How do you differentiate #f(x)= (2 x^2 + 7 x - 2)/ (x - cos x )# using the quotient rule?

1 Answer
Jul 14, 2017

#d/(dx) [(2x^2+7x-2)/(x-cosx)] = color(blue)((4x+7)/(x-cosx) - ((2x^2 + 7x - 2)(1+sinx))/((x-cosx)^2)#

Explanation:

We're asked to find the derivative

#d/(dx) [(2x^2 + 7x - 2)/(x-cosx)]#

Use the quotient rule, which is

#d/(dx) [u/v] = (v(du)/(dx) - u(dv)/(dx))/(v^2)#

where

  • #u = 2x^2 + 7x - 2#

  • #v = x - cosx#:

#= ((x-cosx)(d/(dx)[2x^2 + 7x - 2])-(2x^2 + 7x - 2)(d/(dx)[x-cosx]))/((x-cosx)^2)#

The derivative of #2x^2 + 7x - 2# is #4x + 7# (use power rule for each term):

#= ((x-cosx)(4x+7)-(2x^2 + 7x - 2)(d/(dx)[x-cosx]))/((x-cosx)^2)#

The derivative of #x# is #1# (power rule) and the derivative of #cosx# is #-sinx#:

#= color(blue)(((x-cosx)(4x+7)-(2x^2 + 7x - 2)(1+sinx))/((x-cosx)^2)#

Which can also be written as

#= color(blue)((4x+7)/(x-cosx) - ((2x^2 + 7x - 2)(1+sinx))/((x-cosx)^2)#