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

Question

3436 views around the world

You can reuse this answer
Creative Commons License

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#