I have the function $$f(x)=log(x)$$ and I have found $$f' = 1/x$$
$$f'' = -1/x^2$$
$$f''' = 1*2/x^3$$
and then i find the general: $$f^n = (-1)^n-1 * n-1!/x^n$$
But I need to proof this, can any one help me with this?

Question

I have the function $$f(x)=log(x)$$ and I have found $$f' = 1/x$$
$$f'' = -1/x^2$$
$$f''' = 1*2/x^3$$
and then i find the general: $$f^n = (-1)^n-1 * n-1!/x^n$$
But I need to proof this, can any one help me with this?

geerky42 · Answer

To make general case look little nicer: $f^{(n)}(x) = \dfrac{(-1)^{n-1}(n-1)!}{x^n}\quad\quad n\in\mathbb N$

Not sure how to prove. Can  you help? @Loser66 @SithsAndGiggles

loser66 · Answer

I think we can apply induction here.

badhi · Answer

lets see if its true for n = 1,
$$f^{(1)}(x) = \frac{(-1)^{1-1}(1-1)!}{x^1} = \frac{1\times 0!}{x} = \frac{1}{x}$$

so it is true for $n=1$
assume its true for $n=p$, then

$$f^{(p)} = \frac{(-1)^{p-1}(p-1)!}{x^p}$$

if we differentiate above wrt x 
$$\begin{align}
\frac{df^{(p)}}{dx} &= \frac{(-1)^{p-1}(p-1)!}{x^p}\
f^{(p+1)}&=(-1)^{p-1}(p-1)!\left[\frac{dx^{-p}}{dx}\right]\
&=(-1)^{p-1}(p-1)!\left[(-p)x^{-(p+1)}\right]
\end{align}$$

See if you can simplify it to following
$$f^{(p)} = \frac{(-1)^{p}(p)!}{x^{p+1}}$$

if you can, you have proven that if the expression is true for $n=p$ it is also true for $n=p+1$

since at the begining we have proven that the expression is true for $n=1$ we can conclude that,

$$f^{(n)} = \frac{(-1)^{n-1}(n-1)!}{x^n}$$ is true for $\forall  x  \in \mathbb N$

badhi · Answer

oops @Loser66

loser66 · Answer

oh, yeah, you got it. hehehe

loser66 · Answer

just take derivative of f^(n) to get the answer. it is not hard, right?

badhi · Answer

yeah, thats the main part..  but the process should be followed as well :)