Are there derivatives of the following functions or not? Justify your answer. In the case of a positive answer and calculate the derivatives.
a) $f:(a,b) \to \mathbb R, f(x) = \frac{1}{x-a}$. Calculate $f'(a)$ if it exists.
b)$f: X \to \mathbb R, X={(-1)^n \frac{1}{n}; n \in \mathbb N} \cup {0}, f(x) = x$. Calculate $f'(0)$ if derivative exists.
what i managed to do
a) Yes. $\frac{d}{da}[\frac{1}{x-a}]= - \frac{\frac{d}{da}[x-a]}{(x-a)^2}= - \frac{\frac{d}{da}[x] - \frac{d}{da}[a]}{(x-a)^2}= - \frac{0 – 1}{(x-a)^2}= - \frac{1}{(x-a)^2}$
b)$\frac{\frac{d}{dx}[(-1)^x] x - (-1)^x \frac{d}{x}[x]}{x^2} = \frac{ln (-1)(-1)^x x - (-1)^x1}{x^2}=\frac{ln(-1)x(-1)^x - (-1)^x}{x^ 2}$ for $x = 0$ we have $\lim_{x \to 0}\frac{ln(-1)x(-1)^x - (-1)^x}{x^2}$ does not exist.
Are the answers correct?
Thanks.
I think that before starting any computation, you should come back to the definition.
Question a $f$ is not even defined at $a$. So it does not make sense to speak of its derivative at $a$. Regarding your answer, you're confusing $a$, which is a parameter of the map $f$, and $x$, which is the variable. In fact, it would be better to define $f_a(x) = \frac{1}{x-a}$.
Question b is tricky... To define continuity and derivability at a point $x_0 \in X$, you need to have a topology defined on $X$ and $x_0$ should belong to an open subset $U \subseteq X$. Even if you endow $X$ with the usual distance inherited from the one of $\mathbb R$, $X$ doesn't contain any open subset. Hence you can't speak of the derivative of $f$ at any point and $0$ in particular.