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Math Help - A few basic calculus derivative problems

  1. #1
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    A few basic calculus derivative problems

    1. If f(x)=3(root x)*(x^3-7(root x) +7)

    find f'(x) and f'(4)

    2. If f(x)=9x+2 / x^2+2x-3, find the derivative f'(x).

    3. G(x)=(3x^2-2)(x+2)^2 / (10x^2+8x+2)

    Find G'(2)

    4. f(x)=x^8 h(x)

    h(-1)=3
    h'(-1)=6

    Calculate f(-1)

    I attempted all of these, thought I was doing everything correct but apparently I was not and now im stuck. Any help is greatly appreciated
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  2. #2
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    Edit, I just found number 2 so scratch that one
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  3. #3
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    Quote Originally Posted by sitdownson View Post
    1. If f(x)=3(root x)*(x^3-7(root x) +7)

    find f'(x) and f'(4)
    1. 3\sqrt{x} \cdot (x^3 - 7\sqrt{x} +7)

    Distribute the 3\sqrt{x} to get f(x) = 3x^{3.5} - 21x + 21x^{0.5}

    You can differentiate that one in the standard way - please post any workings you've done if you have any problems
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  4. #4
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    ok great, got numbers 1 and 2 now, still trying to figure out what i did wrong on the rest, but thanks for the help so far!
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  5. #5
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    Why don't you post your attempt at a solution and then someone may help you. We are not answer machines.
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  6. #6
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    Ok sorry about that first time on these forums....

    3. f'(x) 2x +4 (6x) / 20x + 8

    f'(2) (4+4)(12) / 40 +8
    96/48
    2
    Got this wrong

    4. I thought that h(x) was 3x^2 but I realized I was wrong and now I have no idea what to do
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  7. #7
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    #3 ...

    \displaystyle G(x) = \frac{1}{2} \cdot \frac{(3x^2-2)(x+2)^2}{5x^2+4x+1}

    \displaystyle G'(x) = \frac{1}{2} \cdot \frac{(5x^2+4x+1)[(3x^2-2) \cdot 2(x+2) + (x+2)^2 \cdot 6x] - (3x^2-2)(x+2)^2(10x+4)}{(5x^2+4x+1)^2}

    now evauate G'(2)


    #4 ...

    f(x) = x^8 \cdot h(x)

    f'(x) = x^8 \cdot h'(x) + h(x) \cdot 8x^7

    now evaluate f'(-1)
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  8. #8
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    Quote Originally Posted by sitdownson View Post
    3. G(x)=(3x^2-2)(x+2)^2 / (10x^2+8x+2)

    Find G'(2)
    It looks like you have found G(2) rather than G'(2)
    \dfrac{(3x^2-2)(x+2)^2}{10x^2+8x+2} = \dfrac{1}{2} \cdot \dfrac{(3x^2-2)(x+2)^2}{5x^2+4x+1}


    Here you need to use the product rule and quotient rule. First the quotient rule

    u = (3x^2-2)(x+2)^2 and v = 10x^2+8x+2

    To find u' we need to use the product rule - for clarity's sake I'm going to use p and q

    p = 3x^2-2 and q = (x+2)^2


    u' = p'q + q'p = 6x(x+2)^2 + 2(x+2)(3x^2-2)


    Now we have found u' we can use it in the quotient rule for the whole expression:

    u' = 6x(x+2)^2 + 2(x+2)(3x^2-2)
    v' = 20x +8


    g'(x) = \dfrac{vu'-uv'}{v^2} = \dfrac{[(10x^2+8x+2)(6x(x+2)^2 + 2(x+2)(3x^2-2)] - [(3x^2-2)(x+2)^2(20x+8)]}{(10x^2+8x+2)^2}


    g'(2) = \dfrac{(40+16+2)(12(4)^2 + 2(6)(10))] - [(10)(16)(48)]}{(40+16+2)^2} = \dfrac{58 \cdot 312 - 7680}{3364} = \dfrac{2604}{841} \approx 3.096

    Please check my arithmetic
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  9. #9
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    Great thanks for the help!
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