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Chain rule, y^2'=2y*y'?

  1. May 21, 2009 #1
    1. The problem statement, all variables and given/known data
    x2+y2=1
    I want to differentiate this equation. I know that the answer is 2x+2y*y'=0.

    2. Relevant equations
    The chain rule.


    3. The attempt at a solution
    I don't understand how you get 2y*y' from y2. Shouldn't it just be 2x+2y=0?
     
  2. jcsd
  3. May 21, 2009 #2

    Cyosis

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    You're differentiating with respect to x right? Now lets apply the chainrule.

    [tex]
    \frac{dy^2}{dx}=\frac{dy^2}{dy}\frac{dy}{dx}
    [/tex]

    Can you work this out?
     
    Last edited: May 21, 2009
  4. May 21, 2009 #3
    I think that I see it now. Is it....

    2y=2y*y'*1?

    That is pretty much a random shot in the dark, but I see how its 2y*y' because its dy2/dy.
     
  5. May 21, 2009 #4

    Cyosis

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    Sorry but how can you think you see it now when it's a random shot in the dark. It seems you don't really understand how the chain rule works. What is [itex]\frac{dy}{dx}[/itex]?

    No dy^2/dy=2y.

    Do you understand the form I have written the chain rule in in my previous post. If not how do you use the chain rule usually?
     
    Last edited: May 21, 2009
  6. May 21, 2009 #5
    I don't understand it, but thats what I was asking origianaly. Can you explain it since I don't get it? Understood everything up untill this.
     
  7. May 21, 2009 #6

    Cyosis

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    Okay, to check your knowledge of the chain rule do you know how to differentiate, lets say [itex]\exp{x^2}[/itex]?
     
  8. May 21, 2009 #7
    Yeah...........its 2x.
     
  9. May 21, 2009 #8

    Cyosis

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    No it's [itex]2 x \exp x^2[/itex]. Either way if you have a composite function, f(g(x)) then [tex](f(g(x)))'=\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}[/tex]. Have you ever seen this?
     
  10. May 21, 2009 #9
    Yes, I have seen this. The composition Rule. I thought that exp meant example. What function is it. And for example sin(t)10=9*cos(t) right?
     
  11. May 21, 2009 #10

    Cyosis

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    Hehe, no exp means the exponent e. The derivative of [itex]\sin^{10}(t)=10\sin^9(t) \cos(t)[/itex].

    You can view y as a composite function in away as well since we are differentiating y with respect to x, y must be a function of x. Therefore we want to find the derivative of y(x)^2. Try to calculate the derivative now.
     
    Last edited: May 21, 2009
  12. May 21, 2009 #11
    Something like put y2 in for x2 so its (y2)2 so its 2(y2), then what?
     
  13. May 21, 2009 #12

    Cyosis

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    No apply the chain rule I have listed in post 8.
     
  14. May 21, 2009 #13
    I'm lost. I have no idea what to do. I am trying but where does the other functions come from? I usually get everything really easily but then when I run into something like this i'm wondering around in the dark. Can you explain how the numbers work? Like how it all works out and why.
     
  15. May 21, 2009 #14

    Mark44

    Staff: Mentor

    What Cyosis wrote in post 8 is the Chain Rule, using Newton's notation. There is no composition rule, althought the Chain Rule is used for finding the derivative of a function composition. In post 2 Cyosis used a different form of the chain rule, that uses Leibniz notiation.
     
  16. May 21, 2009 #15
    But I still don't get it. This isn't a homework question. You guy can just explain it to me. I need to learn this to be able to continue with my study of calculus. I have tried but I fail to see where the composition part comes in. All I have is y2.
     
  17. May 21, 2009 #16

    Cyosis

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    I am trying to find a form of the chain rule he recognizes.

    Schlynn could you differentiate [itex]\ln x^2[/itex] with a method you're used to? Please show all the steps.

    You differentiate y^2 with respect to x, not y! That's where the chain rule comes into play, it really is [itex]y(x)^2[/itex]. As for just explaining it, I am trying but just giving you the result won't do you any good.
     
  18. May 21, 2009 #17
    Havn't learned the derivative of ln but pretty sure its e right? Because e is the inverse of ln. Like ln(ex)=x. Is is 2e?

    EDIT:
    I reliase that that is wrong now. Because e is the derivative and anti-derivative of e.
     
  19. May 21, 2009 #18

    Cyosis

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    If you haven't learned the derivative of the logarithm ignore my previous post.

    Lets try it again another way!

    We want to find the derivative of [itex]y(x)^2[/itex]. Make the substitution u=y(x), then using the chain rule [itex](y(x)^2)'=(u^2)'*u'=2u*u'=2y(x)y'(x)=2y*y'[/itex].
     
  20. May 21, 2009 #19

    Mark44

    Staff: Mentor

    But the trouble is, y is a function of x, and you want the derivative with respect to x. If you were differentiating with respect to y, it would be easy -- d/dy(y2) = 2y and you're done.

    But what you want is d/dx(y2) which is not 2y.

    To use a better example, let y = (cos(x))3. To get a y value, you have to start with an x value, then evaluate the cosine of that x value, then cube that value. When you take the derivative with respect to x, the chain rule takes the composition into account.

    So dy/dx = d( (cos(x))3)/d( cos(x)) * d(cos(x))/dx

    In words, the derivative of y with respect to x is the derivative of (cos(x))3 with respect to cos(x) times the derivative of cos(x) with respect to x.

    So dy/dx = 3 (cos(x))2 * (-sin(x))

    Is that any clearer?
     
  21. May 21, 2009 #20
    I think this can help you http://www.calculus-help.com/funstuff/phobe.html" [Broken], I recommend you to watch the whole set of explanations, they are very clear :P
     
    Last edited by a moderator: May 4, 2017
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