How do you differentiate #f(x) = x/sinx#?

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Answer 1 · 2015-11-04T09:46:00

#f'(x)=(sinx-xcosx)/(sin^2x)#

you have a function like this

#y=u/v#

Then you have to use this Equation

#y'=(u'*v-u*v')/v^2#

#f(x)=x/(sinx)#

#f'(x)=(x'*sinx-x*sinx')/(sinx)^2#

#f'(x)=(1*sinx-x*cosx)/(sinx)^2=(sinx-xcosx)/(sin^2x)#