Q. 30 - Calculus

Question

Use (a) the h→0 definition of the derivative and then (b) the z→c definition of the derivative to find f'(c) for each function f and value x = c. 

$f\left(x\right)=x^{-1/2},x=9$

Step-by-Step Solution

Answer

(a) $f^{\text{'}}\left(c\right)=-\frac{1}{54}$

(b) $f^{\text{'}}\left(c\right)=-\frac{1}{54}$

Part (a) Step 1. Given information.

Given function is $f\left(x\right)=x^{-1/2}$

We have to find $f^{\text{'}}\left(c\right)$ at $x=9$

Part (a) Step 2. Find the f ' ( c )

We have to find the derivative of the function using h→0 definition,

Therefore,

$\begin{array}{r}\underset{h\to 0}{lim} \frac{f(9+h)-f\left(9\right)}{h}=\underset{h\to 0}{lim} \frac{(9+h{)}^{-\frac{1}{2}}-9^{-\frac{1}{2}}}{h}\\ =\underset{h\to 0}{lim} \frac{9^{-\frac{1}{2}}{\left(1+\frac{h}{9}\right)}^{-\frac{1}{2}}-\frac{1}{3}}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{1}{3}\left(1-\frac{1}{2}\cdot \frac{h}{9}+\frac{3}{8}{h}^2+\cdots \right)-\frac{1}{3}}{h}\\ =\underset{h\to 0}{lim} \frac{-\frac{h}{54}+\frac{1}{8}{h}^2+\cdots }{h}\\ =\underset{h\to 0}{lim} -\frac{1}{54}+\frac{1}{8}h+\cdots \\ =-\frac{1}{54}\end{array}$

Part (b) Step 1. Find f ' ( c )

Find the derivate of the function using $x\to 9$ definition,

$\begin{array}{r}\underset{x\to 9}{lim} \frac{f\left(x\right)-f\left(9\right)}{x-9}=\underset{x\to 9}{lim} \frac{x^{-\frac{1}{2}}-9^{-\frac{1}{2}}}{x-9}\\ =\underset{x\to 9}{lim} \frac{\frac{1}{x^{\frac{1}{2}}}-\frac{1}{3}}{{\left(x^{\frac{1}{2}}\right)}^2-3^2}\\ =\underset{x\to 9}{lim} \frac{-\left(x^{\frac{1}{2}}-3\right)}{\left(x^{\frac{1}{2}}-3\right)\left(x^{\frac{1}{2}}+3\right)}\end{array}$

$\begin{array}{r}=\underset{x\to 9}{lim} \frac{-\left(x^{\frac{1}{2}}-3\right)}{3x^{\frac{1}{2}}\left(x^{\frac{1}{2}}-3\right)\left(x^{\frac{1}{2}}+3\right)}\\ =\underset{x\to 9}{lim} \frac{-1}{3x^{\frac{1}{2}}\left(x^{\frac{1}{2}}+3\right)}\\ =\frac{-1}{3(9{)}^{\frac{1}{2}}\left((9{)}^{\frac{1}{2}}+3\right)}\\ =-\frac{1}{54}\end{array}$