Using differentials, find approximate value of ${(0.009)}^{\frac{1}{3}}$ $(0.009)^(1/3)$ ?

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Answer 1 · 2017-11-28T15:48:30

$0.02083$ $0.02083$ (real value $0.0208008$ $0.0208008$ )

This can be solved with the formula of Taylor:

$f(a+x)=f(a)+xf'(a)+(x^2/2)f''(a)....$

If $f(a)=a^(1/3)$

We will have:

$f'(a)=(1/3)a^(-2/3)$

now if $a=0.008$ then

$f(a)=0.2$ and

$f'(a)=(1/3)0.008^(-2/3)=25/3$

So if $x=0.001$ then

$f(0.009)=f(0.008+0.001)~~f(0.008)+0.001xxf'(0.008)=$

$=0.2+0.001*25/3=0.2083$

Using differentials, find approximate value of (0.009)^(1/3)(0.009)13(0.009)^(1/3)?