Q. 49 - Calculus

Question

Use the definition of the derivative to find $f\text{'}$ for each function $f$ in Exercises 34-59

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Finding the value of f ' ( x )

Derivate using the Chain rule $\frac{d}{dx}\left[f\right(g\left(x\right)]=f\text{'}(g\left(x\right)).g\text{'}(x)$

Rewrite $\sqrt{2x+1}$ as ${(2x+1)}^{\frac{1}{2}}$

Now $f\text{'}\left(g\right(x\left)\right)=\frac{1}{2}{(2x+1)}^{-\frac{1}{2}}$

and $$\begin{gathered} g\left(x\right)=2x+1 \\ g\text{'}\left(x\right)=2 \end{gathered}$$

hence:  $$\begin{gathered} f\text{'}\left(g\right(x\left)\right).g\text{'}\left(x\right)=\frac{1}{2}{(2x+1)}^{-\frac{1}{2}}.2 \\ ={(2x+1)}^{-\frac{1}{2}} \\ =\frac{1}{\sqrt{2x+1}} \end{gathered}$$