Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
If you already have a guess on what the $n$-th derivative might look like after seeing the first few members, you could use induction to prove your claim... –  J. M. Aug 2 '12 at 5:48
add comment

2 Answers

up vote 6 down vote accepted

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.
share|improve this answer
add comment

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 $f^{(n+1)}(x).$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.