How do you differentiate #(1)/ (x^2cosx)# using the quotient rule?

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Answer 1 · 2017-01-23T06:25:04

The answer is #=(xsinx-2cosx)/(xcosx)#

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

#(uv)'=u'v+uv'#

Here, we have

#u=1#, #=>#, #u'=0#

#v=x^2cosx#, #=>#, #v'=2xcosx-x^2sinx#

Therefore,

#(1/(x^2cosx))'=(0*(x^2cosx)-1*(2xcosx-x^2sinx))/(x^2cosx)#

#=(x^2sinx-2xcosx)/(x^2cosx)#

#=(xsinx-2cosx)/(xcosx)#