What is the derivative of sqrt(2x-1)?

Feb 3, 2016

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

Explanation:

This is a case of using the power rule and the chain rule.

First, we note that $\sqrt{2 x - 1}$ can be rewritten as ${\left(2 x - 1\right)}^{\frac{1}{2}}$.
Now we can apply the power rule, where we multiply the function by the exponent then decrease the exponent by one:
$\frac{d}{\mathrm{dx}} {\left(2 x - 1\right)}^{\frac{1}{2}} = \left(\frac{1}{2}\right) {\left(2 x - 1\right)}^{\frac{1}{2} - 1}$
$= \frac{1}{2} {\left(2 x - 1\right)}^{- \frac{1}{2}}$

Then we apply the chain rule, which tells us to multiply this result by the derivative of the "inside" function. In our case, the "inside" function is $2 x - 1$ (because it is inside the square root), and its derivative is simply $2$. Our final derivative now becomes:
$\frac{1}{2} {\left(2 x - 1\right)}^{- \frac{1}{2}} \cdot 2 = {\left(2 x - 1\right)}^{- \frac{1}{2}}$

In radical form, this is $\frac{1}{\sqrt{2 x - 1}}$.