Question #08991

1 Answer
Timber Lin
Dec 20, 2017

#g'(t)=-2cos(2t)*(sec(5-sin(2t)))^2#

Explanation:

#g'(t)=d/dt(tan(5-sin(2t))#
#g'(t)=(sec(5-sin(2t)))^2*d/dt(5-sin(2t))# (tanx derivative and chain rule)
#g'(t)=(sec(5-sin(2t)))^2*(0-cos(2t)*d/dt(2t))# (sinx derivative)

#g'(t)=(sec(5-sin(2t)))^2*(-cos(2t)*2)#

#g'(t)=-2cos(2t)*(sec(5-sin(2t)))^2#

Related questions
Impact of this question
713 views around the world
You can reuse this answer
Creative Commons License