I found an interesting pattern while finding higher order derivatives using the power rule.

I was curious what the 3rd derivative of x^{5} was.

So I used the power rule to find the first derivative and I got f^{1}(x)= 5x^{4} I applied the rule again to get the second derivative and got f^{2}(x)= 20x^{3} I once again used the power rule for the third derivative and found the third derivative was f^{3}(x) = 60x^{2} Now by this time I started to get curious as to what the Nth derivative,fn(x) would be for functions using the power rule. So I began to to work it out algebraically. I started with the general function f(x)= x^{a} and began taking the derivative by continuously using the power rule, and got the following results:

f 1(x)= a x ^{(a-1)}

f 2(x)= a (a-1) x ^{(a-2)}

f 3(x)= a (a-1)(a-2) x^{(a-3)}

f 4(x)= a (a-1)(a-2)(a-3) x^{(a-4)}

f 5(x)= a (a-1)(a-2)(a-3)(a-4) x^{(a-5)}

From these a pattern begins to emerge with, a , as I took the higher order derivatives. Here is the pattern of ,a, throughout each of the derivatives:

a

a (a-1)

a (a-1)(a-2)

a (a-1)(a-2)(a-3)

a (a-1)(a-2)(a-3)(a-4)

As we can see it looks like a pattern similar to a! or a factorial, but this alone is not the pattern we need to a be able to see how this relates to the nth derivative. Well lets look at the derivatives. The second derivative of f(x)= xa is f2(x)= a (a-1) x a-2 ,and from this we see that it is not just a! it is a!/(a-2)! because a! keeps going until (a-some number) = 1 For the second derivative to make sure a! does not go past a (a-1) we use (a-2)! in the denominator because it cancels out all of the other factors past that point. That brings us to the general form of a!/(a-n)!

Now with the a in the exponent x^{a} we see that the a reduces by the the derivative we are on for example the second derivative , f^{2}(x), the exponent is x^{(a-2)} That makes the general form for the exponent x ^{a-n}

To put all this together we get the formula for the nth derivative using the power rule as

f n(x)= a!/(a-n)! x^{(a-n)}