#1**0 **

"*What is the derivative of (3x)/(2x^2-18)?*"

Let u = 3x and v = (2x^{2}-18)^{-1}

Use the chain rule: duv/dx = udv/dx + vdu/dx:

du/dx = 3 dvdx = -(2x^{2}-18)^{-2}4x

d(3x(2x^{2}-18)^{-1})/dx = -12x^{2}(2x^{2}-18)^{-2} + 3(2x^{2}-18)^{-1 }

or (3(2x^{2}-18) - 12x^{2})/(2x^{2}-18)^{2} → -(6x^{2} + 54)/(2x^{2}-18)^{2} → -(3/2)(x^{2} + 9)/(x^{2} - 9)^{2}

Alan
Nov 27, 2018

#2**0 **

Thanks for the response! I used the quotient rule and got the answer (-6x^2-54)/(2x^2-18)^2

I'm not sure where I'm going wrong.

Guest Nov 27, 2018

#5**0 **

I do not believe you need to factor it to be correct. You found the derivative as asked.

ElectricPavlov
Nov 27, 2018