Finding a formula for $f^{(n)} (x)$ if $f(x) = \ln(x-1)$

Question

While working through Stewart's Calculus Late Transcendental 7th edition, I came across this problem:

Find a formula for $f^{(n)} (x)$ if $f(x) = \ln(x-1)$.

Obviously, I calculated the first few derivatives to see if I could spot a pattern:

$$f^{1}(x) = \frac1 {x-1}$$ $$f^{2}(x) = \frac{-1} {(x-1)^2}$$ $$f^{3}(x) = \frac{2}{(x-1)^3}$$ $$f^{4}(x) = \frac{-6}{(x-1)^4}$$ $$f^{5}(x) = \frac{24}{(x-1)^5}$$

I see the pattern of how the numerator is getting multiplied by $-1$ from the first to second derivative, $-2$ from $f^{2}(x)$ to $f^{3}(x)$, then $-3$ from $f^{3}(x)$ to $f^{4}(x)$, etc while the denominator simply increases by one power for each $nth$ derivative.

Despite seeing a pattern, I am at a loss of expressing this as a formula. The answer given by Stewart is:

$$f^{n}(x) = \frac{(-1)^{n-1} (n-1)!} {(x-1)^{n}}$$

Any advice on how to come up with a general formula when you see a pattern, but the formula isn't quite so obvious as with the one above?

Answer 1

You got really far in expressing your answer; this is essentially what you came up with from your textual explanation if you simply separate the negative (breaking down what you've figured out into its parts is a good strategy): $$f^n(x)=\frac{(-1)^{n-1}\prod\limits_{k=1}^{n-1} k}{(x-1)^n}$$ If you put it in a notation like this, you can easily recognize that $\prod\limits_{k=1}^{n-1} k$ is simply a $(n-1)!$ factorial function.

Often times writing down what you figured out (even if it's a series) to visualize it mathematically is a good way to determine a more generalized formula by helping you recognize ways to simplify it.

So, in short:

1. Separate everything you've figured out into parts and attempt to write an equation that uses each of those individual parts to make a whole.
2. Even if one of the individual parts seems silly (like a series), write it out anyway.
3. Simplify your result based on what you come up with.

Answer 2

The key step is that $(-1)^{n-1}(n-1)! = (-(n-1))(-(n-2))\cdots(-2)(-1),$ i.e., you can factor out the $-1$ from the negative exponent that comes down when you differentiate. So in fact it helps in this case to not "multiply out" the factors you bring down right away.

Remember also, once you take the derivative of $\mathrm{log}(x-1),$ you can apply the power rule (and chain rule) for all the higher derivatives.

You can test out the formula also: $$\frac{d}{dx}f^{(n)}(x)=\frac{d}{dx}\left( \frac{(-1)^{n-1}(n-1)!}{(x-1)^n}\right) = (-1)^{n-1}(n-1)!\frac{d}{dx}\left((x-1)^{-n}\right) = (-1)^{n-1}(n-1)!\left( (-n)(x-1)^{-n-1}\right) = \frac{(-1)^{n}n!}{(x-1)^{n+1}}$$

which is exactly what you'd get by using the formula for $f^{(n+1)}(x).$