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
    Oct 28 '20 at 23:24
  • $\begingroup$ @cvgmt : yes, you are correct. $\endgroup$ Oct 29 '20 at 7:55

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? $\endgroup$ Nov 1 '20 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
    Nov 1 '20 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)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.