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Homework Help: Implicit differentiation

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data
    find [tex]\frac{\mathrm{d}y}{\mathrm{d}x}[/tex] where y is defined implicitly as a function of x

    2. Relevant equations

    3. The attempt at a solution


    [tex]\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-\sin(xy)-xy\cos(xy)}{x^2 \cos(xy)}[/tex]

    Apparently the correct answer is [tex]\frac{1-y\cos(xy)}{x}[/tex]

    Thanks !!
  2. jcsd
  3. Jan 2, 2010 #2
    This is a wild guess:


    [tex]\frac{d[sin(xy)]}{dx}=0\Rightarrow cos(xy) \frac{d(xy)}{dx}=0 \Rightarrow cos(xy)[y+x\frac{dy}{dx}]=0 \Rightarrow x\frac{dy}{dx}=-y \Rightarrow \frac{dy}{dx}=-\frac{y}{x}[/tex]

    I have no confidence on this, just a though. And still have y on the right side.
    Last edited: Jan 2, 2010
  4. Jan 2, 2010 #3
    I can easily see that dy/dx=-y/x when sin(xy)=0, so I'm not sure how that helps, but thanks anyway !
  5. Jan 2, 2010 #4


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    What do you mean you are "not sure how that helps"? dy/dx= -y/x is the answer!

    But apparently you have written something incorrectly.
    [tex]\frac{1- ycos(xy)}{x}[/tex] certainly is NOT the derivative of x sin(xy)= x and it does not make sense to me that they would leave that extraneous "x" in the equation.
  6. Jan 2, 2010 #5
  7. Jan 2, 2010 #6
    OK, I've got it now. Seems I've been led astray by the "apparent" answer.
  8. Jan 2, 2010 #7
    Sorry, I am still not sure on this.....

    Is it really valid to cancel x here ?

    How did cos(xy) disappear ?
    And why is my attempt in my initial post wrong ?
  9. Jan 2, 2010 #8
    OK, more on this. Finally I think I have it. Both methods result in -y/x. In the simple method where we cancel x first, we have the proviso that x is not equal to 0 and in the method I wrote initially, my "result" also relies on x not equal to 0 (and also cos(xy) not equal to zero). Once this assumption is made then my result simplifies to -y/x

    Glad that's out of the way.
  10. Jan 2, 2010 #9
    You expand it out, move one term to the right side, cos(xy) is on both side and can be cancel just like x on your original equation. I only wrote down every other step to save me typing all the Latex!!!! Write it out, you'll see.

    One thing still bug me is there is still y on the right side and the question specified y is a function of x so I expect there is no y on the right side!!!!
  11. Jan 2, 2010 #10


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    There's nothing wrong with dy/dx=(-y/x) as the solution to an implicit differentiation. To really find y(x) you could solve that differential equation. Or better yet just look back at sin(xy)=1. That means xy is a constant, right? It really seems odd to have that x on both sides of the problem definition. I suspect it's a typo.
  12. Jan 2, 2010 #11
    I thought about that last night already, [tex]xy=\frac{2n\pi}{2}\Rightarrow y=x\frac{2n\pi}{2}[/tex].

    So [tex]\frac{dy}{dx}=-\frac{2n\pi}{2}[/tex]

    It just don't look like a normal answer to me!!!! Or is it just me??!!
  13. Jan 2, 2010 #12


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    [tex]xy = n\pi[/tex]
    [tex]y = \frac{n\pi}{x}[/tex]
  14. Jan 2, 2010 #13
    I am confussing myself!!!!!

    For [tex]sin(xy)=1 \Rightarrow xy=\frac{\pi}{2}[/tex]

    This is the only value in each cycle of 2\pi. Therefore:

    [tex]xy=\frac{\pi}{2}+2n\pi[/tex] n=0,1,2.....


    Am I getting this? I seem to have a mental block on this!!!!
  15. Jan 2, 2010 #14


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    Actually, you want y=(2n+1/2)*pi/x. It's easy enough to check that dy/dx=-y/x alright. Still a funny problem for implicit differentiation.
  16. Jan 2, 2010 #15
    I know!!!
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