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?

Of course, by taking log and differentiating the Hadamard formula, you obtain a formula for $f'$
which Carlo wrote, but this is not the Hadamard representation of $f'$.

By the way, in the beginning of 20-s century, the question of whether the genus of $f$ is the same
as that of $f'$ was intensively discussed. If I remember correctly, it can be different.