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How do you prove that the derivative of tanh^-1(x/a) is a/(a^2-x^2)? Watch

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    Yeah...question above says it all

    Can't find how to do it in any books
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    do you mean from first principles?

    if not, let y=artanh(x/a)

    then tanh(y)=x/a

    differentiate and rearrange for dy/dx
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    the answer is 3.
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    (Original post by Pheylan)
    do you mean from first principles?

    if not, let y=artanh(x/a)

    then tanh(y)=x/a

    differentiate and rearrange for dy/dx
    Could you explain that a bit further...I know that differentiating tanh(y) would give sech^2(y) but I get stuck after that
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    How does sech^2 relate to tanh?
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    I am an idiot.
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    (Original post by IrrationalNumber)
    How does sech^2 relate to tanh?
    It's the derivative of it
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    (Original post by kellywellydoodle)
    It's the derivative of it
    True, let me try again
    How does sech^2 relate to tanh^2?
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    (Original post by StephenP91)
    Well.

     y = Arctan(x) -> x = Tan(y)

    Differentiate that.

    1 = Sec^{2}(y)\dfrac{dy}{dx}

    \dfrac{1}{Sec^{2}(y)} = \dfrac{dy}{dx}
    tanh not tan
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    (Original post by StephenP91)
    Well.

     y = Arctan(x) -> x = Tan(y)

    Differentiate that.

    1 = Sec^{2}(y)\dfrac{dy}{dx}

    \dfrac{1}{Sec^{2}(y)} = \dfrac{dy}{dx}

    \dfrac{1}{1 + Tan^{2}(y)} = \dfrac{dy}{dx}

    \dfrac{1}{1 + x^{2})} = \dfrac{dy}{dx}
    These are hyperbolic functions Stephen.
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    (Original post by Mr M)
    These are hyperbolic functions Stephen.
    Just noticed. You think'd I'd realise that there is a h infront of the functions. Derp.
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    what id do from my little knowledge of maths is draw the graph and work out the gradient at points, but im a bit of a noob, forgive me
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    (Original post by StephenP91)
    Just noticed. You think'd I'd realise that there is a h infront of the functions. Derp.
    As it happens, your working provides a pretty good hint on how to proceed.
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    (Original post by kellywellydoodle)
    Yeah...question above says it all

    Can't find how to do it in any books
    Couple of ways. You can integrate a/(a^2 - x^2) and get tanh^-1(x/a) + k (best use a hyperbolic substitution). Hence in the case k = 0 this implies d/dx [tanh^-1(x/a)] = a/(a^2 - x^2). Or, differentiate directly as follows:

    Let y = tanh^-1(x/a) ... (thus require dy/dx)
    => tanhy = x/a
    => sech^2y(dy/dx) = 1/a
    => dy/dx = 1/[asech^2y]

    We need to replace sech^2y with a function of x and get the required result.
    tanhy = x/a => 1 - tanh^2y = 1 - x^2/a^2 => sech^2y = 1 - x^2/a^2 (relate tanh^2y to sech^2y using the identity)

    Replace sech^2y with the expression above. Fiddle a bit and voila.
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    (Original post by Mr M)
    As it happens, your working provides a pretty good hint on how to proceed.
    Borderline full solution to be honest.
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    (Original post by Pheylan)
    do you mean from first principles?

    if not, let y=artanh(x/a)

    then tanh(y)=x/a

    differentiate and rearrange for dy/dx
    I think the question requires it from first principles actually :confused:
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    (Original post by kellywellydoodle)
    I think the question requires it from first principles actually :confused:
    use the definition of the derivative
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    (Original post by kellywellydoodle)
    I think the question requires it from first principles actually :confused:
    If it required it from first principles then it would definitely say so. Otherwise, you can use any method you like. Check out my earlier post.
 
 
 
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