How do you differentiate #(xcosx)/ (sinx+x)# using the quotient rule?

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Answer 1 · 2017-10-19T12:43:04

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

Quotient formula: #d/dx(f/g)=(gf^'-fg^')/g^2#

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

#f^'(x)= ((cosx -xsinx)(x+sinx)-xcosx(1+cosx))/(sinx+x)^2 #

#f^'(x)= (xcosx+cosxsinx -x^2sinx-xsin^2x-xcosx-xcos^2x)/(sinx+x)^2 #

or #f^'(x)= (cancel(xcosx)+cosxsinx -x^2sinx-xsin^2x-cancel(xcosx)-xcos^2x)/(sinx+x)^2 #

#f^'(x)= ((cosxsinx -x^2sinx-x(sin^2x+cos^2x)))/(sinx+x)^2 # or

#f^'(x)= ((cosxsinx -x^2sinx-x))/(sinx+x)^2 # [Ans]