calculus 1: differentiation

[Mel Mitch]

SMALL CHANGES...please guide me in the problem:

Given that y=x^-1/3 use calculus to determine an approximate value for 1/0.9^1/3.

 

[Deleted member 4993]

SMALL CHANGES...please guide me in the problem:

Given that y=x^-1/3 use calculus to determine an approximate value for 1/0.9^1/3.

Use Taylor's Theorem

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[soroban]

Hello, Mel Mitch!

We are expected to use differentials . . .

$\displaystyle \text{Given that }\:y\:=\:x^{\text{-}\frac{1}{3}},\;\text{ use calculus to determine an approximate value for: }\:\frac{1}{0.9^{\frac{1}{3}}}$

$\displaystyle \text{We have: }\:f(x) \:=\:x^{\text{-}\frac{1}{3}}$
. . $\displaystyle \text{Then: }\:df \;=\;-\tfrac{1}{3}x^{\text{-}\frac{4}{3}}dx \;=\;-\frac{dx}{3x^{\frac{4}{3}}}$

$\displaystyle \text{Let: }\:x \:=\:1,\;dx \:=\:-0.1$


$\displaystyle \text{Then we have: }\:f(1) \:=\:1^{\text{-}\frac{1}{3}} \:=\:1$

. . $\displaystyle \text{ and: }\;df(1) \;=\;-\frac{\text{-}0.1} {3(1^{\frac{4}{3}}) } \;=\;\frac{1}{30}$


$\displaystyle \text{Therefore: }\;\frac{1}{0.9^{\frac{1}{3}}} \;\;\approx\;\; 1 + \frac{1}{30} \;\;=\;\;\frac{31}{30} \;\;=\;\;\boxed{1.03333...}$


$\displaystyle \left[\text{Actual value: }\:\frac{1}{\sqrt[3]{0.9}} \;=\;1.035744169...\right]$

 

[Mel Mitch]

Thanks Soroban