Manipulating identities I'm having some trouble deriving certain identities. If
$$S(z) = \prod_{i=1}^n (z-z_i)$$
then how can I write 
$$\frac{1}{S(z)}\frac{d^2S}{dz^2} = \sum_{i=1}^n\frac{1}{z-z_i}\sum_{j\neq i}^n\frac{2}{z_i-z_j}$$
and
$$ \frac{1}{S(z)}\frac{dS}{dz}= \sum_{i=1}^n\frac{1}{z-z_i} $$
Truth be told, I'm having some trouble writing the derivative of $S(z)$ in a neat, compact form.

Hint for the second:
$${\frac {{
\frac {d}{dz}}S \left( z \right) }{S \left( z \right) }}={\frac {d}{dz}}\ln  \left( S \left( z \right)  \right) $$
Hint A for the first:
$${\frac {{\frac {d^{2}}{d{z}^{2}}}S \left( z \right) }{S \left( z
 \right) }}={\frac {d^{2}}{d{z}^{2}}}\ln  \left( S \left( z \right) 
 \right) + \left( {\frac {d}{dz}}\ln  \left( S \left( z \right) 
 \right)  \right) ^{2}$$
Hint B for the first:
$${\frac {1}{ \left( z-z_{{i}} \right)  \left( z-z_{{j}} \right) }}={
\frac {1}{ \left( z_{{i}}-z_{{j}} \right)  \left( z-z_{{i}} \right) }
}+{\frac {1}{ \left( z_{{j}}-z_{{i}} \right)  \left( z-z_{{j}}
 \right) }}$$
Where the identities hold for any differentiable function (proof: chain rule).
