Calculus

Power Rule

Now that you've learned some methods of differentiating complex functions, let's take a look at functions of the following form:

f(x)=xn.f\left(x\right)=x^n.

How could we find f(x)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 (it's super cool!) 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:

dxdx=1x0,\frac{dx}{dx}=1x^0,
dx2dx=2x1,\frac{dx^2}{dx}=2x^1,
and\text{and}
dx3dx=3x2.\frac{dx^3}{dx}=3x^2.

And so, we can hypothesize that

dxndx=nxn1\frac{dx^n}{dx}=nx^{n-1}

is true for all n<kn\lt k (This is something we do a lot in induction).Then, we can try to find the derivative of xk.x^k. (Remember that the inductive hypothesis does not apply here, because kk.k\nless k. To do this, we can use the product rule:

xk=xxk1x^k=x\cdot x^{k-1}
dxkdx=(1)(xk1)+(x)(k1)(xk2)\frac{dx^k}{dx}=\left(1\right)\left(x^{k-1}\right)+\left(x\right)\left(k-1\right)\left(x^{k-2}\right)
=xk1+(k1)(xk1)=x^{k-1}+\left(k-1\right)\left(x^{k-1}\right)
=kxk1.=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:

dxndx=nxn1.\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:

(x+y)n=(n0)xny0+(n1)xn1y1++(nn1)x1yn1+(nn)x0yn.\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:

dxndx=limh0(x+h)nxnh\frac{dx^n}{dx}=\lim_{h\to0}\frac{\left(x+h\right)^n-x^n}{h}

Notice that I'm using hh instead of δx.\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,h, just because it's simpler to write.

Let's (partially) expand our binomial, using the binomial theorem:

limh0(n0)xn+(n1)hxn1+h2()xnh\lim_{h\to0}\frac{\binom{n}{0}x^n+\binom{n}{1}hx^{n-1}+h^2\left(\cdots\right)-x^n}{h}
=limh0(n1)hxn1+h2()h=\lim_{h\to0}\frac{\binom{n}{1}hx^{n-1}+h^2\left(\cdots\right)}{h}
=limh0(n1)xn1+h()=\lim_{h\to0}\binom{n}{1}x^{n-1}+h\left(\cdots\right)

Substituting h=0,h=0,

dxndx=nxn1.\frac{dx^n}{dx}=nx^{n-1}.

And we are done!

© 2022 YMath.io owned and operated by Saumya Singhal