How do you differentiate (2x+xe^6x)/(9x-(2x^2)-ln(x)) w.r.t. x?

This problem requires using the quotient rule, product rule and the chain rule. The derivative of the entire thing is ((du/dx)v-(dv/dx)u)/v^2 where u=2x+xe^6x and v=9x-2x^2-lnx. dv/dx is relatively straitforward: 9-4x-(1/x). 2x+xe^6x is less so, because this requires differentiating xe^6x. First notice this is two functions of x times each other, so we can use the product rule: so d/dx(xe^6x)=x(d/dx(e^6x))+e^6x. What is d/dx(e^6x)? We have to use the chain rule here: suppose g=6x, hence d/dg(e^g)xdg/dx=d(e^6x)=6e^6x. So now combining this altogether we know the derivative of the entire thing: ((2+e^6x+xe^6x)(9x-2x^2-lnx)-(2x+xe^6x)(9-4x-1/x))/(9x-2x^2-lnx)^2