Q. 29 - Calculus

Question

Use (a) the $h\to 0$ definition of the derivative and then

(b) the $z\to c$ definition of the derivative to find $f\text{'}\left(c\right)$ for each function f and value $x=c$ in Exercises 23–38.

29. $f\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}, x=9$

Step-by-Step Solution

Answer

$f\text{'}\left(9\right)=\frac{1}{6}$.

Part (a) Step 1. Given information.

A function is given as $f\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ and $x=c=9$.

Part (a) Step 2. Solve f ' ( 9 ) using h → 0 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{h\to 0}{\lim }\frac{f(c+h)-f\left(c\right)}{h} \\ f\text{'}\left(9\right)=\underset{h\to 0}{\lim }\frac{f(9+h)-f\left(9\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{(9+h)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-9^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{9+h}-\sqrt{9}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{9+h}-\sqrt{9}}{h}\times \frac{\sqrt{9+h}+\sqrt{9}}{\sqrt{9+h}+\sqrt{9}} \\ =\underset{h\to 0}{\lim }\frac{9+h-9}{h(\sqrt{9+h}+\sqrt{9})} \\ =\underset{h\to 0}{\lim }\frac{h}{h(\sqrt{9+h}+\sqrt{9})} \\ =\underset{h\to 0}{\lim }\frac{1}{\sqrt{9+h}+\sqrt{9}} \\ =\frac{1}{\sqrt{9+0}+3} \\ =\frac{1}{3+3} \\ =\frac{1}{6} \end{gathered}$$

Part (b) Step 1. Solve f ' ( 9 ) using z → 9 definition of the derivative.

We have

$$\begin{gathered} f\text{'}\left(c\right)=\underset{z\to c}{\lim }\frac{f\left(z\right)-f\left(c\right)}{z-c} \\ f\text{'}\left(9\right)=\underset{z\to 9}{\lim }\frac{f\left(z\right)-f\left(9\right)}{z-9} \\ =\underset{z\to 9}{\lim }\frac{\sqrt{z}-\sqrt{9}}{z-9} \\ =\underset{z\to 9}{\lim }\frac{\sqrt{z}-\sqrt{9}}{{\left(\sqrt{z}\right)}^2-{\left(\sqrt{9}\right)}^2} \\ =\underset{z\to 9}{\lim }\frac{\sqrt{z}-\sqrt{9}}{(\sqrt{z}-\sqrt{9})(\sqrt{z}+\sqrt{9})} \\ =\underset{z\to 9}{\lim }\frac{1}{\sqrt{z}+\sqrt{9}} \\ =\underset{z\to 9}{\lim }\frac{1}{\sqrt{z}+3} \\ =\frac{1}{\sqrt{9}+3} \\ =\frac{1}{3+3} \\ =\frac{1}{6} \end{gathered}$$