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Math Help - Higher Derivatives r''(q)

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

    r''(q) of

    r′= (470)-(2*47*q)-(3*q^2)

    steps please, any help would be wonderful.
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  2. #2
    Newbie driegert's Avatar
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    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.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by youmuggles View Post
    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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