Question

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

Answer 1

`0.02083` (real value `0.0208008`)

Explanation:

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`