Question

How do you find the derivative of `n(t) = 150 - 600/root3(16+3t^2)`?

Answer 1

`n'(t)=1200t/(root(3)(16+3t^2)^2`

Explanation:

writing

`n(t)=150-600(16+3t^2)^(-1/3)`
note that `(150)'=0`
and

`n'(t)=-600*(-1/3)(16+3t^2)^(-2/3)6t` simplifying we obtain

`n'(t)=1200t/root(3)(16+3t^2)^2`