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Math Help - Implicit Differentiation help

  1. #1
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    Implicit Differentiation help

    so I am trying to find \frac {dy}{dx} for x=tan(y)
    here's some of my work
    not sure if I'm right but I'm applying the product rule to find it

    u=tan

    u'=sec^2

    v=y

    v'= \frac {dy}{dx}

    tan \frac {dy}{dx} + sec^2 y=1

    the 1 is from the derivative of x

    then I subtracted sec^2 y from both sides

    which is where I got stuck

    the answer for the problem is supposed to be cos^2 y but I don't know how to get there...
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  2. #2
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    Well, it looks like you did some extra work. The derivative of tan(y) = sec^2(y) dy. Thus we have:

    x = tan(y)

    dx = sec^2(y) dy

    cos^2(y) = dy/dx

    We get to the final answer by dividing both sides by dx and then dividing both sides by sec^2(y).

    I hope this helps.
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  3. #3
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    after doing what you said, I get \frac {1}{sec^2 x} = \frac {dy}{dx}
    I don't get how that is cos^2 y
    is that a trig identity?
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  4. #4
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    sec^2(y) is the same as saying 1/cos^2(y). So, having 1/sec^2(y) is the same as:

    1
    ---
    1
    ----
    cos^2(y)

    We multiply by:

    cos^2(y)
    ----------
    cos^2(y)

    and we get cos^2(y).

    I hope this clears things up!
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