# Iterative derivatives of log

## Homework Statement:

Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.

## Relevant Equations:

n=k
n=k+1
Derivative of lnx is 1/x.
I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?

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BvU
Homework Helper
2019 Award
I don't know where to start
A little googling perhaps ?

PeroK
Homework Helper
Gold Member
Homework Statement: Let f (x) = ln x. Use induction to justify why the ninth derivative to f in x = 1 is f(9) (1) = 40 320 and the fifteenth derivative is f (15) (1) =
87 178 291 200.
Homework Equations: n=k
n=k+1
Derivative of lnx is 1/x.

I have never used induction to justify the derivative to a function, so I don't know where to start. Does anyone have some tips?
You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?

FactChecker
Gold Member
The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.

There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.

A little googling perhaps ?
I tried to google it, but I still did not understand.

You can start by differentiating the function and looking for a pattern in the successive derivatives. Why isn't that obvious?
Induction is one of the things I struggle with in math, that why it was not obvious to me. Thanks for the tips by the way.

The general outline of a basic inductive proof is in two steps:
1) Prove it is true for N=1
2) Prove that, assuming it is true for N=n, prove it is true for N=n+1.

There is another variation where in step #2 you assume that it is true for N##\le##n. That is sometimes necessary. It is just as valid and it gives you more to work with in your proof.
Thank you for the tips and tricks. But how do I prove for n when I don't have a general formula? Do I have to make a general formula for the derivative of lnx?

BvU
Homework Helper
2019 Award
general formula for the derivative of lnx?
For the nth derivative

For the nth derivative
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.

FactChecker
Gold Member
Do you know what the first derivative of ##ln(x)## and of ##x^{-n}## are? That should be all you need.

PS. When you are stuck by an intimidating problem, do what you can do. You might find out that you can do a lot more than you anticipated.

HallsofIvy
Homework Helper
The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$

Kolika28
PeroK
Homework Helper
Gold Member
Yes exactly, I'm sorry for my bad explanation, but that's what I meant. I have a hard time understanding how to find the formula for the nth derivative to lnx.
If you want to find a derivative, you differentiate the function! If you want to find the second derivative, you differentiate the first derivative. And so on.

The first thing I would do is calculate a few derivatives. With f(x)= ln(x), $$f'= 1/x= x^{-1}$$, $$f''(x)= -x^{-2}$$, $$f'''= 2x^{-3}$$, $$f^{IV}= -6x^{-4}$$, $$f^{V}= 24x^{-5}$$, etc.
Did you do that? Looking at that, I see that the sign is alternating so of the form -1 to some power. The coefficient is a factorial: 1= 0!= 1!, 2= 2!l, 6= 3!, 24= 4!, etc. And, finally, the power of x is the negative of the order of the derivative. That is $$f^{(n)}(x)= 0=(-1)^{n+1}(n-1)! x^{-n}$$. It remains to use induction to prove that.
When n= 1, the derivative is $$\frac{1}{x}= x^{-1}$$. The formula gives $$(-1)^2(0!)x^{-1}= x^{-1}$$ so is true for x= 1.
Now assume that for some n= k, $$f^{(n)}(x)= (-1)^{k+1}(k-1)!x^{-k}$$. Then $$f^{(k+1)}= (-1)^{k+1}(k-1)!(-kx^{-k-1})= (-1)^{k+1+ 1}k!x^{-(k+1)}$$
Thank you so much!!! I finally understand it. I really appreciate your help! :)