The statement of the question must be corrected. First, as Carlo pointed out, the Hadamard representation as in the question is not valid for all functions of finite order. The correct Hadamard representation is $$f(z)=z^me^{P(z)}\prod_{n=1}^\infty \left( 1-\frac{z}{z_n}\right) \exp\left(\frac{z}{z_n}+\ldots+\frac{1}{q}\left(\frac{z}{z_n}\right)^q\right).$$ Here $q$ is the genus of the function.
You have to specify whether you are talking about functions of finite order (and thus finite genus) or functions of genus $1$.
Second, what does it mean "to deduce" a representation for the derivative? The derivative of a function of finite order is of finite order, so there is a similar representation for the derivative. To "find" it means to find the zeros of derivative in terms of zeros of the function, and to find the number $q$ and polynomial $P$.
Can you "deduce" the zeros of derivative of a polynomial in terms of zeros of this polynomial?