What is the derivative of #ln(x)/sin(2x)#?

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Answer 1 · 2018-07-10T19:51:08

#f'(x)=(sin(2x)-2xln(x)*cos(2x))/(x(sin(2x))^2)#

#(u/v)'=(u'v-uv')/v^2#
and

#(ln(x))'=1/x#
#(sin(2x))'=cos(2x)*2#

so we get

#(1/x*sin(2x)-ln(x)*cos(2x)*2)/(sin(2x))^2#
multiplying numerator and denominator by #x# we get

#(sin(2x)-2xln(x)*cos(2x))/(x*(sin(2x))^2)#