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I've got function $f(x) = \begin{cases} \frac{ln(1+x)}{x} &\text{for $x\not=0$} \\ 1 &\text{for $x=0$} \end{cases}$

And I need to find the derivative. (also one sided)

I've found that $f'(x)= \frac{x-ln(1+x)(1+x)}{x^2(1+x)}$ for $x\in D_{f}\setminus\{0\}$

and $f'(x)=\lim_{x\to0}\frac{\frac{\ln (1+x)}{x}-1}{x}=...=-\frac{1}{2} for\space x=0$ I've checked it with wolfram alpha and there is different result for $x=0$ ->$f'(0)=0$.

So have I done something wrong or it's on wolfram side?

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$$\lim_{x \to 0}\frac{\frac{\ln(1+x)}{x}-1}{x}=\lim_{x \to 0}\frac{\ln(1+x)-x}{x^2}$$

Now by L'Hospital rule:

$$\lim_{x \to 0}\frac{\ln(1+x)-x}{x^2}=\lim_{x \to 0}\frac{\frac{1}{x+1}-1}{2x}=\lim_{x \to 0}\frac{\frac{-x}{x+1}}{2x}=-\frac{1}{2}$$

You're right.

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Using the Maclaurin series $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ for $\ln(1+x)$, we find that if $|x|\lt 1$ then $$\frac{\ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\cdots$$ if $|x|\lt 1$ and $x\ne 0$. Since $f(0)=1$, the function $f$ has the same Maclaurin series.

Now we can read off all of the derivatives of $f(x)$ at $x=0$: $$f^{(n)}(0)=(-1)^n \frac{n!}{n+1}.$$ In particular, $f'(0)=-\frac{1}{2}$.

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