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## #1 2005-11-09 19:08:01

mikau
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### differentiating

I just completed lesson 30 in my 158 lesson calculus book. I haven't learned too much new stuff yet but from what I've seen so far, all I can say is this.

wigout()
{

Oh my FREAKIN GOSH! Calculus ROXXORS t3h big onez!111;

return sanity;
}

The reasoning and logic is absolutely astounding!

Anyways lets get down to buisiness. I just learned about finding the differential dy. Which you find by taking the derivate of the function and multiplying it by dx.

But a few minutes later, a problem simply tells me to differentiate. Huh?

At first I assumed the meant the same as finding the differantial dy, but it turned out the answer was simply the derivitive of the function. Huh? Does differentiate simply mean "find the derivitive"?

A logarithm is just a misspelled algorithm.

## #2 2005-11-10 04:05:43

ryos
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### Re: differentiating

See my post, "What's a dx?". I'm still not 100% clear on that one, even after they explained it to me.

El que pega primero pega dos veces.

kylekatarn
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wigout()
<SYNTAX ERROR>

## #4 2005-11-10 10:06:02

mikau
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### Re: differentiating

lol!

A logarithm is just a misspelled algorithm.

## #5 2005-11-11 12:05:23

mikau
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### Re: differentiating

Ok I'm tripping again. This happend in trig, a small lack of understanding of a concept occurs and it snowballs as I continue untill I don't know what the devil the book is talking about, and I can no longer move forward. I'm forced to go back, and review and think about for a few days. Darn.

Ok, it seems the problem solving patterns are relatively straightforward but I might as well be learning card tricks if I don't understand what I'm doing. I'm having trouble understanding all this dy dx buisiness. I think i'm having more trouble understanding what the notations stand for.

For I a while I thought dy/dx was just a notation for derivative, but it seems this fraction can be manipulted by multiplying by dx or dividing by dy. Now when we find the differential dy we essentially find the value of the rise, and the rise alone. (rather then the rise over the run) we find this by multiplying the slope (which is the rise over the run) by the run, to cancel out the run and leave the rise. But what is the run? What are we multiplying by? It seems the run could be anything! This is difficult because the slope of the function depends entirely on which point on the line we are talking about, and is dependant on that value of x. So when we multiply by dx or divide by dy, it seems we are multiplying or diving by undefined variables. Very confusing.

Ok look at this problem:

u = (x^2)y    find du

Ok I really don't know what they mean by "du" distance u? From what?

I know how to solve the problem but I really have no idea what I'm doing. :-(

A logarithm is just a misspelled algorithm.

## #6 2005-11-13 11:58:19

mikau
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### Re: differentiating

Hmm... well I reviewed everything I learned about derivities and differentials last night. I think I'm a bit more clear on it.

I think the reason for the confusion is dy/dx is just an expression of a ratio. Therefore even if the ratio is 3/4, dy does not have to be 3 and dx does not have to be 4.

So the value of the indivual values dy and dx is not only infinite (since any ratio can be expressed with any number, OR any denominator) but this ratio is also undefined untill we actually insert an x value into the derivitate of the function, to find the slope of the line at that value of x in the function.

So I suppose its hard to define dy and dx because:

dy/dx = 3/x

Not only is the ratio undefined undefined until we insert a value for x, but even after that the individual values of dy or dx could be any two numbers with a ratio of 3/x.

For instance, if x = 6 the ratio is 1 to 2. So dy could equal 1 if dx equals 2, dy could equal 4 if dx equals 8. I think this double dependance is whats confusing me and preventing me from understanding what values they stand for. Because each individual variable can have any value if the other variables are given the right values.

Also I was looking at some of the manipulations of dy/dx, multiplying the expression by dx or dividing by dy, and it seems while dy and dx seem to be nothing but notations, dx/dx can be cancelled out as well as dy/dy. Yet the manipulations seem to change the form of the notations to change the expression into a differant form. To me, this sounds an awfull lot like unit multipliers.

2 feet. Convert to inches.

2 ft * 12 inches/1 ft    12 inches in one foot. Multiply by 12 inches over 1 foot.
24 inches  ft/ft    ft over ft canels out.

24 inches.

The dy/dx manipulations seem very much like the same thing. Am I correct?

A logarithm is just a misspelled algorithm.

## #7 2005-11-13 16:15:57

MathsIsFun

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### Re: differentiating

#### mikau wrote:

Also I was looking at some of the manipulations of dy/dx, multiplying the expression by dx or dividing by dy, and it seems while dy and dx seem to be nothing but notations, dx/dx can be cancelled out as well as dy/dy. Yet the manipulations seem to change the form of the notations to change the expression into a differant form.

Some examples of that (ie different forms of same equation) would be good. I am interested to see where this leads, coz one day I might try to explain differentiation on the website (with animations maybe!).

