Which of the following is/are correct?
$(A)$ If a function is differentiable then its derivative is a continuous function.
I couldn't find any example which makes derivative, not continuous. So I need help here.
$(B)$ If LHD and RHD does not exist finitely at $x=a$ function is discontinuous at $x=a$
Taking the example of $y=|x|$ it is not differentiable at $x=0$ but is continuous. So this option is incorrect.
So Differentiability $\rightarrow $Continuity but (Not Differentiability) $\nrightarrow$ (Not Continuity )?
$(C)$ Every function which is defined for all $x$ can be thought of as a derivative of some function.
$(D)$Every continuous function(need not to be differentiable) has an antiderivative.
I couldn't find any counter-example for the last two as well, makes them correct or I lack some knowledge please clear my doubts regarding these options.
Best Answer
The standard example for (A) is $$ f(x)=\begin{cases} x^2\sin(1/x) & x\ne0 \\[6px] 0 & x=0 \end{cases} $$ The function is obviously differentiable for $x\ne0$, with $$ f'(x)=2x\sin\frac{1}{x}-\cos\frac{1}{x} $$ It is also differentiable at $0$, because $$ \lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}h\sin\frac{1}{h}=0 $$ However the derivative is not continuous at $0$.
Your example for (B) is incorrect, because your function has one-sided derivatives at $0$. A correct example would be $f(x)=\sqrt[3]{x}$.
For (C) one can consider Darboux's theorem, according to which, if a function is differentiable over an interval, its derivative satisfies the intermediate value property. A function that doesn't satisfy this property cannot be a derivative. Example: the sign function.
For (D), every function that is continuous over an interval has an antiderivative. This is precisely the so-called “fundamental theorem of calculus”.