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calculus 1: differentiation
- Thread starter Mel Mitch
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Mel Mitch said: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.
Hello, Mel Mitch!
We are expected to use differentials . . .
\(\displaystyle \text{We have: }\:f(x) \:=\:x^{\text{-}\frac{1}{3}}\)
. . \(\displaystyle \text{Then: }\:df \;=\;-\tfrac{1}{3}x^{\text{-}\frac{4}{3}}dx \;=\;-\frac{dx}{3x^{\frac{4}{3}}}\)
\(\displaystyle \text{Let: }\:x \:=\:1,\;dx \:=\:-0.1\)
\(\displaystyle \text{Then we have: }\:f(1) \:=\:1^{\text{-}\frac{1}{3}} \:=\:1\)
. . \(\displaystyle \text{ and: }\;df(1) \;=\;-\frac{\text{-}0.1} {3(1^{\frac{4}{3}}) } \;=\;\frac{1}{30}\)
\(\displaystyle \text{Therefore: }\;\frac{1}{0.9^{\frac{1}{3}}} \;\;\approx\;\; 1 + \frac{1}{30} \;\;=\;\;\frac{31}{30} \;\;=\;\;\boxed{1.03333...}\)
\(\displaystyle \left[\text{Actual value: }\:\frac{1}{\sqrt[3]{0.9}} \;=\;1.035744169...\right]\)
We are expected to use differentials . . .
\(\displaystyle \text{Given that }\:y\:=\:x^{\text{-}\frac{1}{3}},\;\text{ use calculus to determine an approximate value for: }\:\frac{1}{0.9^{\frac{1}{3}}}\)
\(\displaystyle \text{We have: }\:f(x) \:=\:x^{\text{-}\frac{1}{3}}\)
. . \(\displaystyle \text{Then: }\:df \;=\;-\tfrac{1}{3}x^{\text{-}\frac{4}{3}}dx \;=\;-\frac{dx}{3x^{\frac{4}{3}}}\)
\(\displaystyle \text{Let: }\:x \:=\:1,\;dx \:=\:-0.1\)
\(\displaystyle \text{Then we have: }\:f(1) \:=\:1^{\text{-}\frac{1}{3}} \:=\:1\)
. . \(\displaystyle \text{ and: }\;df(1) \;=\;-\frac{\text{-}0.1} {3(1^{\frac{4}{3}}) } \;=\;\frac{1}{30}\)
\(\displaystyle \text{Therefore: }\;\frac{1}{0.9^{\frac{1}{3}}} \;\;\approx\;\; 1 + \frac{1}{30} \;\;=\;\;\frac{31}{30} \;\;=\;\;\boxed{1.03333...}\)
\(\displaystyle \left[\text{Actual value: }\:\frac{1}{\sqrt[3]{0.9}} \;=\;1.035744169...\right]\)