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Math Help - How To Get The Derivative of Y...? (Tougher Ones..)

  1. #1
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    How To Get The Derivative of Y...? (Tougher Ones..)

    I checked the answers in the back of the book, and we got different ones.

    Can anyone help?

    #1) y = s*(sqrt(1-s^2) + arccos s
    y' = sqrt(1-s^2) + s/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2)
    y' = sqrt(1-s^2)* (2*(sqrt1-s^2)/(2*(sqrt1-s^2) + s/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2) * (2/2)
    y' = (2(1-s^2) + s - 2)/(2*(sqrt(1-s^2)
    y' = (-2s^2+s)/(2*(sqrt(1-s^2)

    THe answer should be: (-2s^2)/(sqrt(1-s^2)

    #2) y = arccotangent(1/x) - arctangent(x)
    y' = -1/(1+(1/x)^2) * (1/x)' - 1/(1+x^2)
    y' = -1/(1+(1/x))*(-1/x) - 1/(1+x^2)
    y' = 1/(x^2+1) - 1/(1+x^2)
    y' = 0

    These two are wrong. But I tried.
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  2. #2
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    Quote Originally Posted by AlphaRock View Post
    I checked the answers in the back of the book, and we got different ones.

    Can anyone help?

    #1) y = s*(sqrt(1-s^2) + arccos s
    y' = sqrt(1-s^2) - s^2/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2) <-- problem here
    y' = sqrt(1-s^2)* (2*(sqrt1-s^2)/(2*(sqrt1-s^2) + s/(2*(sqrt1-s^2) - 1/(sqrt(1-s^2) * (2/2)
    y' = (2(1-s^2) + s - 2)/(2*(sqrt(1-s^2)
    y' = (-2s^2+s)/(2*(sqrt(1-s^2)

    THe answer should be: (-2s^2)/(sqrt(1-s^2)

    #2) y = arccotangent(1/x) - arctangent(x)
    y' = -1/(1+(1/x)^2) * (1/x)' - 1/(1+x^2)
    y' = -1/(1+(1/x^2))*(-1/x^2) - 1/(1+x^2) <-- problem here
    y' = 1/(x^2+1) - 1/(1+x^2)
    y' = 0

    These two are wrong. But I tried.
    See above in red. The second one corrected itself.
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