/3} - \frac{1}{6}t^{-5/6}$

Answer:

$\frac{2}{3}t^{-1/3} - \frac{1}{6}t^{-5/6}$ (or equivalent forms like $\frac{2}{3\sqrt[3]{t}} - \frac{1}{6\sqrt[6]{t^5}}$)