When I first learnt calculus I was so surprised to learn that there is a meaningful mathematical operator $D$ that$$D(x^n)= n x^{n-1}.$$It seemed to be a very random thing to multiply the exponent by the function and subtract one from the exponent to get the correct result.
Again, this sounded very strange for me when I first learnt about derivatives, I mean if I was asked before I learnt the calculus of $D$ I would have dismissed this as a random useless property.
Back then I thought about this question: is there a mathematical operator $T$ such that$$T(x^n) =f(n)x^{n-1}\;?$$
After $6$ years, now I remembered the question and I modified it as follows: Is there a closed form for the linear operator $T$ defined on the vector space of all vector space of all polynomials and power series be such that$$T(x^n) =f(n)x^{n-1}\quad \ n \ne 0 $$ for any continuous $f$ on $\mathbb{R}$?
After some thought I think the answer is no. Even a simple function like $f(x)=c\ne 0$ is very hard to find and I couldn't find such representation.