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Math Help - derivative with natural logs.

  1. #1
    Junior Member
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    derivative with natural logs.

    just wanted to check to see if i'm doing this right, any step by step guidance is greatly appreciated.

    "find the derivative":
    <br />
f(x) = ln(x^2 + 3x + 1) ln(x^3+1)

    would i start by using chain rule?

    u=  x^2 + 3x + 1

    = 2x +3

    h(u) = lnu = \frac{1}{u}

    = \frac{1}{x^2+3x+1} * 2x+3 = \frac{2x+3}{x^2+3x+1}




    u= x^3 + 1 = 3x^2

    h(u) = lnu = \frac{1}{u}


    = \frac{1}{x^3+1} * 3x^2 = \frac{3x^2}{x^3+1}


    since those are fractions would i use the quotient rule? or just plug them the way they are into product rule?

    with product rule i have:

    f\prime(x) = \frac{2x+3}{x^2+3x+1} * ln(x^3+1) + ln(x^2+3x+1) * \frac{3x^2}{x^3+1}<br />

    help please! im so confused, final exam tomorrow =\
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  2. #2
    Member mohammadfawaz's Avatar
    Joined
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    Lebanon - Beirut
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    Your method is completely correct.
    You're given f(x) = uv, hence f'(x)=u'v+v'u and you have found u' and v' using the chain rule.
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  3. #3
    Junior Member
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    Use the product rule here

    for f=u\times v \implies f' = uv'+vu'

    Also (\ln f(x))' = \frac{f'(x)}{f(x)}
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