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Let$$f(x) = \begin{cases} 0 & \text{for $x$ < 0,} \\ \frac{x}{1+x} & \text{for 0 $\leq$ $x$. } \\ \end{cases}$$The function $f$ is continuous over the entire real line and is differentiable everywhere except at $x=0$.

How did we get to know that $f$ is not differentiable at $x=0$ ?

In General: If any function $f$ is differentiable at $x=x_0$, should it hold true that derivative of $f$ with respect to $x$, that is $f'$, is continuous at $x=x_0$ ?

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up vote 3 down vote accepted

It it was differentiable at $0$, then it would be true that: $$ \lim_{x \to 0+} \frac{f(x)}{x} = \lim_{x \to 0-} \frac{f(x)}{x} = f'(0).$$ Above, the $x\to 0+$ means that we take the limit over the positive values of $x$, and likewise for $x \to 0-$.

Now, these two limits are easy to compute, and unfortunately are different. First, you have: $$ \lim_{x \to 0-} \frac{f(x)}{x} = 0.$$ Secondly, you get with a little more work: $$ \lim_{x \to 0+} \frac{f(x)}{x} = \lim_{x \to 0+} \frac{1}{1+x} = 1.$$ Now, these two values obviously can't be both equal to $f'(x)$. Thus, $f$ is not differentiable at $0$.

As for the latter part: NO, the derivative does not automatically have to be continuous. There is a Wikipedia article that you will surely find relevant. For example, the function $f$ given below is differentiable, but the derivative $f'$ is not continuous at $0$: $$ f(x) = \begin{cases} x^2 \sin \frac{1}{x} \quad& x > 0\\ 0 \quad& x \leq 0 \end{cases}$$ The trick is that the term $x^2$ assures that $f$ goes to $0$ fast enough to have derivative $0$ at $0$, but the term $\sin \frac{1}{x}$ assures that close to $0$ the function has a large slope.

On the other hand, the derivative always has the mean value property, which is known as Darboux Theorem.

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Will you please explain, why if it was differentiable at $0$, then it would be true that:$$\lim_{x \to 0+} \frac{f(x)}{x} = \lim_{x \to 0-} \frac{f(x)}{x} = f'(0). ?$$ – Silent Oct 12 '13 at 9:17
@Sush $$ f'(x_0) = \lim_{h\to0}\frac{f(x_0+h)-f(x_0)}h $$ Let $x_0=0$, $f(x)$ be your function, and compute separately the limit for $h\to0^\pm$. (here, I named the variable $h$ instead of $x$) – AndreasT Oct 12 '13 at 9:23
Well, the definition of the derivative requires that: $f(y) = \lim_{x \to y} \frac{f(x)-f(y)}{x-y}$. We just apply that at $y = 0$, and look at the two ways in which $x$ can approach $0$. – Jakub Konieczny Oct 12 '13 at 9:24
@AndreasT Many many thanks. – Silent Oct 12 '13 at 9:27
@Feanor, thank you so much for clearing all my doubts and giving so nice additional information. – Silent Oct 12 '13 at 9:31

show that $f$ is not diffrentiable at $x=0$ by the definition. $\lim_{x\to 0-}\frac{f(x)}{x}=0$ and $\lim_{x\to 0+}\frac{f(x)}{x}=1$, so $f$ is not differentialbe at $x=0$.

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$$ {\rm f}'\left(x\right) = {\Theta\left(x\right) \over \left(1 + x\right)^{2}}\,, \qquad x\not= 0 $$

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Will you please explain the notation? I am new to calculus. – Silent Oct 12 '13 at 9:12

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