I was curious what the 3rd derivative of

**x**was.

^{5}So I used the power rule to find the first derivative and I got

**f**I applied the rule again to get the second derivative and got

^{1}(x)= 5x^{4}**f**I once again used the power rule for the third derivative and found the third derivative was

^{2}(x)= 20x^{3}**f**Now by this time I started to get curious as to what the Nth derivative,f

^{3}(x) = 60x^{2}^{n}(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**and began taking the derivative by continuously using the power rule, and got the following results:

^{a}**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}

**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)**

**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)= x**is

^{a }**f**,and from this we see that it is not just

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

**we see that the a reduces by the the derivative we are on for example the second derivative ,**

^{a }**f**the exponent is

^{2}(x),**x**That makes the general form for the exponent

^{a-2 }**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}I am not sure if this pattern was already discovered by someone else or not. I have done some research and have not been able to find anything about it online.

This article is so great. Your points are well explained and you are using words in such a way that even non-native english speakers will understand. Thanks.

ReplyDeleteThank you for the comment, The fact that I am self taught in most of the math that I know contributes to my ability to explain things so most can understand. I am also bi-lingual so I understand how important it is to use words closely to how they are defined. I try to avoid "slang" as much as possible.

DeleteI admire the knowledge you show in this article. I just want to bring under the attention that there are many students that can not go to college to pursue a math degree because they they can't afford it. I stumbled upon a site that gives a list of scholarships and grants that are available for students to go back to school. Even if you are an adult with a job and want to complete your degree, going back to school grants are available to you to. I wish you success with a bright future!

ReplyDeleteNice explanation, Higher order derivative is little difficult but interesting while solving them.It's a differential rule(power rule) in calculus.

ReplyDeletePeople are always interested in general formulae for higher order derivatives. I would recommend you improve your knowledge of whats happening in the world of calculus :) . Most of all, keep it up!

ReplyDelete