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If a function is made up of some standard built-in functions such as $\ln$, $\exp$, $\sin$, $\cos$, $\operatorname{abs}$ and the basic operations $+$, $-$, $\times$, $\div$, is it true that this function is differentiable everywhere on its open definition domain, except if its expression involves the $\operatorname{abs}$ function, and then it is differentiable on both open domains where the value in $\operatorname{abs}$ is positive resp. negative?

Edit: By the way, can we say the same thing for infinite derivability, since all the built-in functions except $\operatorname{abs}$ are $C^{\infty}$?

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I'm not sure I understand what you mean by 'made of' , but the rules : f,g differentiable mean that: f+g, fg, f/g $(g\neq 0)$ are differentiable, and, if f is differentiable at values of g(x), then fog is differentiable. –  AQP Mar 4 '12 at 17:20

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up vote 2 down vote accepted

The claim is true, but you have to be careful.

To begin with, if you have two smooth (read: differentiable) functions, say $f$ and $g$ then $f \pm g,\ f \cdot g,\ f \circ g,\ \frac{f}{g}$ are continuous where defined. Since all the functions you mention are smooth, you can apply $+,-, \cdot, \div, \circ$ in order to to freely construct new smooth functions.

As for $\mathrm{abs}$, on the positive and on the negative numbers it is smooth, so if you use it in a formula, the result will be smooth except perhaps where the expression under $\mathrm{abs}$ changes sign.

A word of warning is needed here: When applying the mentioned rules, you should be careful to consider functions only where properly defined. For example, the function given by $$ f(x) = \exp(\frac{1}{2} \log(x^2) )$$ for $x \neq 0$ can be shown to satisfy: $$ f(x) = \exp( \log(|x|) ) = |x|$$ It is often natural to take $f(0) := 0$. However, if we expand domain of $f$ like that, we don't get smoothness at $0$.

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Thanks for answer. In first expression for f, x is not in definition domain, so that f is smooth on its open definition domain, ]-infty,0[U]0,infty[ ; in second expression for f, it is smooth on both domains ]-infty,0[ and ]0,infty[ separated by the value where thing under sqrt is zero --> so this verifies the mentionned 'rule' (I am on the interpreter side, who seeks for domains for running optimization routines in which derivatives are needed) –  antitrust Mar 4 '12 at 18:44
    
How to edit formulas, please? –  antitrust Mar 4 '12 at 18:44
    

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