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I have a function:

$$ f(x) = \frac{\arctan(x)^2}{\ln(1+x^2)} \text{when x $\neq$ 0} \\ a ~~\text{when x =0} $$

Using Taylor series, I have calculated that the function is contiunous at $ a= \frac{1}{3} $. However, I am not sure, how to prove its differentiability. The most apprioproate solution I know is to calculate the leftside and rightside derivatives at $x=0$. For the second case, the constant function, the derivative is $0$. However, for the first case, do I have to use a derivative definition? Can I just calculate the derivative for the compilation of the differentiable function without proving their differentiability? The function looks quite complicated for the use of the definition, so is there any trick for proving differentiability in such a case? The derivative for the first case for $x=0$ is 0, so on the whole $f'(0) = 0$ and the function is $C^1$ class, because we cannot calculate further derivatives in the second case?

I would appreciate your explanation.

Funny
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1 Answers1

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Apply L'Hospital rule twice to get $$\underset{x\to 0}{\text{lim}}\frac{\arctan(x)^2}{\log \left(x^2+1\right)}=1$$ So must be $a=1$ Derivative is $$f'(x)=\begin{cases} \frac{2 \arctan(x) \left(\log \left(x^2+1\right)-x \arctan(x)\right)}{\left(x^2+1\right) \log ^2\left(x^2+1\right)}, \quad x\ne 0\\ 0,\quad x=0 \end{cases}$$ as $$\underset{x\to 0}{\text{lim}}\frac{2 \arctan(x) \left(\log \left(x^2+1\right)-x \arctan(x)\right)}{\left(x^2+1\right) \log ^2\left(x^2+1\right)}=0$$ derivative exists and is continuous at $x=0$.

Raffaele
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