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[tex] x^2=x+x+ ... +x [/tex] (with x terms). Its derivative is [tex] 1+1+ ... +1 [/tex] (also x terms). By this logic, [tex] \frac{d}{dx} x^2=x [/tex] But it doesn't. So what's wrong with this picture?
Does this work for every x? What if x is pi? -1?[tex] x^2=x+x+ ... +x [/tex] (with x terms).
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.You didn't really help much!
You didn't think much. Matt Grime's purpose was to give you something to think about- that's always a great help.You didn't really help much!
That is not the calculation you used before: you said "x^{2}= (x+ x+ ...+ x) (x times)". That is not the same as saying "add [itex]\pi[/itex] to itself 3 times, then add [tex] (\pi - 3) \ast \pi [/tex]." That was d_leet's point. Why do you mean by "[itex]\pi[/itex] added to itself [itex]\pi[/itex] times"?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.
I think the problem itself lies in that there are x terms. I myself find it quite ambiguous but I'm going with it . I'll try and explain:[tex] x^2=x+x+ ... +x [/tex] (with x terms). Its derivative is [tex] 1+1+ ... +1 [/tex] (also x terms). By this logic, [tex] \frac{d}{dx} x^2=x [/tex] But it doesn't. So what's wrong with this picture?
that kind of vaguely definedI think the problem itself lies in that there are x terms. 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: )
My point was not so much that you can't say that "x^{2}= (x+ x+ ...+ x) (x times)" but rather that the derivative of that is not "(1+ 1+ ...+ 1) (x times)". That "(x times)" is another function of x and you haven't "differentiated" that.
That makes sense. Thanks a lot.I think the problem itself lies in that there are x terms. 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: )
You gave no indication of that. We aren't psychicsListen: I have thought about it, contrary to what people may think!
The 'method' you just used, and the method you gave in the first post are completely different.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.