Question

How do you find the derivative of `x(t)=4tan^4(2t)`?

Answer 1

The derivativeis `x'(t)=32tan^3(2t).sec^2(2t)`

Explanation:

We do a chain derivation
`x'(t)=(4tan^4(2t))'=4(4tan^3(2t)).tan(2t)'`
using the following

`(x^n)'=nx^(n-1)`
`(tanx)'=sec^2x`
`(ax)'=a`

So `x'(t)=16tan^3(2t).tan(2t)'`

`=16tan^3(2t).sec^2(2t).(2t)'`

`=16tan^3(2t).sec^2(2t).2`

`=32tan^3(2t).sec^2(2t).`