I Can't Understand dy/dx when you throw in "u" - Tilted Forum Project Discussion Community

tspikes51 · 03-09-2005, 05:12 PM

I just have absolutely no clue exactly what goes on. My prof. only explained derivatives as what you get when you use the power rule, and that dy/dx is just a fancy term for derivative. So then how in the fuck are you supposed to get "u" in there if dy/dx means the exact same thing as derivative??? This guy isn't a good teacher at all and my textbook isn't helping any. Any help would be greatly appreciated.

vinaur · 03-09-2005, 05:31 PM

I assume you are talking about the chain rule.

Let's say that f(x)=5x+2 while g(x)=x^5, so g(f(x))=(5x+2)^5.

To take the derivative, you would have to raise the polynomial to a power, which can be annoying although pascal's triangle (oh and pyramid) make it easier.

So the chain rule says that d (g(f(x)))/dx = d (g(f(x)))/d (f(x)) * d (f(x))/dx, so d (g(f(x)))/dx = 5*(f(x))^4 * 5 , which in turn equals to 5*(5x+2)^4 * 5.

For simplicity f(x) is usually replaced by u, so the derivative sounds something like this: derivative of g(u) is equal to the derivative of g(u)/du times du/dx.

Hope this sort of clarified things up. I know it's kinda hard to follow all the math when it's written this way, but that's the best I can do (well, I could use webassign, but I'm too lazy).

To put it in the simplest terms that I can think, y is a function of x and so is u, so they can be used interchangeably. You treat u as if it was just a regular y.

Slavakion · 03-09-2005, 07:18 PM

Let's take a function. y=3x
The derivative of this can be written as dy/dx = 3. I assume by "u" you mean du/dx, right? Well, the common form of the derivative expression thing is dy/dx because you're usually taking the derivative of y with respect to x. In other words, you're solving for y, and finding the derivative of it.

You could have any other variables in there, like ds/dt or dx/dp. The thing that signifies a derivative is the "d" that precedes the other variable. So we have y, and we have dy. The general form for a derivative (which is what's used in theorems) is du or du/dx. Just sub in y or whatever.

If you're wondering why it's always dy over dx, it's because you're essentially dividing. y=3x would be differentiated into dy=3dx, which would be simplified to dy/dx=3.

There ya go, an unfocused crash course in differential variables. If your teacher really does suck, I suggest going to Borders and picking up an AP study guide for calculus. They'll come with AB and BC in it, so you'll be set for about a semester and a half of calculus. Or you'll be set with your AP class if you're still in high school. Good luck.

tspikes51 · 03-10-2005, 09:55 AM

Thanks for the help. I'm still a bit confused, but I'll take a look at those books.

Stompy · 03-10-2005, 12:30 PM

The only time I've ever used "u" with derivatives was like so:

Find the derivative of 1/(4-x^2)

u = x^2
du/dx = 2x
du = 2x dx

the problem then becomes du/(4-u)^2

Since u is x^2, and du is 2x, it becomes 2x/(4-x^2)^2.

In other words... u substitution. (I skipped a few steps, but you get the idea)

vinaur · 03-10-2005, 12:37 PM

Quote:

Originally Posted by Stompy

The only time I've ever used "u" with derivatives was like so:

Find the derivative of 1/(4-x^2)

u = x^2
du = 2x dx

the problem then becomes du/(4-u)^2

Since u is x^2, and du is 2x, it becomes 2x/(4-x^2)^2.

In other words... u substitution.

The u substitution is more useful when you get to integrations. In diferentiation the u substitution mentioned above is what I was talking about - chain rule.

Stompy · 03-10-2005, 12:47 PM

I haven't had to do u substitution in ages, since Calc 1 (almost a year).

Calc 2 was more about the other advanced methods of integration, and infinite series.

I'm in Calc 3 now, and we're just doing some very basic stuff with vectors. Haven't even touched on integration/differentiation yet.. but I know when we get to it, I'm gonna be a bit rusty.

Slavakion · 03-11-2005, 10:19 AM

Oh, I completely forgot about u substitution. Hmm, hafta remember that for the test.

userasdf1234 · 03-15-2005, 08:21 AM

If your still having trouble you should try The Complete Idiot's Guide to Calculus. The examples in the book are not as difficult as the ones you would find in your textbook but it helps you get the general idea. I found the CIG to Calculs very useful. If you're not willing to actually put up money for the book the library might have it.

