How do you differentiate f(x)= (4x^2+5)(3x-5)  using the product rule?

Mar 12, 2016

$\frac{d}{d x} f \left(x\right) = 36 {x}^{2} - 40 x + 15$
$f \left(x\right) = \left(4 {x}^{2} + 5\right) \left(3 x - 5\right)$
$\frac{d}{d x} f \left(x\right) = {\left(4 {x}^{2} + 5\right)}^{'} \cdot \left(3 x - 5\right) + {\left(3 x - 5\right)}^{'} \cdot \left(4 {x}^{2} + 5\right)$
$\frac{d}{d x} f \left(x\right) = 8 x \left(3 x - 5\right) + 3 \left(4 {x}^{2} + 5\right)$
$\frac{d}{d x} f \left(x\right) = 24 {x}^{2} - 40 x + 12 {x}^{2} + 15$
$\frac{d}{d x} f \left(x\right) = 36 {x}^{2} - 40 x + 15$