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    This one is bothering me, I get a different answer to the back of the book

    y = artanh (x^2)

    This is what I have done:

    tanh y = x^2
    sech^2 y dy/dx = 2x
    (1 - tanh^2 y) dy/dx = 2x
    dy/dx = 2x/(1 - x^4)

    The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

    Can anyone help me here?
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    (Original post by Womble548)
    This one is bothering me, I get a different answer to the back of the book

    y = artanh (x^2)

    This is what I have done:

    tanh y = x^2
    sech^2 y dy/dx = 2x
    (1 - tanh^2 y) dy/dx = 2x
    dy/dx = 2x/(1 - x^4)

    The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

    Can anyone help me here?
    I think the book's wrong :rolleyes:
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    (Original post by Womble548)
    This one is bothering me, I get a different answer to the back of the book

    y = artanh (x^2)

    This is what I have done:

    tanh y = x^2
    sech^2 y dy/dx = 2x
    (1 - tanh^2 y) dy/dx = 2x
    dy/dx = 2x/(1 - x^4)

    The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

    Can anyone help me here?
    You're right. I've marked this one question [34] from Ex. 2A with a cross. I think the rest of the exercise is OK.

    Aitch
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    (Original post by Aitch)
    You're right. I've marked this one question [34] from Ex. 2A with a cross. I think the rest of the exercise is OK.

    Aitch
    ..apart from the answer to question 60 of course...

    [You may have a later edition than mine with the error corrected.]

    Aitch
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    Thanks, thought so
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    (Original post by Womble548)
    This one is bothering me, I get a different answer to the back of the book

    y = artanh (x^2)

    This is what I have done:

    tanh y = x^2
    sech^2 y dy/dx = 2x
    (1 - tanh^2 y) dy/dx = 2x
    dy/dx = 2x/(1 - x^4)

    The back of the book says dy/dx = 2x/(1 - x^2) but surely tanh^2 y = x^4?

    Can anyone help me here?
    Just thought Id add, the way you have done this question is okay but there is a quicker way if you have the differentiation of arctanh x given in yre formula booklet:

    d(arctanh x)/dx = 1/(1 - x^2)

    Therefore, to find y = arctanh (x^2) you simply substitute x^2 for x in the line above and mutiply this by the d(x^2)/dx as we must apply the chain rule:

    d(arctanh x^2)/dx = 1/(1-(x^2)^2) x d(x^2)/dx

    = 1/(1-x^4) x 2x
    = 2x/(1-x^4)

    I think this is the way yre textbook is trying to do the question but they forgot to square the x^2 when substituting it for x in d(arctanh x)/dx = 1/(1 - x^2)
 
 
 
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