let $p_n(x)=x^n$ be a polynomial of degree $n$. I need help to be able to explain to my students why the derivative of $p$ is defined as follows:
$$ p_n'(x)=\frac{dx^n}{dx}=nx^{n-1} $$
for every $n\geq1$.
Note: I'd prefer geometric proofs if any exist.
Edit:I edited the question as it is related to the recent question
Thank you for any help
Look at a cube in $n$-dimensional space, whose side has changing length $x$. Fix the sides meeting at one vertex in their places, so that all motion is motion of $n$ the sides opposite those. Each side has $(n-1)$-dimensional volume $x^{n-1}$. So the total size of the moving boundaries is $nx^{n-1}$.
The size of the boundary times the rate at which the boundary moves is the rate of change of size of the cube.
Therefore $x^n$ changes $nx^{n-1}$ times as fast as $x$ changes.
Differentiation from first principles is great. Draw a curve on the board, then draw the tangent to a point - that's your gradient, you can see that.
Then take $f(x+\delta x)$ and draw that, and the line connecting it to $f(x)$ for some values of $\delta x$ - this 'convinced me' (when I was 15/16) of how differentiation works, because I liked gradient of a line.
It becomes clear that "as $\delta x\rightarrow 0$ we get a better approximation"
We cannot divide by zero, so we cannot just plug in $\delta x =0$ this is where maths got interesting for me. What we did is we put $x^n$ as $f(x)$ (we actually used $x^2$ first, and $x^3$ but it was an AS level class) and this showed you can actually side-step the division by zero by using algebra.
Needless to say this amazed me!
For your $x^n$ it requires the binomial theorem. (I do recommend doing $x^2$ and $x^3$ "manually" first)
We note:
$$\frac{f(x+\delta x)-f(x)}{\delta x}=\frac{(x+\delta x)^n-x^n}{\delta x}$$ We expand: $$=\frac{x^n+nx^{n-1}\delta x+\ldots+nx(\delta x)^{n-1}+(\delta x)^n-x^n}{\delta x}$$ We note the terms cancel to give: $$=\frac{nx^{n-1}\delta x+\ldots+nx(\delta x)^{n-1}+(\delta x)^n}{\delta x}$$ Now we sidestep the division by zero (By factoring out a $\delta x$) $$=\frac{\delta x(nx^{n-1} x+\ldots+nx(\delta x)^{n-2}+(\delta x)^{n-1})}{\delta x}$$ We can do the divide $$=nx^{n-1} x+\ldots+nx(\delta x)^{n-2}+(\delta x)^{n-1}$$ NOW we can substitute $\delta x=0$ - we see: $$=nx^{n-1}$$
My mind was blown.
Use the definition of the derivative: $$\frac{dx^n}{dx}=\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}=\lim_{h\to 0}\frac{x^n+nx^{n-1}h+\cdots+h^n-x^n}{h}=\lim_{h\to 0}(nx^{n-1}+h(\cdots))=nx^{n-1}.$$
When $n$ is a natural number, you can do it with the binomial theorem, as Alec Teal and David Quinn suggested.
When $n$ is a nonzero integer, you can use the previous result and the quotient rule (or some analogue of the quotient rule).
When $n$ is a nonzero rational number, you can do it with implicit differentiation: $y=x^{m/n}$ is the same as $y^n=x^m$, now differentiate implicitly using the previous result and solve for $y'$.
When $n$ is an irrational number, the details get rather messy, and it becomes easier to deal with everything rigorously by taking a detour into the exponential and logarithm functions, and defining $x^n=\exp(n \ln(x))$.
All of this assumes $x$ is positive.
It is the coefficient of $h$ in the linear part of $(x+h)^n$, as a function of $h$. Now by the binomial formula: $$(x+h)^n=x^n+nx^{n-1}h +\text{terms of higher degree in }h.$$ Thus the derivative is $\;nx^{n-1}$.
A zillion calculus textbooks do this by finding $\displaystyle\lim_{h\to0}\frac{(x+h)^n-x^n}{h}$ and expanding the binomial.
I think there's a bit of overkill in that argument: we only need the first two binomial coefficients, $1 \text{ and } n$, not all of them.
So do it like this instead: \begin{align} \lim_{y\to x}\frac{y^n-x^n}{y-x} & = \lim_{y\to x} \frac{(y-x)(y^{n-1} + y^{n-2}x+y^{n-3}x^2 + \cdots + x^{n-1})}{y-x} \\[10pt] & = \lim_{y\to x} \Big( y^{n-1} + y^{n-2}x+y^{n-3}x^2 + \cdots + x^{n-1} \Big) \\[10pt] & = x^{n-1} + x^{n-1} + x^{n-1} + \cdots + x^{n-1}. \end{align}
