Differentiate #f(x)=sinx/-(cos2x)# ?

1 Answer
Sonnhard
Jul 13, 2018

#f'(x)=-(cos(x)cos(2x)+2sin(x)sin(2x))/(-cos(2x))^2#

Explanation:

We Need the Quotient rule

#(u/v)'=(u'v-uv')/v^2#
so we get

#f'(x)=((cos(x)*(-cos(2x))-2sin(x)*(sin(2x)*2)))/(-cos(2x))^2#
so we get

#f'(x)=-(cos(x)cos(2x)+2sin(x)sin(2x))/(-cos(2x))^2#
note that

#(-cos(2x))'=-(-sin(2x))*2=sin(2x)*2#

Related questions
Impact of this question
1460 views around the world
You can reuse this answer
Creative Commons License