How do you differentiate #f(x)=1/(cot(x)) # using the chain rule?

1 Answer
Apr 7, 2016

Answer:

#(cosec^2 x)/(cot^2 x)#

Explanation:

differentiate using the #color(blue)" chain rule " #

#d/dx [ f(g(x)) ] = f'(g(x) . g'(x) #

and the standard derivative D(cotx) = #- cosec^2 x #
#"---------------------------------------------------------------------------"#
rewrite f(x) as f(x) # = (cotx)^-1 #

f(g(x)) =#(cotx)^-1 rArr f'(g(x)) = -1(cotx)^-2 #

and g(x) = # cotx rArr g'(x) = -cosec^2 x #

#rArr f'(x) = -(cotx)^-2 . - cosec^2 x = (cosec^2 x)/(cot^2 x) #