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)#
