Another maths thread

Quote:

Original post by ukdeveloper I would be able to do this fine if it wasn't a reciprocal like that...

That's not a reciprocal, it's called a "quotient". There, now you should know what to do with it. [smile]

Edit: Oh, and good luck trying to find critical points...

Quote:

Original post by ukdeveloper Hi, another maths question from me (God, I hate this subject): Supposing you've got a function: f(x) = x/(x+1) You've got to then differentiate said function, and find the critical points. I would be able to do this fine if it wasn't a reciprocal like that, so how do I rearrange the thing for differentiation? Or do I use something else, like the Product/Quotient Rule? I'm clueless, can't find any hint anywhere on Google, in my notes or textbook, and the lecturer is an idiot. There's a class of 120 struggling because of how bad the teaching is. Can anybody help at all? Thanks in advance, ukdeveloper.

I'd prefer to write the equation like f(x) = x (x+1)^-1. From there you can simply apply the product rule. I prefer the product rule because it is easier to make a silly mistake using the quotient rule, although the quotient rule is entirely appropriate in your example. The only wrinkle is that you are going to end up using the chain rule to get the derivative of (x+1)^-1. However, you'll soon start seeing expressions like that and just writing down the derivative rather than think about the chain rule.

Just to make sure you're on the right track:

Product rule: D[f*g] = D[f]*g + f*D[g]
Quotient rule: D[f/g] = (g*D[f] - D[g]*f)/g²

(here D[f] means you're taking the derivative of some function f)

Quote:

Original post by ukdeveloper Hi, another maths question from me (God, I hate this subject): Supposing you've got a function: f(x) = x/(x+1) You've got to then differentiate said function, and find the critical points. I would be able to do this fine if it wasn't a reciprocal like that, so how do I rearrange the thing for differentiation? Or do I use something else, like the Product/Quotient Rule? I'm clueless, can't find any hint anywhere on Google, in my notes or textbook, and the lecturer is an idiot. There's a class of 120 struggling because of how bad the teaching is. Can anybody help at all? Thanks in advance, ukdeveloper.

Differentiating is matter of applying few simple rules (that can be trivially derived if forgotten). This is done with "quotient rule" (search google). (You might be confused by rule formula using "f" for function, it has nothing to do with your f)

We have:
f(x)=x/(x+1)
applying quotient rule (can be remembered as "bottom times derivative of top minus top times derivative of bottom divided by bottom squared")
we get:
f'(x)=
(edit: solution is hiddeh, if you want quote the post and see.)
BTW, as nice lil project you can write differenciator program using the rules :-)

Quote:

Original post by Dmytry applying quotient rule (can be remembered as "bottom times derivative of top minus top times derivative of bottom divided by bottom squared")

A neat little rhyming jingle I learned to remember it is:

"Low d-high minus high d-low over low squared"

I found it better for me to remember how equation looks, for longer equations. Never undestood why some (or most) people use song/text/whatever mnemonics, actually... for short equations it could be possible to remember that "quotient rule means [that]" in a way how one remembers that, for example, what does "horse" mean (without using mnemonics to remember that it is mammal with bilateral symmetry, four legs, one head, one tail, and so on).

[Edited by - Dmytry on December 6, 2005 9:17:43 AM]

I'd use my TI-89.

Quotient rule is generally easy to remember, but if ever I get stuck:

"vdu minus udv all over v squared"

vdu = Visual Display Unit

Eh, you can easily derive the quotient rule from the definition of the derivative, should you ever get stuck. Or from the product rule, as jjd suggested.

I use the quotient rule song:
lowwwww-deeee-hiiiiii minus hiiii-deeee-lowwwww allova lowwwww-squareddddd.

x/(x+1)

((x+1)*1 - x)/(x+1)^2

1/(x+1)^2

so you can't have x = -1

For derivatives, I like to user Calc 101, a website that can simplify derivatives for you (for free, and with steps!).

Best of luck.

[Edited by - Vasant56 on December 7, 2005 7:36:12 AM]

Quote:

Original post by Vasant56 I use the quotient rule song: lowwwww-deeee-hiiiiii minus hiiii-deeee-lowwwww allova lowwwww-squareddddd.

Math songs rock btw, that was my life saver for Calc 2! The one I always had to remember was: Negative B Plus and Minus Square Root, B Squared Minus 4AC, All over 2A (sung to 'pop goes the weaseal')

Anyways as was said, you can use the quotient rule, but here's an additional tip to use in simplifying problems such as these (using jjd's post):

D[f/g] = (g*D[f] - D[g]*f)/g²

So, you write out:

f = x             D[f] = 1g = x + 1         D[g] = 1g² = (x+1)²

Then you simply fill in the blanks with your formula:
D[f/g] = (g*D[f] - D[g]*f)/g²

And are done! (Well in terms of finding the derivative, not the CP) It takes a lot of practice (trust me, I took Calc 2 twice), but after a while it gets more natural. That and you should get used to this technique, because down the line, you will have to do substitutions where you do this same exact thing to solve otherwise rather impossible problems. ('U' and Trig substituions)

I had a great calculus teacher at high school who had lots of little mneumonics, etc. for remembering stuff. The one she told us for the quotient rule requires you to put on a mock-german accent. Suppose you want the derivative of u(x)/v(x) and u'(x) is the derivative of u(x). Then you say "vil you dash?" (v*u') Always polite to ask before demanding, "you vil dash!" (uv') So the numerator should be v*u' - u*v' -- ask before demanding. There was nothing for the v² on the bottom though [smile].

The reason I suggested the product rule is just that people often subtract the wrong term in the numerator by accident, even though they know better (myself included).

I just used product rule...quotient rule was a pain to do. Bring up the (x+1) term so you get x(x+1)^-1, sometimes this will simplify it a lot.

If you didn't want to use a product rule, you could always add and subtract one from the numerator

(x+1-1)/(x+1)=1-(1/(x+1)) which is easier to differentiate.

Another trick which is especially useful when you are using calculus is to find the partial fraction of the rational function.

The steps are easy: Divide the top by the bottom. If there are any rational functions left then you must make 'simpler' expressions for those.

Perhaps it is better for my buddy.

If you were, say, integrating said function, you would have a hard time, although it's not impossible.

If you had the foresight and didn't know anything about partial fractions you could change the function to:

f(x) = ((x+1)-1) / (x+1)

This would in turn evaluate to: 1 - 1/(x+1).

As you can see, it is much easier to integrate.