For f(x) = x^n, I have to prove f'(x) = 1/n x^{n-1}

[mathusers]

consider: $$f:R \rightarrow R$$, $$f(x) = x^n$$ for some $$n \epsilon N$$. I have the proof that $$f'(x) = nx^{n-1}$$.

PROOF:
-----------------------------------------
We choose an arbitary $$x$$ so,
$$f(x) = x^n$$

$$f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac {(x + h)^n - (x)^n}{h}$$

Now using the fact that $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$$, we get

= $$\lim_{h \to 0} \frac {(x + h)^n - (x)^n}{h} = \lim_{h \to 0} [ (x+h)^{n-1} + (x+h)^{n-2}x + ... + (x+h)x^{n-2} + x^{n-1} ] = nx^{n-1}.$$.

However I need explaining on a few things here:
1) is this proof entirely correct

2) I don't fully understand the line about $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$$. Where does this a and b come from and how do u explain the expansion of $$a^n - b^n$$.

3) Although we are using the definition of the derivative to prove that if f(x) = $$x^n$$ then $$f'(x) = nx^{n-1}$$, does this proof actually directly show that the function f is differentiable? or do u have to use another individual proof? e.g. using limits etc?

4) Do you have an easier proof that $$f'(x) = nx^{n-1}$$?

Cheers :)

 

[Gib Z]

1) No, you made a typo, but it occurred several times so I don't know. When you compute your limit expressions, its just (x+h)^n. Once you apply the power of n, the f disappears. Also, how do you explain why the last limit and nx^(n-1) are equal?

2) That expansion is easy to show. Just expand the RHS, you will see that maybe of the terms will cancel out, so your left with only the 2 on the LHS. "a" and "b" are just any numbers, in this case you can let them equal x and h respectively.

3) To properly show differentiability, you must use "epsilon delta" proofs.

4) Yes.
$$\lim_{h \to 0} \frac {(x + h)^n - (x)^n}{h} = \lim_{h\to 0} \frac{ \left( \left(\begin{array}{c}n\\0\end{array}\right)x^n +\left(\begin{array}{c}n\\1\end{array}\right)x^{n-1}h + \left(\begin{array}{c}n\\2\end{array}\right)x^{n-2}h^2 + \left(\begin{array}{c}n\\3\end{array}\right)x^{n-3}h^3 \cdots +\left(\begin{array}{c}n\\n\end{array}\right)h^n \right) -x^n}{h}$$$$= \lim_{h\to 0} \left( nx^{n-1} + \left(\begin{array}{c}n\\2\end{array}\right)x^{n-2}h^1 + \left(\begin{array}{c}n\\3\end{array}\right)x^{n-3}h^2...\right) = nx^{n-1}$$

 

[disregardthat]

This is correct if n is an integer, right? If n is a non-integer, you can't set $$(\begin{array}{c}n\\2\end{array}\right)$$

Anyway, just set a the constants $$c_0, c_1 ... c_n$$ instead of the $$(\begin{array}{c}n\\c\end{array}\right)$$ Because these constants will be canceled anyway.

 

[Gib Z]

Well if you really want to be like that Jarle :P, then replace the factorials in the definitions of n choose k with Gamma functions.

 

[mathusers]

1) sorry my bad. the f wasnt meant to be there when i computed the limit expresisons. glad you noticed.

2) how is that expansion done though.. it looks like binomial theorem to me? is that correct?

3) I've never come across an "epsilon /delta" proof for differentiation. I've seen it for continuity but for differentiation I've only got the one about ... [ f(x) - f(x0) / x - x0 ],... ? could you please show me a link or any other material that teaches you about the epsilon/delta way for differentiation..

4) is that method just the same as the expansion of a^n - b^n ?

cheers :)

 

[Gib Z]

1) No problem.
2)Yes its binomial theorem.
3)Just search epsilon delta differentiation in google.
4) No. Binomial expansions can deal with $(a-b)^n$, but not a^n - b^n.

 

[mathusers]

4) if binomial expansions can't deal with a^n - b^n then how come (2) is binomally expanded (lol if that's the correct term) .. since we are afterall expanding a^n - b^n?

 

[Gib Z]

I used binomial theorem only to expand the first term, (x+h)^n. I later note that the first term in the expansion cancels out with the 2nd term of the original limit.

 

[CompuChip]

For the epsilon delta proof, you should show that for any $\epsilon > 0$ there exists a $\delta > 0$ such that for all $h, |h| < \delta$
$$\left| \frac{f(x + h) - f(x)}{h} - n x^{n - 1} \right| < \epsilon$$.

You probably don't want to do this, especially if you have no experience with this sort of proofs (I recall my first analysis class when I didn't even see how to prove $$\lim_{x \to a} x = a$$ this way ).

 

[disregardthat]

Well if you really want to be like that Jarle :P, then replace the factorials in the definitions of n choose k with Gamma functions.

Heh, I just thought of that after i posted

 

[uart]

This is correct if n is an integer, right? If n is a non-integer, you can't set $$(\begin{array}{c}n\\2\end{array}\right)$$

Actually the original posters proof (when typo's are corrected) is also only valid for integer n.

Well if you really want to be like that Jarle :P, then replace the factorials in the definitions of n choose k with Gamma functions

No Gibz. The correct way to "fix" it for not necessarily integer n is to use the more general form of the binomal expansion given below.

$$(x+h)^a = x^a \, + \, \frac{a}{1!} x^{a-1} h^1 \, + \, \frac{a(a-1)}{2!} x^{a-2} h^2 \,+\, \frac{a(a-1)(a-2)}{3!} x^{a-3} h^3 \, + \, ...$$

Note that this series only terminates if "a" is integer, in which case it reduces to the more familiar binomial series you posted above.

