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Kindly verify the proof i couldn't find this anywhere. I am fairly new to differentiation so i apologize for mistakes if any... $$\frac{d}{dx}\left(x^n\right)=nx^{n-1}$$ $$\frac{d}{dx}\left(nx^{n-1}\right)=n\left(n-1\right)x^{n-1-1}=n\left(n-1\right)x^{n-2}$$ Hence if differentiated $n$ times: $$\frac{d^n}{dx^n}\left(x^{n}\right)=n!x^{n-1-1-1-...-1}=n!x^{n-n}$$ $$\frac{d^n}{dx^n}\left(x^{n}\right)=n!$$

Thank you

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This question is about calculus, not about discrete-mathematics or implicit differentiation. Your proof yields the correct result, but I think a formal mathematical proof would require the use of mathematical induction. –  Hrodelbert Apr 4 at 14:03
    
My apologies. I could not find the right tag for this. I tried to look up differentiation and proofs but they weren't available. More over i couldn't add them for my reputation is a bit too low hence i looked up the closest words that being mathematics and differentiation. –  Mathbreaker Apr 4 at 14:33
    
No problem, glad to see it fixed. –  Hrodelbert Apr 4 at 14:35
    
Think of the function recursively. –  shortstheory Apr 4 at 18:11

3 Answers 3

up vote 3 down vote accepted

We can do a proof by induction so we start by proving true for n = 1

so we consider x

$\partial_xx = 1 = 1!$ where $\partial_x^n$ is the nth partial derivative with respect to x i.e $ \frac {d^n}{dx^n}$

now we assume true for n

$\partial_x^nx^n=n!$

now we prove true for n+1

$\partial _x ^ {n+1} x ^ {n+1} = \partial_x^n(n+1)\cdot x^n = (n+1)\cdot \partial_x^nx^n$

but we know from our assumption that:

$\partial_x^nx^n=n!$

so if we plug that in we get

$\partial _x ^ {n+1} x ^ {n+1} = n!\cdot(n+1)= (n+1)!$

which is what we wanted

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Or start the induction from $n = 0$. $x^0$ differentiated no times is $1$, which is $0!$. –  Steve Jessop Apr 4 at 18:48
    
I suppose as a technicality, if the question wanted it for all integers or naturals, I should have included the induction for assume true for n and then prove for n-1 as well –  Aaron Apr 5 at 11:49
    
Well, it's not true for negative integers, even if you take "differentiating it -1 times" to mean taking the indefinite integral. So since you proved it for $n >= 1$ there's only one other case to even consider. –  Steve Jessop Apr 5 at 12:01
    
apologies, entirely true, I shouldn't be allowed to comment when I've just woken up –  Aaron Apr 5 at 12:02

The proof should be by induction; the base case, $n=1$, is obvious: $Dx=1=1!$.

Now suppose you know that $D^{n}x^{n}=n!$ and try computing $$ D^{n+1}x^{n+1}=D^n(Dx^{n+1}) $$ (using linearity of derivation).

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I see. Thanks :) Never done induction was just playing around with some formulas i learnt. Starting 11th soon so hope to learn this as well. –  Mathbreaker Apr 4 at 14:15

welcome on SE, you should use TeX to write formulae.

Anyway, the identity $(\frac{d}{dx})^nx^n=n!$ is correct, as well as the way you derived it, but to have a formal proof you probably have to use induction.

Let's suppose this identity holds for $n$ (and you can immediately verify that it holds for $n=1$), then, for $n+1$ we have:

$(\frac{d}{dx})^{n+1}x^{n+1}=(\frac{d}{dx})^{n}(n+1)x^n$

but we already know the n-th derivative of $x^n$ (the $n+1$ in front of it is just a constant) so we can write

$(\frac{d}{dx})^{n+1}x^{n+1}=(n+1)n!=(n+1)!$

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