Question

Find the following for the given function f:

1. $f\text{'}\left(a\right),$where ais in the domain of f, and
2. $f\text{'}\left(3\right)$and$f\text{'}\left(4\right)$.

.$f\left(x\right)=\sqrt{x-2}$

Short answer

(a) The value of$f\text{'}\left(a\right)=\frac{1}{2\sqrt{x-2}}$

(b) The value of$f\text{'}\left(3\right)=\frac{1}{2}andf\text{'}\left(4\right)=\frac{1}{2\sqrt{2}}$ .

Step-by-step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=\sqrt{x-2}$

Step 2. Formula used.

The derivative of a function f at a number,denoted by$f\text{'}\left(a\right),$is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative at a, we have

$\begin{array}{c}f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}Definitionoff\text{'}\left(a\right)\\ =\underset{h\to 0}{\lim }\frac{\left(\sqrt{a+h-2}\right)-\left(\sqrt{a-2}\right)}{h}\times \frac{\left(\sqrt{a+h-2}\right)+\left(\sqrt{a-2}\right)}{\left(\sqrt{a+h-2}\right)+\left(\sqrt{a-2}\right)}f\left(x\right)=\sqrt{x-2}\\ =\underset{h\to 0}{\lim }\\ =\underset{h\to 0}{\lim }\frac{(a+h-2)-(a-2)}{h(\left(\sqrt{a+h-2}\right)+\left(\sqrt{a-2}\right))}Simplify\\ =\underset{h\to 0}{\lim }\frac{1}{(a+h-2)+(a-2)}Cancelh\\ =\frac{1}{\left(\sqrt{a+0-2}\right)+\left(\sqrt{a-2}\right)}\\ =\frac{1}{2\sqrt{a-2}}\end{array}$

Therefore, the derivative of the function ata is$f\text{'}\left(a\right)=\frac{1}{2\sqrt{a-2}}$ .

Step 4. Finding the value of derivative at given point.

We have$f\text{'}\left(a\right)=\frac{1}{2\sqrt{a-2}}$.

Let’s find the value of derivative at $a=3anda=4$.

Substitute the value of$a=3anda=4$ into the$f\text{'}\left(a\right)=\frac{1}{2\sqrt{a-2}}$

$\begin{array}{c}f\text{'}\left(3\right)=\frac{1}{2\sqrt{3-2}}\\ =\frac{1}{2}\\ f\text{'}\left(4\right)=\frac{1}{2\sqrt{4-2}}\\ =\frac{1}{2\sqrt{2}}\end{array}$

Therefore, the value of .$f\text{'}\left(3\right)=\frac{1}{2}andf\text{'}\left(4\right)=\frac{1}{2\sqrt{2}}$