Intergrals/Differentiate

find f 1/(2x+3)^2   d(x)

I assume: Derivative: $\dfrac{d\ \left( \frac{1}{(2 x + 3)^2} \right) }{dx}$

$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{ \dfrac{1}{\left(2 x + 3\right)^2} } \\\\ f'(x) &=& \dfrac{1}{\left(2 x + 3\right)^2} \times \left( \dfrac{0}{1} - \dfrac{2*(2x+3)*2}{(2 x + 3)^2} \right) \\\\ f'(x) &=& \dfrac{1}{\left(2 x + 3\right)^2} \times \left( 0 - \dfrac{4*(2x+3)}{(2 x + 3)^2} \right) \\\\ f'(x) &=& \dfrac{1}{\left(2 x + 3\right)^2} \times \left( - \dfrac{4}{(2 x + 3)} \right) \\\\ \mathbf{f'(x)} &=& \mathbf{-\dfrac{4}{\left(2 x + 3\right)^3}} \\ \hline \end{array}$

Find f 1/(2x+3)^2   d(x)