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I am able to differentiate $x^n$ with respect to $x$ from first principles using the definition of differentiation. Also it seems natural that the gradient of a finite polynomial will be one order lower. However the fact that the coefficient of the derivative of $x^5$ should be $5$, for example, seems less obvious.

Is there a way of showing this result so that it is intuitive?

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    $\begingroup$ You can derive this result from the product rule. Maybe you can transfer intuition about $\frac{d}{dx} x^n$ from the product rule. $\endgroup$ – Reinstate Monica Jan 5 '15 at 23:13
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    $\begingroup$ Perhaps a way is to look at the derivative of $f(x) = x^n$ at $x = 1$. As $n$ increases, we should expect that $f'(1)$ also increases. $\endgroup$ – Simon S Jan 5 '15 at 23:20
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    $\begingroup$ it is the manifestation of the binomial theorem. you can dress the binomial theorem any way you want, e.g. expanding squares, cubes, etc. $\endgroup$ – abel Jan 5 '15 at 23:52
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Picture a cube in $n$-space bounded by the coordinate hyperplanes and other hyperplanes parallel to them at a distance $x$ from them. In the case $n=2$, it's easy to draw the picture: the boundaries are the two coordinate axes and two lines parallel to those, and you're looking at an $x$-by-$x$ square.

The volume of the cube is $x^n$. As $x$ changes, how fast does the volume change?

Use what I call the "boundary rule":

\begin{align} & \text{[size of moving boundary]} \times \text{[rate of motion of boundary]} \\[6pt] = {} & \text{[rate of change of size of bounded region].} \end{align}

There are $n$ moving boundaries, each of size $x^{n-1}$. The rate at which each moves is the rate at which $x$ changes. Hence the rate of change of $x^n$ is $nx^{n-1}$ times the rate of change of $x$.

PS: The boundary rule can be used to prove the product rule if you use a rectangle rather than a square. The two moving boundaries have lengths $f$ and $g$; the rate of motion of each is the rate of change of the other.

PPS: The fundamental theorem also follows from the boundary rule (as noted in comments below). $$ A(x)=\int_a^x f(t)\,dx. $$ The size of the boundary is $f(x)$; the rate at which the boundary moves is the rate at which $x$ changes; hence $\dfrac{dA(x)}{dx}=f(x)$.

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    $\begingroup$ By the way, this "boundary rule" is just a cool way to express the chain rule. $\endgroup$ – fonini Jan 5 '15 at 23:32
  • $\begingroup$ @fonini : I'm not so sure about that. Consider $A(x)=\int_a^x f(u)\,du$. The size of the boundary is $f(x)$ and the rate of motion of the boundary is the rate at which $x$ moves; hence $A'(x)=f(x)$. Is the fundamental theorem also "just a cool way to express the chain rule"? ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 5 '15 at 23:39
  • $\begingroup$ It seem to me that you used the fundamental theorem to calculate $\frac{d}{dx}\int_a^x f=f(x)$, and then used then chain rule (in a way that is algebraically unnecessary, but important for geometric intuition) to calculate $A^{'}(x)=\left(\frac{d}{dx}\int_a^x f\right)\cdot \frac{dx}{dx} = f(x)\cdot 1 = f(x)$. $\endgroup$ – fonini Jan 5 '15 at 23:44
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    $\begingroup$ @fonini : But I was attempting to deduce the fundamental theorem from the boundary rule. $\endgroup$ – Michael Hardy Jan 6 '15 at 0:05
  • $\begingroup$ You're right, this is not just the chain rule. It's just that the chain rule is a specific case of your boundary rule for rectangular regions with one of the dimensions changing with time. $\endgroup$ – fonini Jan 6 '15 at 0:46
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If you think about $x^n$ as the volume, in $n$ dimensions, of a cube of side $n$, you can ask "how does that grow when $x$ increases?" Answer: count the number of sides of dimension $n-1$. For a square in the plane, with one corner fixed at the origin, you have the change in area being produced by the upper and right edges, each multiplied by a thickness $\Delta x$, when you change $x$ to $x + \Delta x$. For a cube in 3-space, you have three sides, each of whose areas is multiplied by $\Delta x$ to get the change in volume. For a segment in $\mathbb R^1$, you have the right hand point moving through a distance $\Delta x$ to get the change in length, and so on.

In general, there are $n$ "sides" of a hypercube in dimension $n$ (with one corner fixed at the origin), corresponding to incrementing each coordinate individually. (This is also where the $n$ in the binomial expansion of $(x + \Delta x)^n$ comes from, of course.)

