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Thread: Hyperbolic Diff helppp

  1. #1
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    Hyperbolic Diff helppp

    Hello
    Find the derivative of the function
    $\displaystyle g(x)=ln(tanh(x))$

    why is the answer
    $\displaystyle 2csch(2x)$?
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Hello
    Find the derivative of the function
    $\displaystyle g(x)=ln(tanh(x))$

    why is the answer
    $\displaystyle 2csch(2x)$?
    You use the chain rule.
    $\displaystyle y=\ln u$ where $\displaystyle u=\tanh x$
    $\displaystyle dy/du=1/u=1/\tanh x=\mbox{coth} x$
    $\displaystyle du/dx=\mbox{sech}^2 x$
    Thus,
    $\displaystyle dy/dx=\mbox{sech}^2 x\cdot \mbox{coth}x$
    Express as fractions and simiplify to get,
    $\displaystyle \frac{1}{\sinh x\cosh x}$
    Multiply numerator and denoimator by 2,
    $\displaystyle \frac{2}{2\sinh x\cosh x}$
    Double angle for hyperbolic functions,
    $\displaystyle \frac{2}{\sinh 2x}=2\mbox{csch} 2x$
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  3. #3
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    Hello, ^_^Engineer_Adam^_^!

    Find the derivative of the function: .$\displaystyle g(x) \:=\:\ln[\text{tanh}(x)]$

    Why is the answer: .$\displaystyle g'(x) \:=\:2\,\text{csch}(2x)$ ?

    You're expected to know these facts for hyperbolic functions:

    . . $\displaystyle \frac{d}{dx}[\text{tanh}(x)] \:=\:\text{sech}^2(x) $

    . . $\displaystyle \text{tanh}(x) \:=\:\frac{\text{sinh}(x)}{\text{cosh}(x)}\qquad\t ext{sech}(x)\:=\:\frac{1}{\text{cosh}(x)}\qquad\te xt{csc}(x) \:=\:\frac{1}{\text{sinh}(x)}$

    . . $\displaystyle 2\,\text{sinh}(x)\text{cosh}(x) \:=\:\text{sinh}(2x)$



    We have: .$\displaystyle g(x)\:=\:\ln[\text{tanh}(x)]$

    Then: .$\displaystyle g'(x) \:=\:\frac{1}{\text{tanh}(x)}\cdot\text{sech}^2(x) \:=\:\frac{1}{\frac{\text{sinh}(x)}{\text{cosh}(x) }}\cdot\frac{1}{\text{cosh}^2(x)} \:=\:\frac{1}{\text{sinh}(x)\text{cosh}(x)}$

    Multiply top and bottom by 2: .$\displaystyle \frac{2}{2\,\text{sinh}(x)\text{cosh}(x)} \;=\;\frac{2}{\text{sinh}(2x)} \;=\;2\,\text{csch}(2x) $

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