Hi again guys,

Two posts from me. But they aren't too hard. For you at least!

Given that $\displaystyle y = x^{-\frac{1}{3}}$ use the calculus to determine an approximate value for $\displaystyle \frac{1}{\sqrt [3]{0.9}}$ Normally with these there is a descernable x and $\displaystyle \delta x$ but I can't see a way to giggle 0.9 into anything useful. Can you?