Higher Derivatives r''(q)

**youmuggles** · February 7th 2010, 10:08 AM

r''(q) of

r′= (470)-(2*47*q)-(3*q^2)

steps please, any help would be wonderful.

**driegert** · February 7th 2010, 10:20 AM

So this is the exact same as taking the derivative of any other function, we can pretend that:

$f(q)=470-94q - 3q^2$

then:

$f'(q)=r''(q)$

and f'(q) is simply:

$f'(q)=r''(q)=94 + 6q$

(unless I did something dumb

). Hope that helps.

**VonNemo19** · February 7th 2010, 10:22 AM

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. $\frac{d}{dx}[u\pm{v}]=\frac{d}{dx}[u]\pm\frac{d}{dx}[v]$

When we apply this to your problem we get

$\frac{d}{dq}[470-94q-3q^2]=\frac{d}{dq}[470]-\frac{d}{dq}[94q]-\frac{d}{dq}[3q^2]$

Now remember that $\frac{d}{dx}[cu]=c\frac{d}{dx}[u]$ where c is constant and $\frac{d}{dx}[c]=0$ .

Applying we get

$\frac{d}{dq}[470]-\frac{d}{dq}[94q]-\frac{d}{dq}[3q^2]=0-94\frac{d}{dq}[q]-3\frac{d}{dq}[q^2]$ .

Now remember that $\frac{d}{dx}[x^n]=nx^{n-1}$ .

Applying this

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

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