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    How would i differentiate y = tan(^-1) (1/X)

    ive done tany = 1/x

    ive got my answer down to (1+x^2)^-1

    when it should be -(1+x^2)^-1

    any help appreciated
    x
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    do you mean  tan^{-1} (\frac{1}{x}) ?
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    (Original post by zulu)
    How would i differentiate y = tan(^-1) (1/X)

    ive done tany = 1/x

    ive got my answer down to (1+x^2)^-1

    when it should be -(1+x^2)^-1

    any help appreciated
    x
    Personally i would do the chain rule with t=1/x. the differential of tan^-1(t) you may already know and if not its in the formula book.
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    Can you post your working please.
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    y=arctanx

    x=tany
    dx/dy = sec^2 x

    Sec^2x=tan^2x + 1

    dx/dy=tan^2x + 1

    dy/dx=1/(tan^2x + 1)=1/(x^2 + 1)

    I get the same answer as you.
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    (Original post by zulu)

    ive got my answer down to (1+x^2)^-1
    That's correct :yep:
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    Please note that in the question the op has set, it is tan^-1(1/x). The above two posts show it for tan^-1(x). The answer book has it right. The way to do this is by the chain rule.
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    In case anyone still needs the solution;

    y = tan^-1(1/x)
    tan y = 1/x
    cot y = x

    differentiate implicitly:
    d/dx (cot y) = d/dx (x)
    -cosec^2(y) dy/dx = 1
    dy/dx = 1/(-cosec^2(x)) = 1/(-cot^2(y)-1)

    since we know cot y = x
    dy/dx becomes 1/(-x^2 - 1) which is the same as 1/-(x^2 + 1) or -(1+x^2)^-1
 
 
 
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