Talk:Derivative (examples)

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Articles for deletion This article was nominated for deletion on 27 May 2006. The result of the discussion was no consensus.

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[edit] frist prcipales dirivitive for power rule =

could an example of the general formulae y = axn be be dirivitiveed by first principals into the power rule y' = nax(n − 1) this would be one of the pest examples that copuld be provided Oxinabox1 13:25, 28 July 2006 (UTC) Oxinabox1 13:25, 28 July 2006 (UTC) Oxinabox1 13:25, 28 July 2006 (UTC)



[edit] What is the dirivitive

Well, what I mean is, what is the derivative of f(x)=x**-a

I forgot how to expand out me brackets...

(x+h)**-a = ?

It would be nice to have some links to bracketed expansion (is that the right term?) here, to help with such problems....


Can I say x**-a = x**1/a ?

I am so illterate... when will the matrix suck my maths brain?

I'm assuming that ** stands for the exponential operator (^ stands for it also)... but x^(-a) equals 1/(x^a), and x^(1/a) equals the a root of x. As for your first question, you can use the power rule to get the derivative of x^-a, this is (if I remember correctly) -ax^(-a-1). Hopefully I know what I'm talking about... ehh. Evil saltine 14:08, 23 Oct 2003 (UTC)

[edit] second derivative of square root

The answer should be negative. There is a very strange comment at the bottom, possibly saying that the sign is a choice. Anyone understand this? --MarSch 12:38, 28 October 2005 (UTC)

I'm the one that placed: " = \frac{1}{4 x \sqrt{x}} (\sqrt{x} has 2 answers that only differ in sign, so it doesn't matter which sign we put in front of the end result.)" under " = \frac{-1}{4 x \sqrt{x}}". Let me explain: ofcourse you can't always choose a sign: -1 is not the same as 1, BUT a square root has always 2 answers (let's call them a1 and a2), with a2=0-a1. For example: As you know, the square root of 16 is 4, so if we call this answer a1 then a2=0-4=-4. You can easily see this is true because -4 times -4 equals 16. This proves that the sign that you put in front of \sqrt{x} doesn't matter and this leads to the fact that the sign on the end-result doesn't matter.

Garo 23:54, 11 January 2006 (UTC)

No. The square root function is defined to be positive. —Keenan Pepper 01:07, 12 January 2006 (UTC)
exactly; for f(x)=\sqrt{x} to be a function, it can not have more then one value of f for every x.
But a square does indeed have more then one root ie.
\begin{align} x^2 - 1&=0 \\ \therefore x&= \pm 1 \end{align}
Hopefully that clears things up --widdma 10:55, 10 April 2007 (UTC)

This example is not complete as it does not remove h from the denominator, thus dividing by zero when the limit is evaluated --widdma 08:38, 10 April 2007 (UTC)

I have now re-written this to flow more logically (to my mind any way) --widdma 10:08, 10 April 2007 (UTC)

[edit] Dividing by 0?

In the first example, is it possible to divide by 0? 85.197.143.234 21:03, 15 December 2005 (UTC)

Nowhere is anything divided by zero. The derivative is the limit of a function of h. The function itself is undefined at h=0, but its limit is defined. —Keenan Pepper 00:13, 16 December 2005 (UTC)
In case you didn't understand, it's like this

f(x) = 5

f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0} \frac{f(x+h)-5}{h} = \lim_{h\rightarrow 0} \frac{(5-5)}{h} = \lim_{h\rightarrow 0} \frac{0}{h} = \lim_{h\rightarrow 0} 0 = 0

And you know, it seems like the first example skips these elementary steps. I'll put them in. We never have to worry about dividing by zero because \frac{0}{h} equals zero, and the limit hasn't been taken yet, so h does not equal zero. It's only after it is reduced to a plain 0 is when the limit is taken, and of course since there are no more variables left, taking the limit doesn't change it --Ķĩřβȳ♥ŤįɱéØ 07:54, 7 September 2006 (UTC)