# How do you differentiate  ln[ (2x^3)-(3x^2)+(7) ]?

$\frac{1}{2 \cdot {x}^{3} - 3 \cdot {x}^{2} + 7} \cdot \left(6 \cdot {x}^{2} - 6 \cdot x\right)$
Using the fact that$\left(\ln \left(x\right)\right) ' = \frac{1}{x}$ and the chain rule we get
$\frac{1}{2 {x}^{3} - 3 {x}^{2} + 7} \cdot \left(6 {x}^{2} - 6 x\right)$