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Math Help - Question about Implicit Differentiation

  1. #1
    Member FalconPUNCH!'s Avatar
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    Question about Implicit Differentiation

    My question is when do you multiply by y' or \frac d{dx} when trying to get the derivative?

    I know that in this problem x^3 + y^3 = 25 that you can do it  3x + 3y (y') = 0

    In more complicated equations I don't know when to use it. For example in this problem

     sin(x) + cos(y) = sin(x)cos(y)

    I don't know where y' belongs. Can anyone help me find out when I have to multiply by y'. I know you are supposed to when you're using the chain rule but I don't know how to do it in that type of problem.

    -FP!
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  2. #2
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    You're not simply multiplying by y'. y' is a component in the chain rule due to the implicit nature of the problem.

    I believe your first problem is incorrect.

    x^3 + y^3 = 25<br />

    <br />
3x^2 + 3y^2 (y') = 0<br />

    Here you are taking the derivative of every term. Because this is implicit, the derivative of y^3 is 3(y)^2 * (y').
    Likewise if your problem was:
    <br />
x^3 + \sin{y^2} = 25<br />
    than,
    <br />
3x^2 + \cos{y^2} * \frac {d(y^2)}{dy} = 0
    [/color]
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  3. #3
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by sixstringartist View Post
    You're not simply multiplying by y'. y' is a component in the chain rule due to the implicit nature of the problem.

    I believe your first problem is incorrect.

    x^3 + y^3 = 25<br />

    3x^2 + 3y^2 (y') = 0<br />

    Here you are taking the derivative of every term. Because this is implicit, the derivative of y^3 is 3(y)^2 * (y').
    Likewise if your problem was:
    <br />
x^3 + \sin{y^2} = 25<br />
    than,
    <br />
3x^2 + \cos{y^2} * \frac {d(y^2)}{dy} = 0
    [/color]
    Yeah I forgot the to square each term. There are problems that I have trouble doing this in though. Problems like the one I mentioned in my OP  sin(x) + cos(y) = sin(x)cos(y)

     cos(x) + (-sin(y)) y' = cos(x)cos(y) + sin(x)(-sin(y)) y'

     (-sin(y)) y' - sin(x)(-sin(y)) y' = cos(x)cos(y) - cos(x)

     y' (-sin(y) - sin(x)(-sin(y)) = cos(x)cos(y) - cos(x)

     y' = \frac {cos(x)cos(y) - cos(x)}{-sin(y) - sin(x)(-sin(y))}

    Is that the correct way to use it?
    Last edited by FalconPUNCH!; October 23rd 2007 at 07:07 PM.
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  4. #4
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    Yes that appears to be correct. I looked over your math, but I didnt write it out myself so I may have overlooked an algebraic error but the method of implicit differentiation appears correct.
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  5. #5
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by sixstringartist View Post
    Yes that appears to be correct. I looked over your math, but I didnt write it out myself so I may have overlooked an algebraic error but the method of implicit differentiation appears correct.
    Ok thank you for your help. I understand it much more clear now.
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  6. #6
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    Quote Originally Posted by FalconPUNCH! View Post
    My question is when do you multiply by y' or \frac d{dx} when trying to get the derivative?

    I know that in this problem x^3 + y^3 = 25 that you can do it  3x + 3y (y') = 0

    In more complicated equations I don't know when to use it. For example in this problem

     sin(x) + cos(y) = sin(x)cos(y)

    I don't know where y' belongs. Can anyone help me find out when I have to multiply by y'. I know you are supposed to when you're using the chain rule but I don't know how to do it in that type of problem.

    -FP!
    Here is one way.

    Remember with respect to what variable you are differentiating.
    If it is with respect to x, then all differentiated variables will end up with dx as the denominator. (Even me, I'm lost in that explanation. ] In actual, it is like this:
    x^3 +y^3 = 25
    Differentiate both sides with respect to x,
    (3x^2)(dx/dx) +(3y^2)(dy/dx) = 0
    (3x^2)(1) + (3y^2)(dy/dx) = 0
    3x^2 +(3y^2)y' = 0
    y' = -3x^2 / 3y^2 = -x^2 /y^2

    ----------------------------------------
    sinX +cosY = sinXcosY

    Differentiate both sides with respect to X,
    cosX dX/dX +(-sinY) dY/dX = sinX[-sinY dY/dX] +cosY[cosX dX/dX]
    cosX -sinY Y' = -sinXsinY Y' +cosXcosY
    -sinY Y' +sinXsinY Y' = cosXcosY -cosX
    Y'[sinXsinY -sinY] = cosXcosY -cosX
    Y' = [cosXcosY -cosX] / [sinXsinY -sinY]
    Y' = [cosX(cosY -1)] / [sinY(sinX -1)]
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