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Leja
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 Posted: Thu Nov 29, 2007 8:36 pm Post subject: Implicit Differentiation and other adventures in Calculus |
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We're learning about implicit differentiation in Calc (which I'm sure you've guessed by the title ^_^), and it isn't so bad. But I'm confused as to the process: take, for example, x^2 + y^2 = 25. Going by the rules I've learned, I come up with 2x+2yy'=0. All I was told was that whenever you take the derivative of the y value, you have to multiply it by y prime.
But why?  |
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Kitty15
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| Posted: Thu Nov 29, 2007 11:31 pm Post subject: |
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Oh gosh. I finish my maths homework only to find someone talking about the same problems I've been forcing my way through?
Okay so you know what to do but you'd like to know why? Personally I just take the rules they give me and use them but I think I can explain this. In a normal differential equation, you would change the y to y' or dy and then differentiate the rest, right? Well implicit differentiation is just another version of that except you can have multiple terms of y. In which case you use the chain rule to, for example, change y^2 to 2y and then you have to differentiate the y which gives y'.
Does that make any sense? I'm not too good at explaining these things... |
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Leja
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| Posted: Fri Nov 30, 2007 9:12 pm Post subject: |
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| In which case you use the chain rule |
*this is one of those ah-hah! I get it now! moments*
Thanks, Heather  |
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| Posted: Fri Nov 30, 2007 10:40 pm Post subject: |
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I was the same way when I first started learning this. Now I really really love implicit differentiation because it makes everything about thirty thousand times easier. It's my favorite way to derive now. The limit process is my least favorite... it sucks.
But yeah, chain rule and all that good stuff. This was definitely not constructive, but I figured I'd throw in this nice little bit of babbling to let you know there are many of us struggling with the same things right now.  |
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Leja
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| Posted: Sat Dec 01, 2007 2:58 am Post subject: |
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Agh, sorry; I've been thinking about this all day (yes. dork. shush.)
How would the chain rule apply here? Like I said before, in y^2=4, if it were x^2=4, the derivative would just be 2x=0. So why does y include a y' when doing the derivative? I mean, in taking the derivative of the y^2 is one term, so shouldn't you get one term for the derivative since doesn't contain anything else? |
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| Posted: Sat Dec 01, 2007 7:43 am Post subject: |
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OMG I love implicit differentiation! 8D <3!!!!
Ok, I'm gonna try and explain it to you using a different shorthand (?) for differentiation. I think the form with primes is Newtonian. So, I don't know what this form is called. You may know it already, but this makes a bit more sense when you use this form. It's spiffy! This would be a bit easier in mathtype, though. Unfortunately the closest thing I have to that is drawing it in paint. Which I may do after I post this. ^-^
This may get a little confusing, because I've been explained/taught about three different ways to do it. XD It may also lack words, because math explains itself to me and so it's hard to transform it into words. I apologize if I just further confuse you!
For this form you write f(x)' as d/dx * f(x).
This means that you are taking the derivative of f(x) with respect to x.
In implicit, this makes:
x^2 + y^2 = 25
d/dx (x^2 + y^2) = d/dx (25)
d/dx (x^2) + d/dx (y^2) = d/dx (25)
!Error
What's going on? Well, you're trying to take the derivative of y with respect to x, but there's no x in y! This is when the chain rule kicks in. You can say that the first function was x^2 and the second one that was inserted into the first was y. Also, what you want when you simplify the derivative is dy/dx to equal the derivative. If I may give a sidebar in the middle of this derivative to show you....
[y = x^2 + 3x +4]
[d/dx (y) = d/dx (x^2) + d/dx (3x) + d/dx (4)]
[dy/dx = 2x + 3]
Basically, dy/dx = f(x)' or y'. ^-~d Now let's continue with our derivative...
2x + 2y*dy/dx = 0
2y*dy/dx = -2x
dy/dx = -2x/2y
dy/dx = -x/y
That's it!
I hope this was helpful and at least halfway apprehensible!
(this whole time I'm hoping Snoink doesn't jump in here before I can post. I'll be very bitter if she does...  ) |
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| Posted: Sat Dec 01, 2007 8:06 am Post subject: |
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| If you love implicit differentiation, just wait until the miracle of partial differentiation. (Seriously). ^_^ |
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| Posted: Sat Dec 01, 2007 4:57 pm Post subject: |
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I'm still confused. How does d/dx(y^2) turn into 2y*dy/dx? I understand the part about using the power rule to come up with 2y, but where does the dy/dx come from? |
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| Posted: Sat Dec 01, 2007 6:40 pm Post subject: |
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It's the chain rule. You can say that y^2 is made up of two equations:
The first = x^2
The second is y
Differentiate the first and you get:
d/dx (x^2) = dx/dx * 2x = 2x
Differentiate the second and you get:
d/dx (y) = dy/dx
Put it together and:
2y * dy/dx
That help? |
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| Posted: Sat Dec 01, 2007 6:50 pm Post subject: |
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Amelia:
Consider: find the derivative of f(x)^2. Then D{f(x)^2}=2f(x)D{f(x)}=2f(x)f'(x).
Saying the implicit differentiation is an application of the chain rule is necessary but not sufficient for understanding what's really going on here. Typically, every equation you've encountered up until this point in your mathematics career has been of the form, y=f(x), where y'=f'(x). This is a so-called "Explicit Function" because y is defined totally in terms of x.
But what if y can't be expressed in terms of x? Then what happens? Well, then we have a function with both x and y as the variables; namely, F(x,y)=0. It might help to think about it like this: F(x,f(x)). Now, we want to know what D{F(x,f(x))} is. So let's use your original equation:
x^2 + y^2 = 25, or, in my notation, F(x,f(x))=x^2+f(x)^2=25.
So we want to know what F'(x,f(x)) is. To do this, since x is the variable, we simply differentiate as usual with respect to x. Thus F'(x,f(x))=2x+D{f(x)^2}. Now we use the above formula and see F'(x,f(x))=2x+2f(x)f'(x)=2x+2yy'=0, which implies (by simple algebraic maneuvers) that y'=-x/y. Thus the slope of F(x,y) at (x,y)=(3,4) is given by y'=-0.75.
Hope that helped some.
Take care!
Brad |
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Leja
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| Posted: Sat Dec 01, 2007 8:38 pm Post subject: |
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| I think I get it. At least for now XD Thanks all. |
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| Posted: Sat Dec 01, 2007 9:48 pm Post subject: |
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| That's the kind of encouraging endorsement I like to hear. =] |
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