If  $$\begin{gathered} f\left(x\right)={\cot }^{-1}(4x^2+10x+7)+{\cot }^{-1}(4x^2+14x+13)+{\cot }^{-1}(4x^2+18x+21) \\ +{\cot }^{-1}(4x^2+22x+31) \end{gathered}$$

then  $f^1\left(0\right)\left[\frac{-185}{16}\right]=\_\_\_\_\_\_\_\_$

$$\begin{gathered} f\left(x\right)={\tan }^{-1}(2x+3)-{\tan }^{-1}(2x+2)+{\tan }^{-1}(2x+4)-{\tan }^{-1}(2x+3) \\ +{\tan }^{-1}(2x+5)-{\tan }^{-1}(2x+4)+{\tan }^{-1}(2x+6)-{\tan }^{-1}(2x+5) \end{gathered}$$

$f\left(x\right)={\tan }^{-1}(2x+6)-{\tan }^{-1}(2x+2)$

$f\text{'}\left(x\right)=\frac{2}{1+{\left(2x+6\right)}^2}-\frac{2}{1+{\left(2x+2\right)}^2}$

$f\text{'}\left(0\right)=\frac{2}{1+{\left(6\right)}^2}-\frac{2}{1+{\left(2\right)}^2}=\frac{2}{37}-\frac{2}{5}$

$f\text{'}\left(0\right)=\frac{10-74}{185}$

$\frac{f\text{'}\left(0\right)\left(-185\right)}{16}=4$