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Math Help - hyperbolic function derivatives!!

  1. #1
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    hyperbolic function derivatives!!

    ok so how exactly would I simplify this function

    sinh(ln(t))
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathlete View Post
    ok so how exactly would I simplify this function

    sinh(ln(t))
    Use the chain rule:
    \frac{d}{dt}sinh(ln(t)) = cosh(ln(t)) \cdot \frac{1}{t}

    -Dan
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  3. #3
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    Before takin' the derivative, observe that

    \sinh x = \frac{{e^x  - e^{ - x} }}<br />
{2}. Now contemplate x=\ln t and take its derivative.
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  4. #4
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    yeah i know how to take the derivative, but it asks me to simplify the function first...so to do that would i just plug in ln(t) into the (e^x-e^(-x))/2??
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Before takin' the derivative, observe that

    \sinh x = \frac{{e^x  - e^{ - x} }}<br />
{2}. Now contemplate x=\ln t and take its derivative.
    Ha! I never saw that.

    -Dan
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  6. #6
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by Krizalid View Post
    Before takin' the derivative, observe that

    \sinh x = \frac{{e^x  - e^{ - x} }}<br />
{2}. Now contemplate x=\ln t and take its derivative.
    Nice, so the derivative is just 1?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by colby2152 View Post
    Nice, so the derivative is just 1?
    No.

    sinh(ln(x)) = \frac{e^{ln(x)} - e^{-ln(t)}}{2}

    = \frac{x - \frac{1}{x}}{2}

    So the derivative will be
    \frac{1}{2} + \frac{1}{2x^2}

    -Dan
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  8. #8
    GAMMA Mathematics
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    Quote Originally Posted by topsquark View Post
    No.

    sinh(ln(x)) = \frac{e^{ln(x)} - e^{-ln(t)}}{2}

    = \frac{x - \frac{1}{x}}{2}

    So the derivative will be
    \frac{1}{2} + \frac{1}{2x^2}

    -Dan
    Whoops, gotta watch where I throw those negative signs!
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