Chapter 4.2, Problem 43E

To determine

To State: True result for $f\text{'}\text{'}(x)=x^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ .

Answer to Problem 43E

(d) $f\text{'}(x)=\frac{3}{2}{x}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6$ .

Explanation of Solution

Given information: Given $f\text{'}\text{'}(x)=x^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ .

Verification:

  $f\text{'}(x)=\frac{3}{2}{x}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6+$

Differentiate with respect to x,

  $\begin{array}{l}f\text{'}\text{'}(x)=\frac{3}{2}\times \frac{d(x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.})}{dx}+\frac{d(6)}{dx}\\ f\text{'}\text{'}(x)=\frac{3}{2}\times \frac{2}{3}{x}^{\frac{-1}{3}}+0\\ f\text{'}\text{'}(x)=x^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}$

Hence, option (d) is true.