How do you differentiate #f(x)= (2 x^2 - x - 6 )/ (x sinx )# using the quotient rule?

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Answer 1 · 2016-01-29T21:14:58

#f'(x)=((2x^2+6)sinx-(2x^3-x^2-6x)cosx)/(x^2sin^2x)#

The quotient rule states that

#d/dx[(g(x))/(h(x))]=(g'(x)h(x)-h'(x)g(x))/[h(x)]^2#

Here, we have

#g(x)=2x^2-x-6#
#h(x)=xsinx#

To differentiate #g(x)#, use the power rule on each term.

#g'(x)=4x-1#

To differentiate #h(x)#, use the product rule.

#h'(x)=sinx+xcosx#

Plug these into the original quotient rule expression.

#f'(x)=((4x-1)(xsinx)-(sinx+xcosx)(2x^2-x-6))/(x^2sin^2x)#

We can distribute and simplify, but the answer can't really get much prettier.

#f'(x)=(4x^2sinx-xsinx-2x^2sinx+xsinx+6sinx-2x^3cosx+x^2cosx+6xcosx)/(x^2sin^2x)#

#f'(x)=((2x^2+6)sinx-(2x^3-x^2-6x)cosx)/(x^2sin^2x)#