Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $x=x_0$ ?

share|improve this question

3 Answers 3

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.

share|improve this answer
    
Will you please explain, why if it was differentiable at $0$, then it would be true that:$$\lim_{x \to 0+} \frac{f(x)}{x} = \lim_{x \to 0-} \frac{f(x)}{x} = f'(0). ?$$ –  Sush Oct 12 '13 at 9:17
1  
@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
1  
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 $0$. –  Feanor Oct 12 '13 at 9:24
    
@AndreasT Many many thanks. –  Sush Oct 12 '13 at 9:27
    
@Feanor, thank you so much for clearing all my doubts and giving so nice additional information. –  Sush 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 $x=0$.

share|improve this answer

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

share|improve this answer
1  
Will you please explain the notation? I am new to calculus. –  Sush Oct 12 '13 at 9:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.