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

Question

1734 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-25T09:44:10

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

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

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

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

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

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