# Using differentials, find approximate value of (0.009)^(1/3)?

Nov 28, 2017

$0.02083$ (real value $0.0208008$)

#### Explanation:

This can be solved with the formula of Taylor:

$f \left(a + x\right) = f \left(a\right) + x f ' \left(a\right) + \left({x}^{2} / 2\right) f ' ' \left(a\right) \ldots .$

If $f \left(a\right) = {a}^{\frac{1}{3}}$

We will have:

$f ' \left(a\right) = \left(\frac{1}{3}\right) {a}^{- \frac{2}{3}}$

now if $a = 0.008$ then

$f \left(a\right) = 0.2$ and

$f ' \left(a\right) = \left(\frac{1}{3}\right) {0.008}^{- \frac{2}{3}} = \frac{25}{3}$

So if $x = 0.001$ then

$f \left(0.009\right) = f \left(0.008 + 0.001\right) \approx f \left(0.008\right) + 0.001 \times f ' \left(0.008\right) =$

$= 0.2 + 0.001 \cdot \frac{25}{3} = 0.2083$