How do you differentiate #f(x)=2^cotx#?

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Answer 1 · 2017-06-05T20:30:04

#f'(x)=-2^cotxln2csc^2x#

use logarithmic differentiation

let#" "y=2^cotx#

#=>lny=cotxln2#

differentiate wrt #x#

#1/y(dy)/(dx)=- csc^2xln2#

#(dy)/(dx)=-yln2csc^2x#

substitute back for #y#

#f'(x)=-2^cotxln2csc^2x#