1. ## Higher Derivatives r''(q)

r''(q) of

r′= (470)-(2*47*q)-(3*q^2)

steps please, any help would be wonderful.

2. 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.

3. Originally Posted by youmuggles
r''(q) of

r′= (470)-(2*47*q)-(3*q^2)

steps please, any help would be wonderful.
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

$\displaystyle 94\frac{d}{dq}[q]-3\frac{d}{dq}[q^2]=-94(1)q^{1-1}-3(2)q^{2-1}=-94q^0-6q^1=-94-6q$