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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    I was working though my problem book for Calculus and I was able to do the simple implicit differentiation questions such as:

    x^2y = 1

    (I don't know how to do derivatives in Latex, somebody help!)

    d/dx ( x^2y) = d/dx (1)

    ( x^2)d/dx(y) + y(2x) = 0

    ( x^2)(dy/dx) + 2xy = 0

    Rearrange and you get:

    dy/dx = -2y/x

    So I could do that and I get the basic idea of what we're doing, but there were two questions that really stumped me.

    (a) xy = tan (xy)

    (b) x^2 = (arctan(y)) / (1 + y^2)

    I am completely lost, any help would be muchly appreciated

    EDIT: for (a) I did manage to get a line that said:

    (x)(dy/dx) + y = sec(xy)

    That on the right track?
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  2. #2
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    Quote Originally Posted by Russ View Post
    I was working though my problem book for Calculus and I was able to do the simple implicit differentiation questions such as:

    x^2y = 1

    (I don't know how to do derivatives in Latex, somebody help!)

    d/dx ( x^2y) = d/dx (1)

    ( x^2)d/dx(y) + y(2x) = 0

    ( x^2)(dy/dx) + 2xy = 0

    Rearrange and you get:

    dy/dx = -2y/x

    So I could do that and I get the basic idea of what we're doing, but there were two questions that really stumped me.

    (a) xy = tan (xy)

    (b) x^2 = (arctan(y)) / (1 + y^2)

    I am completely lost, any help would be muchly appreciated

    EDIT: for (a) I did manage to get a line that said:

    (x)(dy/dx) + y = sec(xy)

    That on the right track?
    First of all, once you get to  \frac{dy}{dx} = \frac{-2y}{x} , you should write  y in terms of  x by rearranging the original equation. So  y = \frac{1}{x^2} , therefore

     \frac{dy}{dx} = \frac{-2\bigg(\frac{1}{x^2}\bigg)}{x} = \frac{-2x^{-2}}{x} = \frac{-2}{x^3}

    Righty.

    Fir the tan one, remember that you must use the chain rule if the argument is not exactly x. Your LHS is fine. But:

     \frac{d}{dx} tan(xy) = \sec^2(xy) \times \frac{d}{dx} (xy) = y\sec^2(xy)
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  3. #3
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    Oh really? Okay. I can do that from now on.

    Hmmm, i'm not sure why I didn't have a sec^2 (xy) in there
    Thanks I will work on this some more
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  4. #4
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    Quote Originally Posted by Russ View Post
    Oh really? Okay. I can do that from now on.

    Hmmm, i'm not sure why I didn't have a sec^2 (xy) in there
    Thanks I will work on this some more
    For the last one:

     x^2 = \frac{\arctan(y)}{1+y^2}

    I would start by multiply both sides by  1+y^2 , which gives@

     x^2(1+y^2) = \arctan(y)

    Hence:

     x^2 + x^2y^2 = \arctan(y)

    Now do your differentiation.

    For the RHS:

     \frac{d}{dx} \arctan(y) = \frac{dy}{dx} \frac{d}{dy} \arctan(y)

    And  \frac{d}{dy} \arctan(y) = \frac{1}{1+y^2}
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  5. #5
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    Ah nice, thanks. I think I was on track for that, but I was trying to use the quotient rule and things were getting complicated.
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  6. #6
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    Quote Originally Posted by Mush View Post

     \frac{d}{dx} tan(xy) = \sec^2(xy) \times \frac{d}{dx} (xy) = y\sec^2(xy)
    Incorrect. the differentiation of xy is y+xy'. Therefore  (tan(xy))' = (sec^2(xy))(y+xy')
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