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How do i differentiate this?

  1. Feb 19, 2004 #1
    x^2 - xy - y^2 = 3
    how do i work out dy/dx?
     
  2. jcsd
  3. Feb 19, 2004 #2

    matt grime

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    either make y a function of x, which is trival but dull or just differentiate everything wrt x and use the chain rule and product rule.

    example, one you can do anyway, suppose xy=1

    then assuming this implicitly defines a function y in terms of x (notice that word implicit, look in your notes for a reference to it)

    then [tex]\frac{d}{dx}xy = \frac{d}{dx}1[/tex] that is to say [tex] (\frac{d}{dx}x)y + x(\frac{d}{dx}y) = 0[/tex]. of course dx/dx=1 so, [tex]y+ x\frac{dy}{dx}=0[/tex] and we see that [tex]\frac{dy}{dx} = -y/x[/tex] recall that y= 1/x, and we see [tex]\frac{dy}{dx} = -1/x^2[/tex] try and apply this idea to you example
     
    Last edited: Feb 19, 2004
  4. Feb 20, 2004 #3
    Implicit Differentiation...


    differentiate:
    x^2 - xy - y^2 = 3

    Implicit Differentiation:
    (d/dx)[x^2 - xy - y^2] = (d/dx)[3]

    (d/dx)[x^2] - (d/dx)[xy] - (d/dx)[y^2] = (d/dx)[3]

     
    Last edited: Feb 21, 2004
  5. Feb 21, 2004 #4

    Zurtex

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    I'm a little confused with the answers above. When differentiating some function of y with respect to x, is it not simply the derivate of the function with respect to y multiplies by the derivative of y with respect to x?

    Such that:

    [tex]x^2 - xy - y^2 = 3[/tex]

    [tex]2x - \left( y + x \frac{dy}{dx} \right) - 2y \frac{dy}{dx} = 0 [/tex]

    [tex]2x - y - x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0 [/tex]

    [tex]x \frac{dy}{dx} + 2y \frac{dy}{dx} + y - 2x = 0 [/tex]

    [tex](x + 2y) \frac{dy}{dx} = 2x - y[/tex]

    [tex]\frac{dy}{dx} = \frac{2x - y}{x + 2y}[/tex]

    Please say if this is wrong somehow, I need the practise.
     
  6. Feb 21, 2004 #5

    matt grime

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    that seems to be a correct assertion. this is called implicit differentiation.
     
  7. Feb 24, 2004 #6

    HallsofIvy

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    Yes, that was, in fact, exactly what matt grime orginally said!:smile:
     
  8. Feb 24, 2004 #7

    Zurtex

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    I didn't really understand the method though and got a bit confused so I wanted to check I knew how to do it, I was unsure why the example of if xy = 1 was given.

    Probably just the way I look at maths at the level I do it, I presume it would differ when I go to university.
     
  9. Feb 24, 2004 #8

    matt grime

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    I gave that example for two reasons: it was easy to rearrange and solve without implicit differentiation, so that you could see that you got the answer you thought you ought to get, and because I didn't want to just solve your homework problem for you, but to prompt you into trying it again for yourself, changind the details as necessary.
     
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