derivatives question what is the derivative of Q^2/400

[Naps]

I'm super rusty on my rules. Can anyone help me out with:

Q^2 / 400

?

[firebirdV]

You have to tell us what you are taking the derivative with respect to.

d/dQ (Q^2/400) and d/dt(Q^2/400) are different.

The first is most likely what you are referring to (unless you are doing physics), and the answer would be Q/200.

[YJM_Rocks]

[loudriver23]

How did you find and dig up my 23 year old geometry exam?

Seriously, I was always good with math, but I never had the patience for geometry.

[Naps]

You have to tell us what you are taking the derivative with respect to.

d/dQ (Q^2/400) and d/dt(Q^2/400) are different.

The first is most likely what you are referring to (unless you are doing physics), and the answer would be Q/200.

Thanks. Yeah, it was d/dQ.

[Ken]

General rule for derivatives of polynomial (x^n)

d/dx (x^n) = n*x^(n-1)

[Naps]

General rule for derivatives of polynomial (x^n)

d/dx (x^n) = n*x^(n-1)

Yeah, I still remembered that. Just wasn't sure about the bottom number. I didn't know if I left it alone or if it became zero (and of course you can't divide by zero). I see now, but I was going by the general rule when doing derivatives and partial derivatives that constants are zero, a variable times a constant is a constant, etc.

In short I was looking at the 400 and wondering if I treated it like a constant.

[Imp]

you're best considering it as (1/400)Q².. anytime you can make a fraction appear as seperate expressions like that, it's usually simpler to do so

[Tor]

you're best considering it as (1/400)Q².. anytime you can make a fraction appear as seperate expressions like that, it's usually simpler to do so

+1.. That's perhaps the easiest way to see it... It makes no difference if you're writing (1/400)Q² or 2Q², math-wise -- the principles used to solve it are the same.

[Naps]

you're best considering it as (1/400)Q².. anytime you can make a fraction appear as seperate expressions like that, it's usually simpler to do so

Thanks, that's the part I was having trouble seeing.

[tone4days]

d/dQ (Q^2/400) = Q/200 + x ... can't forget the constant

[Tor]

d/dQ (Q^2/400) = Q/200 + x ... can't forget the constant

Hm.. Perhaps you're thinking about integral math regarding the constant there..?


If you take the derivative of e.g. f(Q) = Q²/400 + 5, that can be read as Q²/400 + 5*Q^0 as Q^0=1. So the derivative will thus be f'(Q) = 2Q/400 + 0*5^(-1) = Q/200

But if you integrate f'(Q) again, you know that it originally was a constant there, but you aren't "supposed to know", so therefore an unknown constant "jumps" in again for the integral..

[tone4days]

Hm.. Perhaps you're thinking about integral math regarding the constant there..?.

ah, indeed ... i had it backwards..nice catch, thanks
t4d

[Tor]

No problem.

[Ken]

ah, indeed ... i had it backwards..nice catch, thanks
t4d

Geez Bill,

You're slipping in your old age.

[tone4days]

Geez Bill,

You're slipping in your old age.

when it comes to calc, i done slipped a long time ago
in fact, my wife and i are still in negotiations about who has to relearn calc next year with my son ... i think i am gonna lose on calc but will make her relearn chemistry

[Dan-ola]

Why are there no zingers from newking or sososomething in this thread?

Or an Odie reply like "d/dQ (Q^2/400) = Q/200 + x = CHOWDER"

Sorry, I sucked at math, didnt mean to interrupt one of the few threads here with thinking involved

Dano

[Tor]

Or an Odie reply like "d/dQ (Q^2/400) = Q/200 + x = CHOWDER"

.....that made me laugh out loud.