Why is the nth derivative of x to the n power equal to n factorial?

In summary: The Attempt at a Solution There is no general answer to this problem, as it depends on the function. However, one method that might work is to divide the order of the derivative by 4 and take the fraction from the whole number. For example, for the derivative of sin(x) with respect to x, f(4n)(x) would be cos(x) and f4n(x) would be sin(x).
  • #1
madah12
326
1

Homework Statement


proving the nth derivative of x to the n power is n factorial


Homework Equations





The Attempt at a Solution


proving it for n=1
d^(1)x^1/dx = 1!=1 (a)
d/dx x^1 =1 (b)
a=b therefore at n=1 it is true
supposing it is true for n=k
then d^(k)x^k/dx = k!
verifying if it holds for n=k+1 and = (k+1)!

d^(k+1)x^(k+1)/dx = d/dx (d^(k)x^(k+1)/dx)
=d/dx(d^k/dx [x^k*x])=d/dx ([d^k/dx x^k * x]+[x^k d^k/dx x])
=d/dx ([k! * x]+[x^k * 0]
=d/dx x*k!=k!
this doesn't equal (k+1)! ,why?
 
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  • #2
hi madah12! :smile:

(try using the X2 icon just above the Reply box :wink:)
madah12 said:
=d/dx(d^k/dx [x^k*x])=d/dx ([d^k/dx x^k * x]+[x^k d^k/dx x])

No, the chain rule doesn't work for dk/dxk unless k = 1.

(You'd have to use a binomial theorem coefficients version. :yuck:)

So try doing the d/dx and the dk/dxk in the other order! :wink:
 
  • #3
I didn't use the chain rule I used the product rule and why doesn't it work?
 
  • #4
oops!

oops! :redface:

i meant the product rule!

because (ab)' = a'b + ab',

so (ab)'' = (a'b)' + (ab')'

= a''b + a'b' + a'b' + ab'' = a''b + 2a'b' + ab''

and so on (and you can see how the binomial theorem coefficients are coming in)! :wink:
 
  • #5
ok then dk+1/dx ( xk+1)=
dk/dx (d/dx x^(k+1))=
dk/dx(k+1 x^k)=
k+1 * dk/dx(x^k)= k+1* k! = (k+1)!
right?
 
  • #6
also since I don't want to make a new thread when I was trying to make one for sin(x) I noticed the pattern but don't know how to formulate it if f(x) = sin(x)
f'(x)=cos(x) f''(x)=-sin(x) f'''(x)=-cos(x) f''''(x) = sin (x)
and keeps looping every derivative with power of four the thing one method i found out is to divide the order of the derivative by 4 and take the fraction from the whole number like 22 derivative = (22+2)/4=5+ 2/4= -sin(x)
but this is useless for making a proof any help?
 
  • #7
madah12 said:
ok then dk+1/dx ( xk+1)=
dk/dx (d/dx x^(k+1))=
dk/dx(k+1 x^k)=
k+1 * dk/dx(x^k)= k+1* k! = (k+1)!
right?

What you've written is true, but as a proof it doesn't really work, because it relies on you first proving the binomial thing …

and even I only proved it for k = 2, not for general k.

For a proper proof, you need (as I said before) to do the d/dx first …

d/dx (x*xk) = x*kxk-1 + 1*xk. :wink:
madah12 said:
f'(x)=cos(x) f''(x)=-sin(x) f'''(x)=-cos(x) f''''(x) = sin (x)
and keeps looping every derivative with power of four the thing one method i found out is to divide the order of the derivative by 4 and take the fraction from the whole number like 22 derivative = (22+2)/4=5+ 2/4= -sin(x)
but this is useless for making a proof any help?

I don't understand why you think that's not a proof …

f''''(cosx) = cosx, so f(4n)(cosx) = cosx, for any whole number n.
 
  • #8
hmm can you clarify more on the last one? I mean do you mean that f4n=cos
f2n=-cos and f3n=sin? but that doesn't work
 
Last edited:
  • #9
No, I only mean it for 4n.

For 4n + 1 etc, you need to keep differentiating after reaching 4n.

(btw, we always enclose the power of a derivative in brackets, as in f(4n)(x), to distinguish it from f4n(x), which means (f(x))4n :wink:)
 
  • #10
ok thanks I got it
 

1. What is an Nth derivative?

An Nth derivative is a mathematical concept that involves taking the derivative of a function multiple times. It represents the rate of change of a function at a particular point, with N representing the number of times the derivative has been taken. For example, the second derivative is the rate of change of the first derivative, and the third derivative is the rate of change of the second derivative.

2. How is the Nth derivative calculated?

The Nth derivative can be calculated using the power rule, product rule, quotient rule, or chain rule, depending on the complexity of the function. For example, the Nth derivative of a polynomial function can be found by repeatedly applying the power rule, while the Nth derivative of a composite function can be found using the chain rule.

3. What is the purpose of using Nth derivatives?

Nth derivatives are useful in mathematical analysis and engineering, where they can be used to analyze the behavior of functions. They can help determine the concavity, inflection points, and critical points of a function, as well as its maximum and minimum values. Nth derivatives are also important in solving differential equations, which are used to model various natural phenomena.

4. What is induction and how is it related to Nth derivatives?

Induction is a mathematical proof technique where a statement is shown to be true for all natural numbers by proving its truth for the first number (usually 0 or 1) and then showing that if it is true for one number, it is also true for the next number. Induction is related to Nth derivatives because it can be used to prove the validity of formulas involving Nth derivatives, such as the power rule or the product rule.

5. Can Nth derivatives be used for all types of functions?

Nth derivatives can be used for most types of functions, including polynomial, exponential, logarithmic, and trigonometric functions. However, there are some special cases where the Nth derivative may not exist, such as at discontinuities or singularities. In these cases, alternative methods may need to be used to analyze the function.

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