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #8 2005-11-13 18:06:54

mikau
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### Re: differentiating

Some examples of that (ie different forms of same equation) would be good. I am interested to see where this leads, coz one day I might try to explain differentiation on the website (with animations maybe!).

By that, are you saying your not sure what I mean?

Btw, did you ever teach calculus?

Last edited by mikau (2005-11-13 18:14:24)

A logarithm is just a misspelled algorithm.

## #9 2005-11-13 18:47:20

MathsIsFun

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### Re: differentiating

I just want to pick apart a nice example, and was thinking you might have one to work when you said "Yet the manipulations seem to change the form of the notations to change the expression into a differant form." We could analyze the different forms. (I didn't teach calculus, I just studied a few subjects)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #10 2005-11-18 12:47:53

mikau
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### Re: differentiating

Well I mean like implicit differentiation.

x^2 - y^2 = 9

differentiate

2x dx - 2y dy = 0

We can now find dy/dx by dividing the equation by dx.

2x dx/dx = 2y dy/dx

dx/dx cancels out,

2x = 2y dy/dx

Divided both sides by 2y.

2x/2y = dy/dx

simplified

dy/dx = x/y.

You can also find dx/dy by dividing by dy after we differentiate.

original equation x^2 - y^2 = 9

differentiated:

2x dx - 2y dy = 0

Divide by dy (we're trying to find dy/dx)

2x dx/dy -2y dy/dy = 0

dy/dy cancels out.

2x dx/dy - 2y = 0

solved for dx/dy and simplified.

dx/dy = y/x.     (note dy/dx was x/y and dx/dy was y/x)

We can also find dy/dt or dx/dt (whatever dt is, lol) by dividing by dt after we differentiate.

original equation:

x^2 - y^2 = 9

differentiated

2x dx - 2y dy = 0

Divide by dt.

2x dx/dt  - 2y dy/dt = 0

Now we can solve for either dy/dt or dx/dt depending on which one we want. We'll solve for dy/dt.

2 dy/dt = 2 dx/dt

dy/dt = (2 dx/dt)/2

I'm not sure what dt stands for at this point, I just know its something to do with time. I'm sure I'll soon find out what its for.

At first I thought dy/dx was just a notation, this is what I meant by saying the form of the expression can be changed by multiplying or dividing by dy, dx, or dt. Its kind of like unit multiplyers I suppose. I believe I"m getting the hang of it.

A logarithm is just a misspelled algorithm.

## #11 2005-11-18 12:52:44

mikau
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### Re: differentiating

One thing that confused me initially is that I would differentiate a term like U^2 like this:

2u dx.

I thought it was always dx that was used (and dy for the vertical). This would especially confuse me in implicite differentiation. I couldn't understand how to differentiate y because if you differentiate y^2 you get

2y dy

This would drive me up a wall.

(gotta go for a bit, more on this when I get back.)

A logarithm is just a misspelled algorithm.

## #12 2005-11-18 13:26:59

mikau
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### Re: differentiating

Back. Ok, anyways, when you differentiate you take the derivitive and have to multiply by d( ) and you insert the variable in the parenthesis.

For instance:

y = x^4

differentiate:

dy = 4x^3  dx

now pay attention, mortals, if we have this function:

y = u^4

differentiate:

dy = 4u^3 du        notice! We used du instead!

likewise, if we have this function:

k = s^4

differentiated:

dk = 4s^3 ds

In my oppinion, often the best way to demonstate something is by visual presentation, and allow the student to inspect if to look for patterns, rather then the teacher having fun listening to himself talk about the reasoning behind it, proudly displaying how smart he is, and how it makes PERFECT sense to him. (...what? lol, just kidding). I think its a good idea to get the message across with pictures and written examples. But thats just my oppinion.

A logarithm is just a misspelled algorithm.

## #13 2005-11-21 14:17:58

mikau
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### Re: differentiating

So you going to pick it apart, MathIsFun or did I just do it for you?

A logarithm is just a misspelled algorithm.

## #14 2005-11-21 20:24:42

Ashkar
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### Re: differentiating

Some basics:

dy/dx means changes in y corresponding to change in x.
x, y are two parameters( this could be anything time, speed,distance etc.. )and y always be a function of x.

Eg:
y=x^4
differentiate by x we, get
dy/dx=4x^3
dy=4x^3 dx

same way
k=s^4
dk/ds=4s^3
dk=4s^3 ds
and so on...
it is possible to diff based on a third parameter also.

luv
ash

## #15 2005-11-21 23:22:06

MathsIsFun

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### Re: differentiating

Thanks Ashkar ... that is the kind of thing we are looking at.

Mikau ... good examples, give me a day or so to respond, it has been chaos here.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #16 2005-11-22 03:45:22

mikau
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### Re: differentiating

Take your distance/rate. No hurry here.

Also differentiating by "u" substituion ROXXORS T3H BIG ONEZ!!11

A logarithm is just a misspelled algorithm.