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Finding the second derivative of (arcosh(x))^2 watch

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    This problem has stumped me for hours now, and I'm getting rather frustrated with it!

    This is what I'm getting:

     f(x) = (arcosh(x))^2

    \frac{df(x)}{dx} = 2arcosh(x).(x^2 -1)^{-1/2}

    \frac{d^2f(x)}{dx^2} = \frac{2(4x^2-1)^{1/2}}{(x^2-1)^{1/2}} - \frac{2xarcosh(x)}{(x^2-1)^{3/2}}

    That's using the product rule. I get something different with the quotient rule.

    What's worse is that my textbook's answer of,

    - \frac{2xarcosh(x)}{(x^2-1)^{3/2}}

    doesn't agree with the horrendous thing WolframAlpha throws up: http://www.wolframalpha.com/input/?i...28Arccosh%29^2

    Can anyone please help!?
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    (Original post by StandardCarpet)
    This problem has stumped me for hours now, and I'm getting rather frustrated with it!

    This is what I'm getting:

     f(x) = (arcosh(x))^2

    \frac{df(x)}{dx} = 2arcosh(x).(x^2 -1)^{-1/2}

    \frac{d^2f(x)}{dx^2} = \frac{2(4x^2-1)^{1/2}}{(x^2-1)^{1/2}} - \frac{2xarcosh(x)}{(x^2-1)^{3/2}}

    That's using the product rule. I get something different with the quotient rule.

    What's worse is that my textbook's answer of,

    - \frac{2xarcosh(x)}{(x^2-1)^{3/2}}

    doesn't agree with the horrendous thing WolframAlpha throws up: http://www.wolframalpha.com/input/?i...28Arccosh%29^2

    Can anyone please help!?
    I get:

    \frac{2}{x^2-1}-\frac{2x\text{arcosh}(x)}{(x^2-1)^{3/2}}

    by product and quotient rules.
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    (Original post by ghostwalker)
    I get:

    \frac{2}{x^2-1}-\frac{2x\text{arcosh}(x)}{(x^2-1)^{3/2}}

    by product and quotient rules.
    That tallys with Wolfram's derivation for x=2, so I'm gonna assume my textbook's just been misprinted without the first bit.

    Would it be much to ask your working from the first to second derivation? I'd really appreciate it.
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    (Original post by StandardCarpet)
    That tallys with Wolfram's derivation for x=2, so I'm gonna assume my textbook's just been misprinted without the first bit.

    Would it be much to ask your working from the first to second derivation? I'd really appreciate it.
    Quotient or product rule? State your preferance (I won't do both).
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    (Original post by ghostwalker)
    Quotient or product rule? State your preferance (I won't do both).
    Product rule would be awesome. Thanks.
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    (Original post by StandardCarpet)
    ...
    so, starting from:

    \frac{df(x)}{dx} = 2arcosh(x).(x^2 -1)^{-1/2}

    we have

    \frac{d^2f(x)}{dx^2} = 2.(x^2 -1)^{-1/2}.(x^2 -1)^{-1/2}+2arcosh(x).(x^2 -1)^{-3/2}.2x.(-1/2)

     = 2.(x^2 -1)^{-1}-2xarcosh(x).(x^2 -1)^{-3/2}

    Or to put it more neatly.

     = \frac{2}{(x^2 -1)}-\frac{2xarcosh(x)}{(x^2 -1)^{3/2}}
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    Note: just spotted the power of the denominator of that second fraction should be positive not negative.
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    (Original post by Tla)
    Note: just spotted the power of the denominator of that second fraction should be positive not negative.
    Yep, I was in the middle of correcting it, and corrected the wrong one, thus having to correct it again. Ho, hum!
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    (Original post by ghostwalker)
    Yep, I was in the middle of correcting it, and corrected the wrong one, thus having to correct it again. Ho, hum!
    :p:
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    (Original post by ghostwalker)
    so, starting from:

    \frac{df(x)}{dx} = 2arcosh(x).(x^2 -1)^{-1/2}

    we have

    \frac{d^2f(x)}{dx^2} = 2.(x^2 -1)^{-1/2}.(x^2 -1)^{-1/2}+2arcosh(x).(x^2 -1)^{-3/2}.2x.(-1/2)

     = 2.(x^2 -1)^{-1}-2xarcosh(x).(x^2 -1)^{-3/2}

    Or to put it more neatly.

     = \frac{2}{(x^2 -1)}-\frac{2xarcosh(x)}{(x^2 -1)^{3/2}}
    Thank you so much!!!!
 
 
 
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