Consider a function


When calling it with some argument, like derivative[2], instead of $\cos(x)$ it gives me $\partial_{2}\sin(2)$. Is it possible to define a function in one string that first takes the derivative and then returns it at the given point? And to generalize this to an arbitrary function f, i.e.

  • 1
    $\begingroup$ Derivative[1][f][x], here f is a pure function. I think you want a function such as derivative[f_,x_,x0_] ? $\endgroup$
    – cvgmt
    Commented Oct 28, 2020 at 23:24
  • $\begingroup$ @cvgmt : yes, you are correct. $\endgroup$ Commented Oct 29, 2020 at 7:55

2 Answers 2


I think simplest and cleanest way would be to use

derivative[f_] := f'

E.g. if you evaluate


you get the pure function

Cos[#1] &

and therefore



  • $\begingroup$ Oh this is much more elegant than my solution! Does this generalise to arbitrary derivatives, e.g. derivative[f_] := f^(1,0) for f[x_,y_] = Cos[x]Sin[y] for example? $\endgroup$ Commented Nov 1, 2020 at 12:42
  • $\begingroup$ For that case you might want to take a look at Derivative. You could say derivative[f_] := Derivative[1,0][ f ]; Beware that in that case you need too invoke derivative[ Cos[#1] Sin[#2]& ], (or g[x_,y_] := Cos[x]Sin[y]; derivative[g] ) because your function takes 2 arguments and mathematica needs to know which one is the first argument and which one the second. $\endgroup$
    – Gert
    Commented Nov 1, 2020 at 21:23

I think the issue here is one of the order of operations. Consider instead:

derivative[x_] := ( D[Sin[y],y] /. y -> x )

And, more generally,

derivative[f_, x_] := ( D[f[y], y] /. y -> x)

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