find the derivative $$\cfrac{(x-6)(x^2+4x)}{x^3}$$ and $$\left(\frac{q^6+4}{2q}\right)\left(\frac{q^8+ 6}q\right)$$
Okay so Im reviewing for a test I have tomorrow and these two derivative question come up ive been using the quotient rule and I know I have to use the quotient rule for theses but Im not sure how to approach these problems. Any help, maybe because it not in the general function/function format im panicking and over thinking it .   

most often it is better to simplify before taking the derivative or use more complicated rules. here we are going to use just the power rule 
$$\frac{d}{dx}\left(x^n\right) = nx^{n-1}.$$ 
first we simplify the quotient $$y = \frac{(x-6)(x^2 + 4x)}{x^3} = \left(x^3+4x^2-6x^2-24x\right)x^{-3} = 1-2x^{-1}-24x^{-2} $$  so $$\frac{dy}{dx} = 2x^{-2}+48x^{-3}=\frac{2(x+24)}{x^3}$$
i will let you do the second one.
