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

May 14, 2015

Using the chain rule!

Let's name your function $y$, as $y = \sqrt{x + 1}$

Let's consider $u = x + 1$ and now derive $y = \sqrt{u}$ (which is the same as $y = {u}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{du}} = \left(\frac{1}{2}\right) \cdot {u}^{- \frac{1}{2}} = \frac{1}{2 {u}^{\frac{1}{2}}}$

However, we have already stated that $u = x + 1$. So,
$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

and

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} \text{ (chain rule)}$

So

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \cdot {\left(x + 1\right)}^{\frac{1}{2}}} \cdot \left(1\right) = \frac{1}{2 \sqrt{x + 1}}$