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

[madah12]

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?

 

[tiny-tim]

hi madah12!

(try using the X² icon just above the Reply box )

=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 d^k/dx^k unless k = 1.

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

So try doing the d/dx and the d^k/dx^k in the other order!

 

[madah12]

I didn't use the chain rule I used the product rule and why doesn't it work?

 

[tiny-tim]

oops!

oops!

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)!

 

[madah12]

ok then d^k+1/dx ( x^k+1)=
d^k/dx (d/dx x^(k+1))=
d^k/dx(k+1 x^k)=
k+1 * d^k/dx(x^k)= k+1* k! = (k+1)!
right?

 

[madah12]

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?

 

[tiny-tim]

ok then d^k+1/dx ( x^k+1)=
d^k/dx (d/dx x^(k+1))=
d^k/dx(k+1 x^k)=
k+1 * d^k/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*x^k) = x*kx^k-1 + 1*x^k.

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.

 

[madah12]

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

 

[tiny-tim]

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 )

 

[madah12]

ok thanks I got it