If given a function $f$ that I am supposed to demonstrate is derivable, is it enough for me to try to compute its derivative $f'$ and if the result is continuous, can I then state that the original function $f$ is derivable?
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$\begingroup$In most cases you would distinguish two situations:
- A region where $f$ is clearly derivable because it is a composition of elementary functions known to be derivable (polynmials, exponentials, trigonometric functions,...)
- Some points where it is not clear what happens, and you have to show that the derivative exists using the definition.
Let me give an example: $f(x)=x^2\sin\dfrac{1}{x}$ if $x\ne0$, $f(0)=0$. $f$ is derivable in $\mathbb{R}\setminus\{0\}$ because it is a composition and product of derivable functions. What hapens at $x=0$? Using the definition of derivative, you can show that $f'(0)=0$, s0 that $f$ is derivable in $\mathbb{R}$. Observe hat $f'$ is not continuous at $x=0$ (the limit as $x\to0$ of $f'(x)$ does not exist.) Continuity of $f'$ is not a necessary condition for $f$ to be derivable. Functions with a continuous derivative are called $C^1$ functions.
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