# What is the derivative of sqrt(2x+1)?

Mar 30, 2016

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

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{ chain rule}}$

 d/dx [f(g(x) ] = f'(g(x)) . g'(x)

rewrite $\sqrt{2 x + 1} \text{ as } {\left(2 x + 1\right)}^{\frac{1}{2}}$

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

and g(x) = 2x + 1 → g'(x) = 2

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

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