How do you find the derivative of f(x)=sqrt(1+3x^2)?

Jan 16, 2017

$f \left(x\right) = \sqrt{1 + 3 {x}^{2}} \implies f ' \left(x\right) = \frac{3 x}{\sqrt{1 + 3 {x}^{2}}}$

Explanation:

First remember $\sqrt{x} = {x}^{\frac{1}{2}}$, then we can write
$f \left(x\right) = \sqrt{1 + 3 {x}^{2}} = {\left(1 + 3 {x}^{2}\right)}^{\frac{1}{2}}$

and remember the chain rule

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

So, let

$h \left(x\right) = \sqrt{x}$

and

$g \left(x\right) = 1 + 3 {x}^{2}$

Then we can write

$f \left(x\right) = \left(h \circ g\right) \left(x\right) = h \left(g \left(x\right)\right) = \sqrt{g \left(x\right)}$

then

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

which by the power rule $\left(\left({x}^{n}\right) ' = n {x}^{n - 1}\right)$

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

then we can plug in g(x)

$h ' \left(g \left(x\right)\right) = \frac{1}{2} {\left(1 + 3 {x}^{2}\right)}^{- \frac{1}{2}}$

and

$g \left(x\right) = 1 + 3 {x}^{2} \implies g ' \left(x\right) = \left(1\right) ' + \left(3 {x}^{2}\right) ' = 0 + 6 x = 6 x$

by the fact that the derivative of sums is the sum of derivatives
and the power rule

then since,

$f \left(x\right) = \left(h \circ g\right) \left(x\right) \implies f ' \left(x\right) = h ' \left(g \left(x\right)\right) g ' \left(x\right)$

we can plug in our results

$f ' \left(x\right) = 6 x \left(\frac{1}{2} {\left(1 + 3 {x}^{2}\right)}^{- \frac{1}{2}}\right) = 3 x {\left(1 + 3 {x}^{2}\right)}^{- \frac{1}{2}} = \underline{\frac{3 x}{\sqrt{1 + 3 {x}^{2}}}}$