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

  • Thread starter schlynn
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  • #1
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Homework Statement


x2+y2=1
I want to differentiate this equation. I know that the answer is 2x+2y*y'=0.

Homework Equations


The chain rule.


The Attempt at a Solution


I don't understand how you get 2y*y' from y2. Shouldn't it just be 2x+2y=0?
 

Answers and Replies

  • #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?
 
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  • #3
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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.
 
  • #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]?

but I see how its 2y*y' because its dy2/dy
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?
 
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  • #5
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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.
 
  • #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]?
 
  • #7
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Yeah...........its 2x.
 
  • #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?
 
  • #9
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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?
 
  • #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.
 
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  • #11
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Something like put y2 in for x2 so its (y2)2 so its 2(y2), then what?
 
  • #12
Cyosis
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No apply the chain rule I have listed in post 8.
 
  • #13
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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.
 
  • #14
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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?
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.
 
  • #15
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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.
 
  • #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.

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.
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.
 
  • #17
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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.
 
  • #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].
 
  • #19
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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.
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?
 
  • #20
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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
 
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  • #21
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The equation for the chain rule is (un)'*u'? Why didn't you say that? It makes since now. The dy/dx notation is confusing. I know that it's y with respect to x but it's not as easy to visualize as u'.

If I am still mistaken I still don't get it. Can you explain it with out y(x) and dy/dx? Just use y' and y2. The video on that site helped but I already knew that. Doing this with trig functions is super easy.
 
  • #22
Cyosis
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Why didn't I say that? First of all it is the worst form of the chain rule, it's not very clear. For example the first accent means differentiation with respect to u and the second one with respect to x. Secondly I asked you multiple times how you used the chain rule yourself.
 
  • #23
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Well not that I see how it works. Can someone help me explain the underlying math? I want to understand HOW. Not just a equation. I know that you guys have been trying to do that all along but maybe a different approach?
 
  • #24
Cyosis
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  • #25
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Thank you all for being patient with me. Now I can learn more.
 

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