# How do you find the derivative of 1/([16x+3]^2)?

Write your function as: $f \left(x\right) = {\left(16 x + 3\right)}^{-} 2$
$f ' \left(x\right) = - 2 {\left(16 x + 3\right)}^{- 3} \cdot 16 = - \frac{32}{16 x + 3} ^ 3$