I am trying to prove that the derivative of 1/x exists. Is my proof correct? Please help me!
Proof.
According to the definition of the derivative.
It is necessary to analyze two cases:
Case numer 1: x is not equal to 0.
For this case f(x), f(a), x, and a exist because 1/x is a continuous function everywhere except when x=0 because x is a polynomial and 1/x is undefined only when the denominator is equal to 0, what happens only when x=0.
Therefore the derivative exists when x is not equal to 0.
Case numer 2: x is equal to 0.
Let analyze the graph of 1/x.
1/x is discontinuous at x=0 because the left limit and right limit of 1/x is not the same.
As continuity is a necessary condition for differentiability, 1/x is not differentiable at x=0. Q.E.D.
$\endgroup$ 101 Answer
$\begingroup$For the case $a \neq 0$ we have that
$$ \lim_{x \to a} \frac{f(x) - f(a)}{x-a} = \lim_{x \to a} \frac{\frac 1 x - \frac 1 a}{x-a} = \lim_{x \to a} \frac{\frac{a - x}{ax}}{x-a} = \lim_{x \to a} -\frac{1}{ax} = -\frac{1}{a^2}.$$
If $a = 0$ then $f$ is not continuos and hence not differentiable.
I hope it helps you :)
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