 

[HallsofIvy]

consider: $$f:R \rightarrow R$$, $$f(x) = x^n$$ for some $$n \epsilon N$$. I have the proof that $$f'(x) = nx^{n-1}$$.

PROOF:
-----------------------------------------
We choose an arbitary $$x$$ so,
$$f(x) = x^n$$

$$f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac {(x + h)^n - (x)^n}{h}$$

Now using the fact that $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$$, we get

= $$\lim_{h \to 0} \frac {(x + h)^n - (x)^n}{h} = \lim_{h \to 0} [ (x+h)^{n-1} + (x+h)^{n-2}x + ... + (x+h)x^{n-2} + x^{n-1} ] = nx^{n-1}.$$.

However I need explaining on a few things here:
1) is this proof entirely correct

2) I don't fully understand the line about $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$$. Where does this a and b come from and how do u explain the expansion of $$a^n - b^n$$.

3) Although we are using the definition of the derivative to prove that if f(x) = $$x^n$$ then $$f'(x) = nx^{n-1}$$, does this proof actually directly show that the function f is differentiable? or do u have to use another individual proof? e.g. using limits etc?

4) Do you have an easier proof that $$f'(x) = nx^{n-1}$$?

Cheers :)

To answer the last question (the others have been dealt with), for n a positive integer, you can use the product rule and induction:
If n= 1, then f(x)= x so f'(x)= 1= 1*x⁰= nx^n-1
so the statement is true for n= 1.

Assume the statement is true for n= k: If g(x)= x^k then g'(x)= kx^k-1. Now, let f(x)= x^k+1= x(x^k)= xg(x). By the product rule, f'(x)= (x)'g(x)+ x(g'(x))= 1(x^k)+ x(kx^k-1= x<sup>k+ kxk= (k+1)xk which is the correct formula for xk+1.

By induction, then, if f(x)= xn, f'(x)= nxn-1 for any positive integer n

It is easy to see that the derivative of x0 is 0(x-1)= 0 since x0= 1 is a constant.

To find the derivative of x-n, write it as 1/xn and use the quotient rule.

To find the derivative of xr where r is not an integer, use logarithmic differentiation. (You could use the product and quotient rules for r rational and reserve logarithmic differentiation for irrational powers.)

Putting all of that together, if f(x)= xr, then f'(x)= r xr-1 for r any real number.</sup>

 

[disregardthat]

Actually the original posters proof (when typo's are corrected) is also only valid for integer n.



No Gibz. The correct way to "fix" it for not necessarily integer n is to use the more general form of the binomal expansion given below.

$$(x+h)^a = x^a \, + \, \frac{a}{1!} x^{a-1} h^1 \, + \, \frac{a(a-1)}{2!} x^{a-2} h^2 \,+\, \frac{a(a-1)(a-2)}{3!} x^{a-3} h^3 \, + \, ...$$

Note that this series only terminates if "a" is integer, in which case it reduces to the more familiar binomial series you posted above.

Nice, uart :) I was thinking about that and revisited the thread and whoosh, an answer!
But is the binomial expansion defined for numbers that are not integers?

 

[Gib Z]

No Gibz. The correct way to "fix" it for not necessarily integer n is to use the more general form of the binomal expansion given below.

$$(x+h)^a = x^a \, + \, \frac{a}{1!} x^{a-1} h^1 \, + \, \frac{a(a-1)}{2!} x^{a-2} h^2 \,+\, \frac{a(a-1)(a-2)}{3!} x^{a-3} h^3 \, + \, ...$$

Note that this series only terminates if "a" is integer, in which case it reduces to the more familiar binomial series you posted above.

Yup sorry I only accounted for non negative reals, forgot about negative numbers :P

 

[uart]

...
But is the binomial expansion defined for numbers that are not integers?

Sure it is. For an example of it's use for non-integer powers consider sqrt(1-x), for |x| <= 1,

$$(1-x)^{1/2} = 1 \, - \, \frac{1}{2} x \, - \, \frac{1}{8} x^2 \, - \, \frac{1}{16} x^3 \, - \, \frac{1\times 1 \times 3 \times 5} {2^4 \times 4!} x^4 \, + \, ...$$

 

[uart]

Yup sorry I only accounted for non negative reals, forgot about negative numbers :P

What you said about replacing factorial(") with Gamma("+1) in calculating the binomial coefficients is also a valid way to generalize it. The thing I was picking on is that this by itself isn't enough, you also have to make it an infinite series instead of terminating at $$\left(^n_n\right)$$, that's all.It's quite interesting that when you do generalize the binomial coefficients using Gamma() to replace factorial() that you can use the recurrance relation of Gamma to cancel out all references to Gamma. You can actually evaluate

$$\left(\begin{array}{c}v\\k\end{array}\right) = \frac {\Gamma(v+1)} {k! \, \Gamma(v+1-k)}$$

without ever evaluting Gamma. (And of course you just end up with the more explicit form of the expansion that I posted).

 

[disregardthat]

when will the serie of $$(1-x)^{\frac{1}{2}$$ end? is it an infinite serie?

 

[Gib Z]

Yes it is an infinite series, and its binomial series is identical to its taylor series.

 

[Werg22]

Regarding the proof of the differential of a power function, using a proof by induction is far more succinct. If the derivative of $$x^{k-1}$$ gives $$(k - 1)x^{k-2}$$, then, using the product rule, the derivative of [tex]x^{k}[/itex] can be written as $$(k - 1)x^{k-1} + x^{k-1}$$, which in turn simplifies to $$kx^{k-1}$$. The derivative of $$f(x) = x^0$$ is $$f'(x) = 0*x^{0-1}$$; the proof is complete.