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    I need to show that the differential of 1/cosh^2(x) = -2sinh(x)cosh^3(x).

    There has already been a couple of mistakes on this so this could be another mistake. Is it possible to get that differential?

    Shouldn't the cosh^3(x) be to the power of negative 3?


    I tired by rearranging cosh(2x) = 2cosh^2(x) 1-1 and by dividing sinh^2(x) + cosh^2(x) = 1 by cosh^2(x) and then substituting that in and trying to differentiate but I can't get the minus sign in front of the answer and the cosh to have a positive power of 3.


    Is it possible to do this then?

    Thank you
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    Let  u = cosh(x)

    Then  \frac{d}{dx} (\frac{1}{cosh^2(x)}) = \frac{d}{du}(u^{-2}) . \frac{du}{dx}

     \frac{du}{dx} = sinh(x)

    So the differential becomes:  -2(cosh(x))^{-3} . sinh x

    Seems the book may have a typo
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    (Original post by Spungo)
    Let  u = cosh(x)

    Then  \frac{d}{dx} (\frac{1}{cosh^2(x)}) = \frac{d}{du}(u^{-2}) . \frac{du}{dx}

     \frac{du}{dx} = sinh(x)

    So the differential becomes:  -2(cosh(x))^{-3} . sinh x

    Seems the book may have a typo
    See thats what I got one time, so thanks for your help
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    (Original post by Spungo)
    Let  u = cosh(x)

    Then  \frac{d}{dx} (\frac{1}{cosh^2(x)}) = \frac{d}{du}(u^{-2}) . \frac{du}{dx}

     \frac{du}{dx} = sinh(x)

    So the differential becomes:  -2(cosh(x))^{-3} . sinh x

    Seems the book may have a typo
    Sorry, I think I spotted a mistake in your work. Is du/dx not -sinh(x),making the whole answer positive?
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    (Original post by claret_n_blue)
    Sorry, I think I spotted a mistake in your work. Is du/dx not -sinh(x),making the whole answer positive?
    No. unlike its trigonometric counterpart, the derivative of cosh is just sinh. if you are unconvinced, refer to the definition and differentiate.
 
 
 
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