#(log_2(x^2sinx^2))'#
first we need to shift the base into natural logs as these are just made for calculus
so using #log_a b = (log_c b)/(log_c a)#
so
#log_2(x^2sinx^2) = (ln (x^2sinx^2))/(ln 2) = 1/(ln 2)ln (x^2sinx^2)#
then we use another standard result namely that
#(ln f(x))' = 1/(f(x))*f'(x)#
so #( 1/(ln 2)ln (x^2sinx^2))' = 1/(ln 2)( ln (x^2sinx^2))' #
#= 1/(ln 2)1/ (x^2sinx^2) (x^2sinx^2)' \\qquad square#
for # (x^2sinx^2)'# we use the product rule
# (x^2sinx^2)' = (x^2)'sinx^2 + x^2 (sinx^2)' #
#= 2x sinx^2 + x^2 (sinx^2)' \qquad star#
for #(sinx^2)'# we use the chain rule so
#(sinx^2)' = cos x^2 (2x) = 2x cos x^2#
pop that back into #star # to get
# (x^2sinx^2)' = 2x sinx^2 + x^2 * 2x cos x^2 #
#\implies (x^2sinx^2)' = 2x sinx^2 + 2x^3 cos x^2 #
pop that back into #square# for
#( 1/(ln 2)ln (x^2sinx^2))' = 1/(ln 2)1/ (x^2sinx^2) (x^2sinx^2)'#
#= 1/(ln 2)1/ (x^2sinx^2) (2x sinx^2 + 2x^3 cos x^2)#
this can be simplified in any number of ways, so i'm going for brevity
#= 2/(ln 2) ( 1/x + x cot x^2)#
