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Math Help - derivative of cosh

  1. #1
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    derivative of cosh

    I have this question and i figured an answer but I am not sure if its completely right... Someone help?


    the question is

    if f(x) = cosh(ln(x)), then show that f ' (x) = 1/2( (x^2 - 1) / x^2)

    The way i went about it was i turned the last part into

    • 1/2 ( ((x^2)/x^2) - ( 1/x^2) )


    • 1/2(1- (1/x^2)


    then i took the derivative of f(x)

    f ' (x) = sinh(lnx) (1/x)

    = (e ^ lnx - e ^ -lnx) / 2 * 1/x

    = (x - x^ - 1) / 2 * 1/x

    = 1/2 (1 - x^ -2)

    = 1/2 ( 1 - (1 / x^2) )

    Thanks for the help...
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  2. #2
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    Quote Originally Posted by Cucc1137 View Post
    I have this question and i figured an answer but I am not sure if its completely right... Someone help?


    the question is

    if f(x) = cosh(ln(x)), then show that f ' (x) = 1/2( (x^2 - 1) / x^2)

    The way i went about it was i turned the last part into

    • 1/2 ( ((x^2)/x^2) - ( 1/x^2) )

    • 1/2(1- (1/x^2)

    then i took the derivative of f(x)

    f ' (x) = sinh(lnx) (1/x)

    = (e ^ lnx - e ^ -lnx) / 2 * 1/x

    = (x - x^ - 1) / 2 * 1/x

    = 1/2 (1 - x^ -2)

    = 1/2 ( 1 - (1 / x^2) )

    Thanks for the help...
    Your solution is correct.
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  3. #3
    Member Nacho's Avatar
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    Santiago, Chile
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    the answer is right, however you can use the definition of cosh before

    <br />
\cosh \left( {\ln x} \right) = \frac{{\exp \left\{ {\ln x} \right\} + \exp \left\{ { - \ln x} \right\}}}<br />
{2} = \frac{{x + \frac{1}<br />
{x}}}<br />
{2}<br />

    there you donīt use de the quotient rule
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