How do you differentiate #f(x)=sec(4x)/tan(4x)#?

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Answer 1 · 2017-11-23T09:46:23

#f'(x)=-4csc(4x)cot(4x)#

#sec(4x)=1/cos(4x)#

So, #f(x)=((1/cos(4x)))/tan(4x)=1/(cos(4x)tan(4x))#

#tan(4x)=sin(4x)/cos(4x)#

#cos(4x)*sin(4x)/cos(4x)=(cos(4x)sin(4x))/cos(4x)#

#(cancel(cos(4x))sin(4x))/cancel(cos(4x))=sin(4x)#

#f(x)=1/sin(4x)=u/v#

#u=1#
#v=sin(4x)#

#u'=0#
#v'=4cos(4x)#

#f'(x)=(vu'-uv')/v^2=>(0sin(4x)-1(4(cos(4x))))/sin^2(4x)#

#=-(4cos(4x))/sin^2(4x)#

#=-4/sin(4x)*cos(4x)/sin(4x)#

#=-4/sin(4x)cot(4x)#

#1/sin(4x)=csc(4x)#

#f'(x)=-4csc(4x)cot(4x)#