Question

Use the definition of the derivative to find $f\text{'}$for each function $f$in Exercises 39-54

$f\left(x\right)=\frac{1}{\sqrt{2x+1}}$

Short answer

The value of$f\text{'}\left(x\right)=-{(2x+1)}^{-\frac{3}{2}}$

Step-by-step solution

Step 1. Given information

The given function$f\left(x\right)=\frac{1}{\sqrt{2x+1}}$

Step 2. Finding the value of f'(x)

We know that $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

Given $f\left(x\right)=\frac{1}{\sqrt{2x+1}}$

Then role="math" localid="1648288671514" $f(x+h)=\frac{1}{\sqrt{2a+2h+1}}$

$$\begin{gathered} f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{\sqrt{2x+2h+1}}}-{\displaystyle \frac{1}{\sqrt{2x+1}}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2x+1}-\sqrt{2x+2h+1}}{h\sqrt{2x+2h+1}\sqrt{2x+1}} \\ =\underset{h\to 0}{\lim }\frac{(\sqrt{2x+1}-\sqrt{2x+2h+1})(\sqrt{2x+1}+\sqrt{2x+2h+1)}}{h\sqrt{2x+2h+1}\sqrt{2a+1}(\sqrt{2x+1}+\sqrt{2x+2h+1})} \\ =\underset{h\to 0}{\lim }\frac{(2x+1)-(2x+2h+1)}{h\sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1}+\sqrt{2x+2h+1})} \\ =\underset{h\to 0}{\lim }\frac{-2h}{h\sqrt{2x+2h+1}\sqrt{2x+1}(\sqrt{2x+1}+\sqrt{2x+2h+1})} \\ Puttingh=0 \\ =\frac{-2}{\sqrt{2x+1}\sqrt{2x+1}(\sqrt{2x+1}+\sqrt{2x+1})} \\ =\frac{-1}{{(2x+1)}^{{\displaystyle \frac{3}{2}}}} \\ =-{(2x+1)}^{-\frac{3}{2}} \end{gathered}$$

Hence, the value of$f\text{'}\left(x\right)=-{(2x+1)}^{-\frac{3}{2}}$