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Izzhov

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- Thread starter Izzhov
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Izzhov

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- #2

matt grime

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- #3

Izzhov

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I think it's valid because the derivative of the derivative of the sum of any number of functions equals the sum of the derivatives of the functions, and multiplication is just repeated addition.

I don't think it's valid because when I plug [tex] x^2 [/tex] into the formula [tex] \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex] it comes out as [tex] 2x [/tex].

I think that it isn't valid, because the book said so, and plugging it into the definition of a derivative is more fundamental (and therefore less likely to be inaccurate) than the rule for the derivative of the sum of functions. However, I can't find the flaw in the first reasoning.

I don't think it's valid because when I plug [tex] x^2 [/tex] into the formula [tex] \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex] it comes out as [tex] 2x [/tex].

I think that it isn't valid, because the book said so, and plugging it into the definition of a derivative is more fundamental (and therefore less likely to be inaccurate) than the rule for the derivative of the sum of functions. However, I can't find the flaw in the first reasoning.

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- #4

Izzhov

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You didn't really help much!

- #5

hrc969

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You need to think about when you can apply the rule for derivatives of sums.

- #6

d_leet

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[tex] x^2=x+x+ ... +x [/tex] (with x terms).

Does this work for every x? What if x is pi? -1?

- #7

matt grime

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You didn't really help much!

In my mind this is homework. People only get homework help if they actually state what they've done in an attempt. Examine the first premise: that x^2 is x added up x times. That is clearly nonsense.

- #8

Gib Z

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- #9

HallsofIvy

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You didn'tYou didn't really help much!

You might also think about how x+ x+ x (exactly 3 times) is different from (x+ x+ ...+ x) (x times). Each is a function of x in what way?

- #10

Izzhov

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Listen: I *have* thought about it, contrary to what people may think! If x is [tex] \pi [/tex], for example, you just add [tex] \pi [/tex] three times and then add [tex] (\pi - 3) \ast \pi [/tex] and you get [tex] \pi^2 [/tex]. I don't see the problem with applying this method to *any* real number.

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HallsofIvy

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Listen: Ihavethought about it, contrary to what people may think! If x is [tex] \pi [/tex], for example, you just add [tex] \pi [/tex] three times and then add [tex] (\pi - 3) \ast \pi [/tex] and you get [tex] \pi^2 [/tex]. I don't see the problem with applying this method toanyreal number.

My point was not so much that you

- #12

Mr.4

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I think the problem itself lies in that there are x

Since there are f(x) 'x' terms where f(x) is a function of x. In that we would have to differentiate f(x)*x using product rule which would result: xf'(x)+f(x).

Thus we cannot directly apply derivatives of sums becuase the number of such sums is in turn a function of x.

(Well, now! I don't understand that myself!:tongue2: )

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- #13

HallsofIvy

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that kind of vaguely definedI think the problem itself lies in that there are xterms. I myself find it quite ambiguous but I'm going with it . I'll try and explain:

Since there aare f(x) 'x' terms where f(x) is a function of x. In that we would have to differentiate f(x)*x using fromduct rule which would result: xf'(x)+f(x).

Thus we cannot directly apply derivatives of

Actually pretty well stated.sums becuase the number of such sums is in turn a function of x.

(Well, now! I don't understand that myself!:tongue2: )

- #14

Mr.4

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LOL, I like to make up stories!

- #15

Izzhov

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My point was not so much that youcan'tsay that "x^{2}= (x+ x+ ...+ x) (x times)" but rather that the derivative of that isnot"(1+ 1+ ...+ 1) (x times)". That "(x times)" is another function of x and you haven't "differentiated" that.

I think the problem itself lies in that there are xterms. I myself find it quite ambiguous but I'm going with it . I'll try and explain:

Since there are f(x) 'x' terms where f(x) is a function of x. In that we would have to differentiate f(x)*x using product rule which would result: xf'(x)+f(x).

Thus we cannot directly apply derivatives of sums becuase the number of such sums is in turn a function of x.

(Well, now! I don't understand that myself!:tongue2: )

That makes sense. Thanks a lot.

- #16

matt grime

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Listen: Ihavethought about it, contrary to what people may think!

You gave no indication of that. We aren't psychics

If x is [tex] \pi [/tex], for example, you just add [tex] \pi [/tex] three times and then add [tex] (\pi - 3) \ast \pi [/tex] and you get [tex] \pi^2 [/tex]. I don't see the problem with applying this method toanyreal number.

The 'method' you just used, and the method you gave in the first post are completely different.

- #17

Izzhov

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- #18

matt grime

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- #19

Gib Z

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- #20

gammamcc

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lim as h-> 0 [f(x+h)-f(x)]/h.

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