Question 762179
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First you need to accept (or read and understand the proof of) the theorem that says


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<b>Difference of Two n th Powers</b>
<p>If a and b are real numbers, and n is a positive integer, then

	<p><ul>a<sup>n</sup> <font face = "Courier">-</font> b<sup>n</sup>  =  (a<font face = "Courier">-</font>b)(a<sup>n<font face = "Courier">-</font>1</sup> + a<sup>n<font face = "Courier">-</font>2</sup>b + a<sup>n<font face = "Courier">-</font>3</sup>b<sup>2</sup> + . . . + ab<sup>n<font face = "Courier">-</font>2</sup> + b<sup>n<font face = "Courier">-</font>1</sup>).
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See <a href="http://www.proofwiki.org/wiki/Difference_of_Two_Powers">Difference of Two Powers</a> if you need to see the proof of the above.


Start with *[tex \LARGE f(x)\ =\ x^n]


Then since *[tex \LARGE f'(x)\ =\ \lim_{h\,\rightarrow\,0}\,\frac{f(x\ +\ h)\ -\ f(x)}{h}]  we can substitute and get:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\,\rightarrow\,0}\,\frac{(x\ +\ h)^n\ -\ x^n}{h}]


Apply the Difference of Two n-th Powers Theorem:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\,\rightarrow\,0}\,\frac{\[x\ +\ h\ -\ x\]\[(x+h)^{n-1}\,+\,(x+h)^{n-2}x\,+\,\cdots\,+\,x^{n-1}\]}{h}]


Simplify the numerator:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\,\rightarrow\,0}\,\frac{h\[(x+h)^{n-1}\,+\,(x+h)^{n-2}x\,+\,\cdots\,+\,x^{n-1}\]}{h}]


Cancel the *[tex \LARGE h]


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \lim_{h\,\rightarrow\,0}\,\left((x+h)^{n-1}\,+\,(x+h)^{n-2}x\,+\,\cdots\,+\,x^{n-1}\right)]


Since we don't have *[tex \LARGE h] in the denominator any longer, take the limit:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \left(x^{n-1}\,+\,x^{n-2}x\,+\,\cdots\,+\,x^{n-1}\right)]


Simplify


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ \left(x^{n-1}\,+\,x^{n-1}\,+\,\cdots\,+\,x^{n-1}\right)]



Then, since there are *[tex \LARGE n] terms:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ f'(x)\ =\ nx^{n-1}]


Wa alaykumu s-salam


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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