$nth$ derivative of log function The question is , what is nth derivative of $\log(4-x^2)$? I know how to solve general derivatives, but after I encountered with this I am completely bewildered.
 A: For $\ln\lvert 4-x^2\rvert=\ln\lvert x-2\rvert+\ln\lvert x+2\rvert$, the first derivative is
$$\frac1{x-2}+\frac1{x+2},$$
hence an easy induction shows the $n$-th derivative is
$$(-1)^n(n-1)!\biggl(\frac1{(x-2)^n}+\frac1{(x+2)^n}\biggr).$$
A: First, one should use the logarithm properties:

$$ \log(4-x^2) = \log(2 + x) + \log(2 - x)$$

Then, one works on the logarithm derivatives:

 $$D(\log(2 - x)) = -\frac{1}{2-x}, \quad D_2(\log(2-x)) = -\frac{1}{(2-x)^2}$$

and so on.
Then, the solution would be (not 100% sure, calculations done pretty quickly)

 $$ D_n(\log(4-x^2)) = (-1)^{n-1}\frac{(n-1)!}{(2+x)^n} -\frac{(n-1)!}{(2-x)^n} $$

A: As it's said over there:
$$log(4-x^2)=log((x+2)(x-2))=log(x-2)+log(x+2)$$
Then derivate it's simple: taking the first one, for example:
$$ (log(x+2))'= 1/(x+2)$$
Now time to do some calculus, just use the normal derivation rules.
And look, every step, you make the product of the last exponent you got, that's why we use factorial numbers$$$$
the n-esime derivative(without sign): $$(n-1)!/(x+2)^n$$
And make ake some calculus in order to know when it's positive or negative. With this one it's easy, with negative x you must look that negative sign can disappear with the one you obtain derivating $-x$
