# How do you find the derivative of sqrt(3x+1)?

Jan 11, 2017

First off, recall that the derivative of $\sqrt{u} = \frac{1}{2 \sqrt{u}}$. If you don't remember that, simply change the square root to a $+ \text{1/2}$ exponent and use the power rule: $\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$.

Now, in your expression, $u = 3 x + 1$. Since $u$ is a function of $x$, you must incorporate the chain rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

for $f \left(u\right) = \sqrt{u \left(x\right)} = \sqrt{3 x + 1}$.

So, taking the derivative, we have:

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[\sqrt{3 x + 1}\right]} = \frac{1}{2 \sqrt{u \left(x\right)}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= \frac{1}{2 \sqrt{3 x + 1}} \cdot \frac{d}{\mathrm{dx}} \left[3 x + 1\right]$

$= \textcolor{b l u e}{\frac{3}{2 \sqrt{3 x + 1}}}$