# How do you find the derivative of sqrt(x+1)?

$\setminus \frac{d}{\mathrm{dx}} \setminus \sqrt{x + 1} = \frac{1}{2 \setminus \sqrt{x + 1}}$

#### Explanation:

Differentiating w.r.t. $x$ by using chain rule as follows

$\setminus \frac{d}{\mathrm{dx}} \setminus \sqrt{x + 1}$

$= \setminus \frac{d}{\mathrm{dx}} {\left(x + 1\right)}^{\frac{1}{2}}$

$= \frac{1}{2} {\left(x + 1\right)}^{\frac{1}{2} - 1} \setminus \frac{d}{\mathrm{dx}} \left(x + 1\right)$

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

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

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

Jul 12, 2018

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

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$\text{here } y = \sqrt{x + 1} = {\left(x + 1\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2}} \times \frac{d}{\mathrm{dx}} \left(x + 1\right)$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{2 \sqrt{x + 1}}$