# Question #3df2a

Jun 1, 2017

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

#### Explanation:

Assuming the question meant "differentiate" using the chain rule rather than "derive" the chain rule.

Differentiate $\sqrt{{x}^{2} + 4}$

Note that the chain rule states that for a composition of functions:
$h \left(x\right) = f \left(g \left(x\right)\right)$
The derivative of $h \left(x\right)$ would be:
$h ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

First, we note that:
$\sqrt{{x}^{2} + 4} = {\left({x}^{2} + 4\right)}^{\frac{1}{2}}$

In this problem, we apply the chain rule.
We notice that in this case, $f \left(x\right) = \sqrt{x}$ and $g \left(x\right) = {x}^{2} + 4$, so:
$\frac{d}{\mathrm{dx}} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} = \frac{1}{2} {\left({x}^{2} + 4\right)}^{- \frac{1}{2}} \left(2 x\right)$
$\frac{d}{\mathrm{dx}} {\left({x}^{2} + 4\right)}^{\frac{1}{2}} = \frac{x}{\sqrt{{x}^{2} + 4}}$