# [✓] Meaning of & in output - derivative ?

GROUPS:
 Hello everyone. What is the meaning of & in output in the following code? g[x_] := x^9 Derivative[1][g] 9 #1^8 & Thank you for your help.
7 days ago
4 Replies
 Valeriu Ungureanu 2 Votes See Pure Functions in the Wolfram Documentation
 Valeriu Ungureanu 2 Votes Accordingly to the Wolfram Documentation: body& - a pure function in which arguments are specified as # or #1, #2, #3, etc. $1 + #$ is the body of pure function and the symbol $&$ marks/highlights the pure function. $#$ is an argument of pure function.A pure function may be used as any other function with arguments specified in square brackets, so the following code seems to be correct: 1 + # &[#] And the result is In[1]:= 1 + # &[#] Out[1]= 1 + #1 So, In[2]:= 1 + # == 1 + # &[#] Out[2]= True i.e. the answer to your question is "Yes".