### Calculus

Topics within Calculus

## The $x^n$ Derivative

Now that you've learned some methods of differentiating complex functions, let's take a look at the following function: $$f\left(x\right)=x^n.$$ How could we find  $f'\left(x\right)$ ?

It turns out that there's two ways to do this. We could use either induction or the binomial theorem (or some third method that you can email me about :)). I'm going to go through the induction strategy first, because it's a bit simpler. If you don't like induction or haven't learned it yet (who is teaching you Precalculus?) you can skip ahead to the other method. Otherwise, let's go!

#### Induction

In order to use induction, we need an inductive hypothesis and a few base cases to support it. We can easily compute the following base cases by hand: $$\frac{dx}{dx}=1x^0,$$ $$\frac{dx^2}{dx}=2x^1,$$ $$\text{and}$$ $$\frac{dx^3}{dx}=3x^2.$$ And so, we can hypothesize that $$\frac{dx^n}{dx}=nx^{n-1}$$is true for all $n\lt k$ (This is something we do a lot in induction).Then, we can try to find the derivative of  $x^k.$ (Remember that the inductive hypothesis does not apply here, because $k\nless k.$ To do this, we can use the product rule: $$x^k=x\cdot x^{k-1}$$ $$\frac{dx^k}{dx}=\left(1\right)\left(x^{k-1}\right)+\left(x\right)\left(k-1\right)\left(x^{k-2}\right)$$ $$=x^{k-1}+\left(k-1\right)\left(x^{k-1}\right)$$ $$=kx^{k-1}.$$ And we are done! Because we've proved that the pattern begins, and we've also proved that it's possible to get from a base case to a non base case while holding the pattern, we've proved that: $$\frac{dx^n}{dx}=nx^{n-1}.$$

#### Binomial Theorem

The proof of this derivative is more commonly done using the Binomial theorem, which says that: $$\left(x+y\right)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\cdots+\binom{n}{n-1}x^1y^{n-1}+\binom{n}{n}x^0y^n.$$ This, as you will see soon, turns out to be quite a useful definition for us.To calculate the derivative this time, we'll be using a first principle limits approach. So, let's set up our limit: $$\frac{dx^n}{dx}=\lim_{h\to0}\frac{\left(x+h\right)^n-x^n}{h}$$ Notice that I'm using $h$ instead of  $\delta x.$ It's just a different name I gave it, and you can call it whatever you want. I generally like calling it  $h,$ just because it's simpler to write.

Let's (partially) expand our binomial, using the binomial theorem: $$\lim_{h\to0}\frac{\binom{n}{0}x^n+\binom{n}{1}hx^{n-1}+h^2\left(\cdots\right)-x^n}{h}$$ $$=\lim_{h\to0}\frac{\binom{n}{1}hx^{n-1}+h^2\left(\cdots\right)}{h}$$ $$=\lim_{h\to0}\binom{n}{1}x^{n-1}+h\left(\cdots\right)$$

Substituting  $h=0,$ $$\frac{dx^n}{dx}=nx^{n-1}.$$ And we are done!