Answer
f'(x)=18$x^{2}$-20x-9
Work Step by Step
The product rule states that if f(x)=h(x)g(x), then f'(x)=g'(x)h(x)+g(x)h'(x). We can find the derivative of the function f(x)=(3x-5)($2x^{2}-3$) by setting g(x)=3x-5 and h(x)=$2x^{2}-3$, and applying the product rule.
f'(x)=$\frac{d}{dx}$[3x-5]($2x^{2}-3$)+(3x-5)$\frac{d}{dx}$[$2x^{2}-3$]
$\frac{d}{dx}$[3x-5]=3, using the power rule
$\frac{d}{dx}$[$2x^{2}-3$]=4x, using the power rule
Therefore,
f'(x)=(3)($2x^{2}-3$)+(3x-5)(4x)
=(6$x^{2}$-9)+(12$x^{2}$-20x) (Simplify)
=18$x^{2}$-20x-9 (Simplify)
