I am trying to find the general form for the $n$-th derivative of $f(x)=\log\left(\frac{1+x}{1-x}\right)$.

I have rewritten the original formula as: $\log(1+x)-\log(1-x)$ for my calculations. I have calculated the derivatives up to $5$:

$$\begin{align} f^{(1)}(x)&=\frac{1}{x+1} - \frac{1}{x-1}\\ f^{(2)}(x)&= \frac{1}{(x-1)^{2}} - \frac{1}{(x+1)^{2}}\\ f^{(3)}(x)&=\frac{2}{(x+1)^{3}} - \frac{2}{(x-1)^{3}}\\ f^{(4)}(x)&=\frac{6}{(x-1)^{4}} - \frac{6}{(x+1)^{4}}\\ f^{(5)}(x)&=\frac{24}{(x+1)^{5}} - \frac{24}{(x-1)^{5}}\\ \end{align}$$

I have noticed that the exponent in the denominator corresponds to $n$, and the positions of $x+1$ and $x-1$ switch each time. I have also noticed that in the 3rd and 4th derivative, the numerator times the exponent in the denominator equals the numerator for the next derivative.

I understand that the pattern is probably quite obvious, but I am just having a little issues piecing this all together to derive the general formula. Any help would be much appreicated, thank you.

  • $\begingroup$ The numerator is $(n-1)!$ $\endgroup$
    – GoodDeeds
    Jan 27, 2016 at 14:53
  • $\begingroup$ It might be useful the factorial notation... $\endgroup$
    – AndreasT
    Jan 27, 2016 at 14:53

3 Answers 3


Hint If you derivate $g(x)=\frac{1}{(x-a)}$ n times [keep in mind that this is the n+1 derivative in your case, the pattern is easy:

$$ \begin{align} g(x) & =(x-a)^{-1} \\ g'(x) & =(-1)(x-a)^{-2} \\ g''(x) & =(-1)(-2)(x-a)^{-3} \\ & \,\,\, \vdots\\ g^{(n)}(x) & =(-1)(-2)\cdots(-n)(x-a)^{n+1} \end{align} $$

Now you can rewrite $$(-1)(-2)\cdots(-n)=(-1)^n n!$$



  • $\begingroup$ It will be good if u plz mention reason for downvoting. $\endgroup$
    – user268307
    Jan 27, 2016 at 15:01
  • 7
    $\begingroup$ I'm not your downvoter, but it would improve your Answer if you explained how you matched this to the pattern given in the Question, or better, if you shared a proof (possibly sketched or outlined) of the correctness of the formula. One-liners rarely make good Answers, but in this case what you've said is true. $\endgroup$
    – hardmath
    Jan 27, 2016 at 15:26

Let $f(x)=h(x)-g(x)$, where $h(x)=\log(1+x)$ and $g(x)=\log(1-x)$

Now, $f^{(n)}(x)=h^{(n)}(x)-g^{(n)}(x)$

$h(x)=\log(1+x)$,$\ $ $g(x)=\log(1-x)$

$h^{(1)}(x)$$=$$1\over {1+x}$ , $\ $$g^{(1)}(x)$$=$$(-1) \frac{1} {1-x}$$=$$\frac{1} {x-1}$

$h^{(2)}(x)$=$(-1) \frac{1} {(1+x)^2}$,$\ $ $g^{(2)}(x)=$$(-1) \frac{1} {(x-1)^2}$

$h^{(3)}(x)$=$(-1)^2 (2\times1)\frac{1} {(1+x)^3}$,$\ $ $g^{(3)}(x)$=$(-1)^2 (2\times1)\frac{1} {(x-1)^3}$

Thus on generalizing the pattern, we get

$h^{(n)}(x)$=$(-1)^{(n-1)} (n-1)!\frac{1} {(x+1)^n}$,$\ $ $g^{(n)}(x)$=$(-1)^{(n-1)} (n-1)!\frac{1} {(x-1)^n}$



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