03-09-2005, 05:12 PM	#1 (permalink)
tspikes51 Fuckin' A Location: Lex Vegas	I Can't Understand dy/dx when you throw in "u" I just have absolutely no clue exactly what goes on. My prof. only explained derivatives as what you get when you use the power rule, and that dy/dx is just a fancy term for derivative. So then how in the fuck are you supposed to get "u" in there if dy/dx means the exact same thing as derivative??? This guy isn't a good teacher at all and my textbook isn't helping any. Any help would be greatly appreciated. __________________ "I'm telling you, we need to get rid of a few people or a million." -Maddox

03-09-2005, 05:31 PM	#2 (permalink)
vinaur Insane	I assume you are talking about the chain rule. Let's say that f(x)=5x+2 while g(x)=x^5, so g(f(x))=(5x+2)^5. To take the derivative, you would have to raise the polynomial to a power, which can be annoying although pascal's triangle (oh and pyramid) make it easier. So the chain rule says that d (g(f(x)))/dx = d (g(f(x)))/d (f(x)) * d (f(x))/dx, so d (g(f(x)))/dx = 5(f(x))^4 5 , which in turn equals to 5(5x+2)^4 5. For simplicity f(x) is usually replaced by u, so the derivative sounds something like this: derivative of g(u) is equal to the derivative of g(u)/du times du/dx. Hope this sort of clarified things up. I know it's kinda hard to follow all the math when it's written this way, but that's the best I can do (well, I could use webassign, but I'm too lazy). To put it in the simplest terms that I can think, y is a function of x and so is u, so they can be used interchangeably. You treat u as if it was just a regular y. Last edited by vinaur; 03-09-2005 at 05:34 PM..

03-10-2005, 09:55 AM	#4 (permalink)
tspikes51 Fuckin' A Location: Lex Vegas	Thanks for the help. I'm still a bit confused, but I'll take a look at those books. __________________ "I'm telling you, we need to get rid of a few people or a million." -Maddox

03-10-2005, 12:30 PM	#5 (permalink)
Stompy Banned from being Banned Location: Donkey	The only time I've ever used "u" with derivatives was like so: Find the derivative of 1/(4-x^2) u = x^2 du/dx = 2x du = 2x dx the problem then becomes du/(4-u)^2 Since u is x^2, and du is 2x, it becomes 2x/(4-x^2)^2. In other words... u substitution. (I skipped a few steps, but you get the idea) __________________ I love lamp. Last edited by Stompy; 03-10-2005 at 12:36 PM..

03-10-2005, 12:47 PM	#7 (permalink)
Stompy Banned from being Banned Location: Donkey	I haven't had to do u substitution in ages, since Calc 1 (almost a year). Calc 2 was more about the other advanced methods of integration, and infinite series. I'm in Calc 3 now, and we're just doing some very basic stuff with vectors. Haven't even touched on integration/differentiation yet.. but I know when we get to it, I'm gonna be a bit rusty. __________________ I love lamp.

03-09-2005, 07:18 PM	#3 (permalink)
Slavakion Mjollnir Incarnate Location: Lost in thought	Let's take a function. y=3x The derivative of this can be written as dy/dx = 3. I assume by "u" you mean du/dx, right? Well, the common form of the derivative expression thing is dy/dx because you're usually taking the derivative of y with respect to x. In other words, you're solving for y, and finding the derivative of it. You could have any other variables in there, like ds/dt or dx/dp. The thing that signifies a derivative is the "d" that precedes the other variable. So we have y, and we have dy. The general form for a derivative (which is what's used in theorems) is du or du/dx. Just sub in y or whatever. If you're wondering why it's always dy over dx, it's because you're essentially dividing. y=3x would be differentiated into dy=3dx, which would be simplified to dy/dx=3. There ya go, an unfocused crash course in differential variables. If your teacher really does suck, I suggest going to Borders and picking up an AP study guide for calculus. They'll come with AB and BC in it, so you'll be set for about a semester and a half of calculus. Or you'll be set with your AP class if you're still in high school. Good luck.

03-11-2005, 10:19 AM	#8 (permalink)
Slavakion Mjollnir Incarnate Location: Lost in thought	Oh, I completely forgot about u substitution. Hmm, hafta remember that for the test.

03-15-2005, 08:21 AM	#9 (permalink)
userasdf1234 Tilted	If your still having trouble you should try The Complete Idiot's Guide to Calculus. The examples in the book are not as difficult as the ones you would find in your textbook but it helps you get the general idea. I found the CIG to Calculs very useful. If you're not willing to actually put up money for the book the library might have it.