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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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continuously differentiable function –  pedja Jan 22 '12 at 9:27
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I guess you mean differentiable. If the derivative exists at some point, then the function is differentiable there. If the derivative exists and is continuous, the function is said to be continuously differentiable. –  aelguindy Jan 22 '12 at 9:33

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up vote 2 down vote accepted

In most cases you would distinguish two situations:

  1. A region where $f$ is clearly derivable because it is a composition of elementary functions known to be derivable (polynmials, exponentials, trigonometric functions,...)
  2. 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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