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My problem is the following: Are there any differentiable functions on $\Bbb R$ for which we don't know or can't find an explicit expression for the derivative? So is approximating the derivative numerically the only choice?

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    $\begingroup$ Given a differentiable function we can always compute the derivative by difference quotients. Sometimes we have to. $\endgroup$ – daw Dec 20 '17 at 8:02
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    $\begingroup$ in process of proving anything is differentiable you calculate the derivative. You calculate the left hand derivative and the right hand derivative and if they are equal that is the derivative and the function is differentiable. $\endgroup$ – Sonal_sqrt Dec 20 '17 at 8:02
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    $\begingroup$ There are differentiable functions for which we can't write an expression for the function itself (but for obvious reasons it's rather difficult to mention an example). In those cases we can't find an expression for the derivative either. $\endgroup$ – Arthur Dec 20 '17 at 8:13
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    $\begingroup$ @BadEnglish If you are ok, you can set as solved. Thanks! $\endgroup$ – gimusi Dec 20 '17 at 21:09
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    $\begingroup$ @BadEnglish math.meta.stackexchange.com/questions/3286/… $\endgroup$ – gimusi Dec 22 '17 at 13:00
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If you mean that we can't find an explicit expression of the first derivative, then yes, they do exist. Take $f:\Bbb R\to \Bbb R$ as

$$f(x)=\int_0^x \left(\int_0^t e^{\frac{-y^2}{2}}dy\right)dt.$$ Then, it can be proved that

$$f'(x)= \int_0^xe^{\frac{-y^2}{2}}dy.$$ However, this function cannot be expressed in terms of elementary functions. That being said, you can always approximate the derivative numerically. Hope that helps.

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There's a neat trick involving dual numbers (a number system that has a nonzero infinitesimal element $\epsilon$ such that $\epsilon^2=0$), that lets you numerically calculate derivatives of any function that you can numerically calculate in the first place, to machine precision. This procedure is no less exact than determining an analytical expression for the derivative and computing the value of that expression numerically.

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  • $\begingroup$ looks amazing, thanks $\endgroup$ – Bad English Feb 21 '18 at 16:02
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For functions on $\Bbb R$: derivability $\iff$ differentiability.

NOTE That’s not the case for functions of several variables for which only one direction is true: differentiability $\implies$ derivability.

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  • $\begingroup$ What do you mean by "derivability"? I don't see what your first statement is supposed to mean and why it should be true. $\endgroup$ – Luke Dec 20 '17 at 10:51
  • $\begingroup$ it means that exists finite the limit $\frac{f(x+h)-f(x)}{h}$ for $h\to 0$ $\endgroup$ – gimusi Dec 20 '17 at 11:32
  • $\begingroup$ I see...but I guess that was not what the question asked for, because he asked for explicit expressions, see the answer by Magnusseen. $\endgroup$ – Luke Dec 20 '17 at 13:07

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