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

Jun 3, 2016

$\frac{d}{d x} \left(x \cdot \sqrt{x + 1}\right) = \frac{3 x + 2}{2 \cdot \sqrt{x + 1}}$

#### Explanation:

d/(d x)(x*sqrt(x+1))=?

$\frac{d}{d x} \left(x \cdot \sqrt{x + 1}\right) = {x}^{'} \cdot \sqrt{x + 1} + {\left(\sqrt{x + 1}\right)}^{'} \cdot x$

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

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

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

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

$\frac{d}{d x} \left(x \cdot \sqrt{x + 1}\right) = \frac{3 x + 2}{2 \cdot \sqrt{x + 1}}$