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Thread: 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
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    Be careful when differentiating, you have:

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

    When it should be:

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

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

    This will get you an equation for $\displaystyle y'$ in terms of $\displaystyle y$ and $\displaystyle x$, and then use the (x,y) point you have given to find $\displaystyle y'$ 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; Jan 1st 2010 at 07:55 PM.
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  4. #4
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    Hello, Milou!

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

    Find $\displaystyle y'.$
    First, eliminate the fractions . . .

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

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

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

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


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


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


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


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


    Substitute this into my answer and simplify.

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

    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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