Question

If $f\left(x\right)=\sqrt{2x+3}$, Then $f^{\prime }\left(x\right)=$___

• A. $\sqrt[3]{3(2x+3{)}^2}$
• B. $\frac{1}{2\sqrt{(2x+3)}}$
• C. $\frac{1}{\sqrt{2x+3}}$
• D. $\frac{2}{\sqrt{2x+3}}$

Answer 1

The correct option is C

$\frac{1}{\sqrt{2x+3}}$

$f\left(x\right)=\sqrt{2x+3}{f}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}{f}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\sqrt{2(x+h)+3}-\sqrt{2x+3}}{h}$
Multiply the numerator and denominator with
$(\sqrt{2x+2h+3}+\sqrt{2x+3})f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{(2x+2h+3-2x-3)}{h(\sqrt{2x+2h+3}+\sqrt{2x+3})}{f}^{\prime }\left(x\right)={lim}_{h\to 0}\frac{2h}{h}.{lim}_{h\to 0}\left(\frac{1}{\sqrt{2x+2h+3}+\sqrt{2x+3}}\right)f^{\prime }\left(x\right)=2.\frac{1}{2\sqrt{2x+3}}=\frac{1}{\sqrt{2x+3}}$