What is the derivative of sqrt(4x² + 1)?

Mar 20, 2018

$\frac{d}{\mathrm{dx}} = \frac{4 x}{\sqrt{4 {x}^{2} + 1}}$

Explanation:

The chain rule has you treat functions as follows:

$f \left(x\right) = g {\left(x\right)}^{n}$

$f ' \left(x\right) = \left(n \cdot g {\left(x\right)}^{n - 1}\right) \cdot g ' \left(x\right)$

You treat anything inside of the exponent as a second function, and you first take the derivative of the exponent, and then multiply by the derivative of the inner function.

For this problem:

$g \left(x\right) = 4 {x}^{2} + 1$

$f \left(x\right) = \sqrt{g \left(x\right)} = {\left(g \left(x\right)\right)}^{\frac{1}{2}}$

Now, lets take our derivatives:

$g ' \left(x\right) = 8 x$

$f ' \left(x\right) = \frac{1}{2} {\left(g \left(x\right)\right)}^{- \frac{1}{2}} \cdot g ' \left(x\right)$

$f ' \left(x\right) = \left(\frac{1}{2}\right) \frac{1}{g \left(x\right)} ^ \left(\frac{1}{2}\right) \cdot g ' \left(x\right)$

$f ' \left(x\right) = \left(\frac{1}{2}\right) \frac{1}{\sqrt{4 {x}^{2} + 1}} \cdot \left(8 x\right)$

color(red)(f'(x)=(4x)/sqrt(4x^2+1)