Question

use differentials to approximate the value of (26)^(1/3)

Answer 1

2.962496060 round it up

Answer 2

I'd rather just say that \(\sqrt[3]{26} \approx \sqrt[3]{27} \approx 3\)

Answer 3

Why use differentials?

Answer 4

Use differentials? As in approximate it with a taylor or maclaurin series?

Answer 5

Haha - there's no kind of variable to use Taylor Series. :P

Answer 6

I was thinking something like this, and plug in 26... http://www.wolframalpha.com/input/?i=power+series+of+%28x%29%5E%281%2F3%29+at+27

Answer 7

\(f(x+\Delta x)\sim f(x)+f'(x)\Delta x \)
Use \(f(x)=x^{1/3}\) and put \(x=27\) and \(\Delta x=-1\)