posted
Okay, I know there are a lot of people on this forum who are really really good at math. I'm decent- I got a 108% in precal last year. That's incredible for me. But I say I'm decent because there's one thing I don't understand.
Implicit differentiation.
I'm completely, hopelessly, helplessly lost. Does anyone out there know of a way to get me to understand this quicker?
Posts: 1591 | Registered: Jul 2005
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posted
Not sure if you've seen it, but Wikipedia has an article that deals with it a bit.
I can't think of a good way to explain it online though, but if I have time later and someone hasn't beaten me to it I can try.
Posts: 187 | Registered: Jan 2005
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posted
You are probably making it more difficult than it is. The basic premise is that we can find derivatives of functions even why y isn't completely seperated out. The key is remembering how to treat a function with y; basically you treat y like any function of x using the chain rule. Depending on the question you may want to solve for dy/dx or find the derivative for a spefic value.
Here is a quick example:
sin(2x-y2) = 17y d/dx(sin(2x-2y)) = d/dx(17y) cos(x-y)d/dx(2x-2y) (chain rule) = 17(dy/dx) cos(x-y)(2-2dy/dx) = 17(dy/dx) multiply through 2cos(x-y)-2dy/dx(cos(x-y)) = 17(dy/dx) move dy/dx's to one side -2dy/dx(cos(x-y)) - 17(dy/dx) = -2 cos(x-y) divide out leaving dy/dx dy/dx = -2cos(x-y)/(-2cos(x-y)-17)
The whole key is remembering the chain rule and getting dy/dx's on one side.
posted
Yeah. It's like KPhysicsGeek said. You just differentiate each side with resepect to the variables. Then you try and get all the dy/dx's to one side. It's not too bad.
Posts: 1960 | Registered: May 2005
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posted
If it helps, remember that y is actually y(x). So when you differentiate a function with y in it, you're actually differentiating a function within a function. So the derivative of [y(x)]^2 is 2y(x)*y'(x), or 2y(x)*dy/dx depending on whether you're using Liebnitz or Newton notation. You're using the chain rule here, as others have mentioned.
Once you've done the differentiation, you can solve the resulting equation for dy/dx to find the derivative of y.
Posts: 3546 | Registered: Jul 2002
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quote:Originally posted by genius00345: It's probably a little thing I like to call 'extra credit'.
Yes but then that begs the question, 108% of what? Certainly not possible points or maximum points. 108% of what?
Posts: 12591 | Registered: Jan 2000
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posted
It should be pointed out that you can't always get an explict expression for dy/dx even doing implicit differentiation. (ie you can't always solve for dy/dx)
Posts: 1621 | Registered: Oct 2001
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quote:It's probably a little thing I like to call 'extra credit'.
If enough extra credit is being given to average 108%, then a sickening amount of grade-inflation is going on.
Unlike a lot of teachers, I am not totally opposed to extra credit. But extra credit should not be capable of raising your average by as many as eight points.
posted
The class was weighted- as is Calculus, Physics, all AP classes, Honors English 3... you get the drift. My actual "grade" was a 98% of the maximum points, WITHOUT the 10% weight that they add to the report card. My precal teacher was the best teacher I've ever had- she's one of the NHS advisers, and has been teacher of the year quite a few times. She's incredible at what she does- where other teachers have a hard time getting students to understand things, she just got them to click with me. Keep in mind, most of my grades cluster around 97%-99%, without weights, and I always have a 100% in band, because it's a participation grade. But my teachers don't offer extra credit... I just did really well in precal.
Posts: 1591 | Registered: Jul 2005
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posted
Actually, that website really helped. Our teacher explained it in about 5 minutes, but never really told us when to use the d/dx, when to find a regular derivative, and nothing made sense. I think I get it now. Thanks.
Posts: 1591 | Registered: Jul 2005
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quote:Originally posted by KPhysicsGeek: You are probably making it more difficult than it is. The basic premise is that we can find derivatives of functions even why y isn't completely seperated out. The key is remembering how to treat a function with y; basically you treat y like any function of x using the chain rule. Depending on the question you may want to solve for dy/dx or find the derivative for a spefic value.
Here is a quick example:
sin(2x-y2) = 17y d/dx(sin(2x-2y)) = d/dx(17y) cos(x-y)d/dx(2x-2y) (chain rule) = 17(dy/dx) cos(x-y)(2-2dy/dx) = 17(dy/dx) multiply through 2cos(x-y)-2dy/dx(cos(x-y)) = 17(dy/dx) move dy/dx's to one side -2dy/dx(cos(x-y)) - 17(dy/dx) = -2 cos(x-y) divide out leaving dy/dx dy/dx = -2cos(x-y)/(-2cos(x-y)-17)
The whole key is remembering the chain rule and getting dy/dx's on one side.
posted
A Calculus-based anecdote of sorts: Three friends from my Calculus BC AP class (as well as the Physics C AP class) and I started a boy band known as the Gr4duates (the 4 was because we graduated in 2004). Our hit song was "Integrate My Heart," although the lyrics have long escaped me now. The Gr4duates really didn't go very far though, aside from doing group karaoke (singing Backstreet Boys and N'Sync songs) at our Project Graduation.
Posts: 1960 | Registered: May 2005
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posted
At functions' party, everybody is having fun. You can see Square Root and Addition grooving all around, Logarithm is boozing with some friends, Cosine is chatting some girls up. But there, in a dark corner, Exponential is sitting all by himself, sad and blue, his eyes fixed on the ground. Tangent and Arc Tangent approacch him and say: "Come on, what you're doing there! It's a party, you gotta have fun! Just integrate with the others!". Exponential glances them even more depressed and replies: "and how exactly would that change things?"
Posts: 1621 | Registered: Oct 2001
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quote:Originally posted by Icarus: Unlike a lot of teachers, I am not totally opposed to extra credit. But extra credit should not be capable of raising your average by as many as eight points.
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