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

1 Answer
Oct 25, 2016

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).#