r''(q) of
r′= (470)-(2*47*q)-(3*q^2)
steps please, any help would be wonderful.
So this is the exact same as taking the derivative of any other function, we can pretend that:
$\displaystyle f(q)=470-94q - 3q^2$
then:
$\displaystyle f'(q)=r''(q)$
and f'(q) is simply:
$\displaystyle f'(q)=r''(q)=94 + 6q$
(unless I did something dumb ). Hope that helps.
Remember that 1. $\displaystyle \frac{d}{dx}[u\pm{v}]=\frac{d}{dx}[u]\pm\frac{d}{dx}[v]$
When we apply this to your problem we get
$\displaystyle \frac{d}{dq}[470-94q-3q^2]=\frac{d}{dq}[470]-\frac{d}{dq}[94q]-\frac{d}{dq}[3q^2]$
Now remember that $\displaystyle \frac{d}{dx}[cu]=c\frac{d}{dx}[u]$ where c is constant and $\displaystyle \frac{d}{dx}[c]=0$.
Applying we get
$\displaystyle \frac{d}{dq}[470]-\frac{d}{dq}[94q]-\frac{d}{dq}[3q^2]=0-94\frac{d}{dq}[q]-3\frac{d}{dq}[q^2]$.
Now remember that $\displaystyle \frac{d}{dx}[x^n]=nx^{n-1}$.
Applying this
$\displaystyle 94\frac{d}{dq}[q]-3\frac{d}{dq}[q^2]=-94(1)q^{1-1}-3(2)q^{2-1}=-94q^0-6q^1=-94-6q$