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

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Answer 1 · 2017-11-28T15:48:30

#0.02083# (real value #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#