## Calculus (3rd Edition)

f'(x)=14$x^{6}$-24$x^{5}$+15$x^{4}$-24$x^{3}$
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)=(3$x^{4}$+2$x^{6}$)(x-2) by setting g(x)=3$x^{4}$+2$x^{6}$ and h(x)=x-2, and applying the product rule. f'(x)=$\frac{d}{dx}$[3$x^{4}$+2$x^{6}$](x-2)+(3$x^{4}$+2$x^{6}$)$\frac{d}{dx}$[x-2] $\frac{d}{dx}$[3$x^{4}$+2$x^{6}$]=12$x^{3}$+12$x^{5}$, using the power rule $\frac{d}{dx}$[x-2]=1, using the power rule Therefore, f'(x)=(12$x^{3}$+12$x^{5}$)(x-2)+(3$x^{4}$+2$x^{6}$)(1) =(12$x^{4}$+12$x^{6}$-24$x^{3}$-24$x^{5}$)+(3$x^{4}$+2$x^{6}$) (Simplify) =14$x^{6}$-24$x^{5}$+15$x^{4}$-24$x^{3}$ (Simplify)