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Math Help - implicit help

  1. #1
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    implicit help

    how would you implicit differentiate this?
    x + y = tan(x - y)
    I can't isolate y'
    thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cubs3205 View Post
    how would you implicit differentiate this?
    x + y = tan(x - y)
    I can't isolate y'
    thanks
    what was the answer you got. tell us and we will show you how to isolate the y'
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  3. #3
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    i simplified to
    1 + y' = sec(x - y)^2(1 - y')
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cubs3205 View Post
    i simplified to
    1 + y' = sec(x - y)^2(1 - y')
    the square is on the secant, not the (x - y)

    1 + y' = \sec^2 (x - y) (1 - y')

    you don't have many options here, what can we do with this. there are brackets on the right, multiply them out, we get:

    1 + y' = \sec^2 (x - y) - y' \sec^2 (x - y)

    we want to solve for y', so get all the temrs with y' on one side of the equation

    \Rightarrow y' + y' \sec^2 (x - y) = \sec^2 (x - y) - 1

    y' is a common term on the left, factor it out:

    y' \left( 1 + \sec^2 (x - y) \right) = \sec^2 (x - y) - 1

    divide both sides by 1 + \sec^2 (x - y):

    \Rightarrow y' = \frac {\sec^2 (x - y) - 1}{\sec^2 (x - y) + 1}

    now this answer has a nice symmetry to it, but if you don't care about that, you can simplify the numerator on the right:

    \Rightarrow y' = \frac {\tan^2 (x - y)}{\sec^2 (x - y) + 1}
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