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

Feb 5, 2016

f'(x)=1/(2\sqrt{(x+1)}
Given $f \left(x\right) = \setminus \sqrt{x + 1}$. I'm sure you're familiar with the fact that if $y = {x}^{n}$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = n {x}^{n - 1}$ where $n$ may be any real number and that if $g \left(y\right)$ is a function on$f \left(x\right)$ such that $g \left(f \left(x\right)\right)$, then derivative of that is g'(f(x))*f'(x))
So that means we can rewrite the function as $f \left(x\right) = {\left(x + 1\right)}^{\frac{1}{2}}$ (using the law of indicies here).
So, differentiating it, $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = \frac{1}{2} {\left(x + 1\right)}^{\frac{1}{2} - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right) = \frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2}}$