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Math Help - Calc Implicit Differentiation Problem?

  1. #1
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    Calc Implicit Differentiation Problem?

    Okay, so I'm trying to do my Calc homework and for some reason, I'm getting stuck on a differentiation problem.

    So, it's basically (sin(x)/(x))+(sin(y)/(y))=C, where C is some constant. Obviously the constant would disappear, so I get: (after simplifying)
    ((xcos(x)-sin(x))/(x^2))+((y'cos(y')-sin(y'))/(y'^2))=0

    My problem is getting a y' on one side of the equation and the rest on the other side so that I can solve it. (Provided I actually differentiated it correctly above...)

    I do have one point on the curve, (pi/3, pi/4) if that helps at all, but I need to get it to the point where I can plug in a value for x and get the y value, and I'm kinda stuck.

    I would really appreciate any help. :-)

    Thank you!
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  2. #2
    Member eXist's Avatar
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    Be careful when differentiating, you have:

    \frac{xcos(x)-sin(x)}{x^2}+\frac{y'cos(y')-sin(y')}{y^2}=0

    When it should be:

    \frac{xcos(x)-sin(x)}{x^2}+\frac{y'cos(y)y-sin(y)y'}{y^2}=0

    Now, solve the about for y' by moving everything with it to one side, then factoring it out, then divide by what remains.

    This will get you an equation for y' in terms of y and x, and then use the (x,y) point you have given to find y' at that point.
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  3. #3
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    Thanks!

    Okay.... I think I get it. Thanks!!!

    Edit:

    Ahh - so I forgot about the inside / outside thing when differentiating.
    Last edited by Milou; January 1st 2010 at 07:55 PM.
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  4. #4
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    Hello, Milou!

    Given: . \frac{\sin x}{x} + \frac{\sin y}{y} \:=\:C . [1]

    Find y'.
    First, eliminate the fractions . . .

    Multiply by xy: \;\;y\sin x + x\sin y \:=\:Cxy

    Differentiate implicitly: . y\cos x + y'\sin x + x\cos y\cdot y' + \sin y \;=\;Cxy' + Cy

    Re-arrange terms: . y'\sin x + y'x\cos y - Cxy' \;=\;Cy - y\cos x - \sin y

    Factor: . y'(\sin x + x\cos y - Cx) \;=\;Cy - y\cos x - \sin y


    Therefore: . y' \;=\;\frac{Cy - y\cos x - \sin y}{\sin x + x\cos y - Cx}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This answer looks different from yours, which has no C.


    From [1], we have: . C \;=\;\frac{y\sin x + x\sin y}{xy}


    Substitute this into my answer and simplify.

    And we get: . y' \;=\;\frac{y^2(\sin x - x\cos x)}{x^2(y\cos y - \sin y)}

    This is answer you should get, using the Quotient Rule.

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  5. #5
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    Thank you!

    It's so nice of you to help me.

    I like your method - fractions confuse me, so that really helps. I'll try it for myself right now.

    Thanks!!!
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