OK, I've spent about 10 hours trying to do this now and I'm pissed off, so what I'm hoping is that someone will take pity on me and explain to me (don't do it, cause then its not my work) how to find a derirative of this formula

A = 40xCosy - 0.5.x^2.Cosy +1600.Siny.Cosy - 40xsinycosy + 0.25.x^2.Siny.Cosy

I need to find (dA)/(dx) so I can substitute in various values of y(an angle) and find the max tp.(I can't substitute in before deriving, as I'm gonna write a program to go through every y value between -90 and +90. => The programming link that justifys me posting it here).

Any help would be appreciated.

You need vladimir_l.

Either that or a ti-89 :)

No, really no, you don't need vladimir_l.

What is "the max tp", and what are you trying to do exactly?

What you seem to be wanting is the partial derivative of A with respect to x. To find it, you just treat y as a constant and differentiate as normal with respect to x. (The notation is actually dA/dx where the d's are lowercase Greek deltas. My math 209 prof pronounces this "di-A di-x", which starts to sound really funny after a whole bunch of partials in a row. Thing Jerrell from goats - "Diediediediedie, etc.")

i.e. in your case treat all the sin y and cos y terms as fixed constants and you just have a polynomial in x, which is easy to differentiate.

The 1600*sin(y)*cos(y) term drops out because it doesn't vary with x.

Does that help? Partial derivatives are really simple once you see how they work, but I didn't see them until first year university calculus. There may be an easier way than cycling through all y values.

it's kinda easy, since most of the terms with x only are either k*x, where k is a constant or k*x^n, again k beeing an constant and n > 0.

k*x becomes k in the derivate
k*x^n becomes k*n*x^(n-1)

cos(y) or sin(y) is a constant, since it doseen't vary with x (as rnd already mentioned)

you can deriver terms that are added/subtracted on their own, eg
f(x) + g(x) becomes simply f'(x) + g'(x)

the whole thing becomes a bit harder if you want to deriver towards y, though :D
hope this helps a bit :)

hehe.. wait till you have to differentiate and integrate in multiple dimensions ^^ My friend is studying to be an accountant and he had to study it...

I couldn't quite work out when in accounting you might need n-dimensionsal differentiation. Only thing I could think of is if your trying to predict values in the future.

I can't wait till multi-dimensional differentiation, I'm hoping to do a joint computers/maths masters degree in the UK's best science university(/me prays that he'll get in).

I sure Psion can help with this. He said my math course are easy (some time ago on IRC and since I've got a photographic memory for useless facts, I still remember) and this is exactly the stuff we need to be able to do.
Basically it's simply applying rules. constants are droped, XeN becomes NxXeN-1, sin becomes cos, cos becomes -sin (I think). But this is the stuff I'm planning on studying on as soon as I get the time. (which has to be in the very near futur, since every course I have seems to included at least 1 formula with derivatives and/or integration.)

I'm pretty sure I have the right formula to begin with but it still won't differentiate:(I changed it a little since yesterday)
0 <= x <= 80
-90 < T < 90

A = 80xCosT - x^2CosT + 1600CosTSinT - 40xCosTSinT +0.25x^2CosTSinT

My (dA)/(dx) (which I don't think is right) is:

80CosT - 2xCosT - 40CosTSinT + 0.5xCosTSinT

Anyone see whats wrong?

At first I thought "My" was some new term "M of y."

Your derivative is correct. Why do you say it is wrong?

- taubz

Hmm, Back to the calculations to see if I've substituted something in wrong.

" 80CosT - 2xCosT - 40CosTSinT + 0.5xCosTSinT "
After checking the manual I believe it should be something like

80 (-sinT) - 2x (-sinT) - 40( (-sinT) x sinT + (-cosT) x cosT ) + 0.5 ( (-sinT) x sinT + (-cosT) x cosT )

because:

D(sin X) = cos X; D(cos X) = -sin X; D(f+g) = D(f) + D(g);
D(f x g) = Df x g + Dg x f

Sorry RedX, but it isn't.

It is. Well, Perhaps not in this dimension...

Everything after "because" is correct. Since I have it out of a course book (thick, boring).

RedX, you're trying to find dA/dT, while rdd is trying to find dA/dx. They're different quantities so you'll get different answers. Incidentally, you're answer is almost right, but you've got an extra minus sign in the "(-cos T)(cos T)" terms.

And try not to think of calculus as just "rules to apply", you need to understand where the stuff is coming from and why it works if you want to grasp it properly. Once you've derived all the basic identities, then you can just use the rules :)

And multivariable calculus looks scary at the start, but it's really no different than single variable, you just integrate or differentiate multiple times. Not the funnest thing in the world, but not hard once you wrap your brain around it.

Revised Formula:

A = ((80xCosT - x^2CosT)/2) + 1600CosTSinT + 40xCosTSinT + ((x^2CosTSinT)/4)

therefore

dA/dx = 80CosT - 2xCosT + 80CosTSinT + ((xCosTSinT)/2)

And yet it doesn't work.

isn't ((80xCosT - x^2CosT)/2) = 40xcos(t) - 0.5x^2cos(t), and thus the derivate 40cos(t) - xcos(t) ?

and 40xcos(t)sin(t) -> 40cos(t)sin(t) (your formula says 80 instead of 40... maybe a typo)

rrrr ... Still no pleasure.

Oh and just so you can see where I am coming from heres how I got the formula - www.mp3charts.4t.com/Math

Any terms seperated by addition or subtraction can be taken aside and derivated seperately, then recombined using the same addition or subtraction, IE:

(3x^3+10x^2-4x-6)'
=
(3x^3)' + (10x^2)' - (4x)' - (6)'
=
6x^2+20x-4

I think I've come up with a solution (dA)/(dT) instead of (dA)/(dX) - Both would meet my requirements, but I was looking solely at X instead of T.

/me is very happy.

1 more problem(Then I've this thing solved)

A = (800 . Sin(180/n))) / (n . Sin^2(90/n))

How do you differentiate that?

Why don't you read an introductory calculus textbook so you can be self sufficient?

Yeah, you could probly find some useful tutorials, ebooks ect on Internet.

Quatient Rule:

F(x) = A/B
F'(x) = (BA'-B'A)/B^2

So in your problem:

A = 800Sin(180/n)
B = nSin(90/n)^2

BUT, before you can use the quotient rule, you have to use the product rule on both the numerator and denominator, because you are multiplying two terms together in both cases, BUT before you even do that you have to take care of the division inside the Sine functions, look it up on google.

...or you could get Mathmatica and let it do it for you

Or Maple 7.....

*coughKaZaAcoughcough*

"Why don't you read an introductory calculus textbook so you can be self sufficient? "

I have, and I've been reading through it, but this work is due in on Wednesday, and if I know how to differentiate it, then I'll be able to prove something.

After poting my message by the way, I did use the quotient rule + the product rule.

And Thanks for the coughing, and the division inside the sine thing.

What I came up with was (((((nSin^2(90/n))x144000Cos(180/n)) - 124000Sin(180/n)Sin(90/n)Cos(90/n)))/n^2)/(n^2Sin^4(90/n)) - but I think thats wrong. Don't bother checking it for me, I'll get my teacher or Mapel to take a look at it.