Answer is f'(x)=d/dx{log(tanx)}=1/tanx×sec²x=sec²x/tanx=1/cos²x/sinx/cosx=2/2cosx.sinx=2/sin2x=2.cosec2x so f'(x)=2.cosec2x - Study24x7
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Vishal kumar shared post from Defence exams
26 Apr 2025 01:04 AM study24x7 study24x7

Answer is
f'(x)=d/dx{log(tanx)}
=1/tanx×sec²x
=sec²x/tanx
=1/cos²x/sinx/cosx
=2/2cosx.sinx
=2/sin2x
=2.cosec2x
so
f'(x)=2.cosec2x

Defence exams
26 Apr 2025 12:56 AM study24x7 study24x7

the differential co-efficient of log(tanx) is ?

Defence
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