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Arguing with infinitesimals and using Newton's binomial: $$\frac {1}{h}((x+h)^n-x^n)=\frac{1}{h}(x^n+nhx^{n-1}+h^2(\ldots )-x^n)= nx^{n-1} + h(\dots ).$$

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  • $\begingroup$ You could use a factorisation identity to get a more ‘ algebraic’ proof… $\endgroup$ – Bernard Jan 5 '15 at 23:28
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I will try to give intuition about the product rule and apply it to your case. Suppose you are taking the derivative of $f_1 \cdot f_2 \cdot \ldots f_n$ at $x$. This is roughly $\frac{1}{\Delta}$ times

$$f_1(x+\Delta) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) - f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x).$$

Now we break up this expression into a telescoping sum:

$$\big( f_1(x+\Delta) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) - f_1(x) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) \big) + \big( f_1(x) \cdot f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) - f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x+\Delta) \big) + \ldots + \big( f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x+\Delta) - f_1(x) \cdot f_2(x) \cdot \ldots \cdot f_n(x) \big),$$

with each term in the outer sum successively replacing one $x+\Delta$ with $x$. This step introduces $n$ outer terms, which is where the factor of $n$ will come from. Simplifying, we get:

$$ (f_1(x+\Delta) - f_1(x)) f_2(x+\Delta) \cdot \ldots \cdot f_n(x+\Delta) + f_1(x) (f_2(x+\Delta) - f_2(x)) \cdot \ldots \cdot f_n(x+\Delta) + \ldots + f_1(x) f_2(x) \cdot \ldots \cdot (f_n(x+\Delta) - f_n(x)),$$

which is approximately equal to

$$ (f_1(x+\Delta) - f_1(x)) f_2(x) \cdot \ldots \cdot f_n(x) + f_1(x) (f_2(x+\Delta) - f_2(x)) \cdot \ldots \cdot f_n(x) + \ldots + f_1(x) f_2(x) \cdot \ldots \cdot (f_n(x+\Delta) - f_n(x)).$$

When all the $f_i$s are equal this is simply $n (f(x+\Delta) - f(x)) f^{n-1}(x)$.

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Solomonoff's Secret nailed the intuition by suggesting the product rule!

$$\frac{d}{dx}x^n = \frac{d}{dx}[\underbrace{(x)(x) ... (x)}_\text{n times}] \\ = \frac{dx}{dx}{\underbrace{(x)(x) ... (x)}_\text{n-1 times}} + (x)\frac{dx}{dx}\underbrace{(x)(x) ... (x)}_\text{n-2 times} + (x)(x)\frac{dx}{dx}\underbrace{(x)(x) ... (x)}_\text{n-3 times} + ... \\ = (1){\underbrace{(x)(x) ... (x)}_\text{n-1 times}} + (x)(1)\underbrace{(x)(x) ... (x)}_\text{n-2 times} + (x)(x)(1)\underbrace{(x)(x) ... (x)}_\text{n-3 times} + ... \\ =\underbrace{x^{n-1} + x^{n-1} + x^{n-1} + ... + x^{n-1}}_\text{n times} \\ = nx^{n-1}.$$

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    $\begingroup$ Intuition? More like a formal trick. $\endgroup$ – user2345215 Jan 5 '15 at 23:40
  • $\begingroup$ Multiply $x$ by itself $n$ times and apply the generalized product rule to get $n$ terms all the same. It's a clear, and I think very intuitive, way to see where the $n$ comes from. $\endgroup$ – John Jan 5 '15 at 23:48
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It comes from the very definition of the derivative in this case and the remarkable identity for $x'^n-x^n$. Indeed the derivative is the limit, as $x'\to x$ of $$\frac{x'^n-x^n}{x'-x}=\sum_{k=0}^{n-1}x'^{n-k}x^k.$$ Each term in this sum tends to $x^{n-1-k}x^{k}=x^{n-1}$, and there are $n-1$ terms.

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  • $\begingroup$ Is this really much of an intuitive argument? OP stated that he could derive the result from the definition just fine. $\endgroup$ – theage Jan 5 '15 at 23:28
  • $\begingroup$ What's intuitive, in my opinion, is the answer to the question: ‘why a coefficient equal to $n$?’ Everyone knows the factorisation brings $n$ terms, whatever they are. $\endgroup$ – Bernard Jan 5 '15 at 23:31
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If $f(x) = x^n$, $\ln f(x) = n \ln x$. Since $(\ln f(x))' =\frac{f'(x)}{f(x)} $ and $(\ln f(x))' =(n \ln x)' =\frac{n}{x} $, we get $f'(x) =f(x)(\ln f(x))' = x^n \frac{n}{x} =n x^{n-1} $.

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  • $\begingroup$ better to use $\ln |x|$ $\endgroup$ – GFauxPas Jan 6 '15 at 1:12
  • $\begingroup$ But then I would have to think more. $\endgroup$ – marty cohen Jan 6 '15 at 3:20
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Take 1, and go on integrating.. $ x/1, x^2/2!, x^3/3! $..

Then take $x^n$, go on differentiating.. $ n x^{n-1}, n (n-1) x^{n-2} $

So there is an order with the factorial when degree is changing both in integration and

differentiation. Intuitivly in the process $ n^ {th}$ derivative should have $n$ at the

head or coefficient as it is built up to that degree from unity.

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