What is the derivative of sqrt(4+x^2) ?

Apr 2, 2016

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

Explanation:

Change the function $f \left(x\right) = \sqrt{4 + {x}^{2}}$ into something more manageable by the laws of indices: $\sqrt{u} = {u}^{\frac{1}{2}}$

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

This is approachable with the chain rule, which states that if $f \left(x\right) = g \left(h \left(x\right)\right)$ then $f ' \left(x\right) = h ' \left(x\right) g ' \left(h\right)$, where $h$ is just shorthand for $h \left(x\right)$.

With $h \left(x\right) = 4 + {x}^{2}$ and $g \left(x\right) = {h}^{2} \left(x\right)$ and simple differentiation,

$h ' \left(x\right) = 2 x$
$g ' \left(h\right) = \frac{1}{2} h {\left(x\right)}^{- \frac{1}{2}}$
$= \frac{1}{2} {\left(4 + {x}^{2}\right)}^{- \frac{1}{2}}$
 = 1/(2sqrt(4 + x^2)

Multiplying these two together as $h ' \left(x\right) g ' \left(h\right)$,

$\frac{2 x}{2 \sqrt{4 + {x}^{2}}} = \frac{x}{\sqrt{4 + {x}^{2}}}$