differentiate. g(t) = (t - √t) / t^(1/9) g(t) =

Explanation:

Step1: Rewrite the function

Rewrite $g(t)=\frac{t - \sqrt{t}}{t^{1/9}}=\frac{t - t^{1/2}}{t^{1/9}}=t^{1 - 1/9}-t^{1/2 - 1/9}=t^{8/9}-t^{7/18}$.

Step2: Apply power - rule

The power - rule for differentiation is $(x^n)'=nx^{n - 1}$. For $y = t^{8/9}$, $y'=\frac{8}{9}t^{\frac{8}{9}-1}=\frac{8}{9}t^{-1/9}$. For $y = t^{7/18}$, $y'=\frac{7}{18}t^{\frac{7}{18}-1}=\frac{7}{18}t^{-11/18}$.

Step3: Find $g'(t)$

$g'(t)=\frac{8}{9}t^{-1/9}-\frac{7}{18}t^{-11/18}$.

Answer:

$\frac{8}{9}t^{-1/9}-\frac{7}{18}t^{-11/18}$