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Math Help - Derivative of a Natural Log

  1. #1
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    Derivative of a Natural Log

    I have the equation:

    s(t) = ln((t^2 + 1)/(t^2 - 1)) + 6lnt

    on the interval 1.1 < t < 10.

    How do I find the derivative of this? If someone could show me step-by-step, I'd really appreciate it as I'm having troubles with this one.

    Thanks!
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  2. #2
    Moo
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    Hello,

    Use the chain rule :

    You have to find the derivative for ln((t^2 + 1)/(t^2 - 1))=ln(u(t))

    Its derivative will be \frac{u'(t)}{u(t)}

    So let's calculate u'(t).
    This will be calculated with the quotient rule :

    u(t)=\frac{t^2+1}{t^2-1}=\frac{p(t)}{q(t)}

    u'(t)=\frac{p'(t)q(t)-p(t)q'(t)}{(q(t))^2}

    Here, p'(t)=2t and q'(t)=2t

    Hence u'(t)=\frac{2t(t^2-1)-2t(t^2+1)}{(t^2-1)^2}=\frac{2t^3-2t-2t^3-2t}{(t^2-1)^2}=\frac{-4t}{(t^2-1)^2}


    s'(t)=\underbrace{\frac{1}{u(t)}}_{\frac{t^2-1}{t^2+1}} u'(t)+\frac{6}{t}

    s'(t)=\frac{t^2-1}{t^2+1} \frac{-4t}{(t^2-1)^2}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi

    Otherwise, you can also split the log into two parts :

    f(t) = \ln\frac{t^2+1}{t^2-1}+6\ln t=\ln (t^2+1)-\ln(t^2-1)+6\ln t
    and then differentiate each term:
    f'(t)=\frac{2t}{t^2+1}-\frac{2t}{t^2-1}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}=\ldots
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  4. #4
    Moo
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    I was obsessed with the calculus...
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    Thanks both of you.

    Now, from that derivative, how do I find the maximum and minimum values for the function?

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    Thanks both of you.

    Now, from that derivative, how do I find the maximum and minimum values for the function?

    set the derivative equal to zero and solve for t. you may need further tests (first or second derivative test) to classify the points as maximum or minimum
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    Moo
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    Maximum and minimum values are points where the derivative is null.

    Make the fractions in the derivative have the same denominator, then look the values of x for which the numerator is null.

    To know if it's a maximum or a minimum, you can do a table of signs, or study the second derivative (which can be tricky)
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    I was obsessed with the calculus...
    calculus or calculation ?
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    Okay, I'm trying to solve for the t value but I can't.

    s'(t) = (6t^4 - 4t^2 - 6)/(t)(t^2 + 1)(t^2 - 1)

    Where do I go from here?
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  10. #10
    Moo
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    Quote Originally Posted by flyingsquirrel View Post
    Avec le calcul direct de la dérivée

    Is it better ?
    I'm still not used to the difference between the two concepts...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    Okay, I'm trying to solve for the t value but I can't.

    s'(t) = (6t^4 - 4t^2 - 6)/(t)(t^2 + 1)(t^2 - 1)

    Where do I go from here?
    ok, assuming that is the right answer, set the numerator equal to zero. note that it is quadratic in t^2
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  12. #12
    Moo
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    Quote Originally Posted by Jeavus View Post
    Okay, I'm trying to solve for the t value but I can't.

    s'(t) = (6t^4 - 4t^2 - 6)/(t)(t^2 + 1)(t^2 - 1)

    Where do I go from here?
    The derivative will be equal to 0 if the numerator is equal to 0 (and the denominator different of 0, or else it'll be undefined).

    So substitute : u=t²
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