Find the derivative of each of these functions, 9. \( y=\left(3 x^{2}+6\right)(2 x-1) \) 10. \( f(x)=e^{x}(2 x+\sin x) \) 11. \( y=\frac{4 x}{x^{2}+1} \) 12. \( y=\frac{x-2}{2 x^{2}-\cos x} \) 13. \( f(x)=\sin 5 x \) 14. \( f(x)=\log 2 x \) 15. \( f(x)=\left(x^{3}+2 x\right)^{27} \)

To find the derivatives of each function, we will use various differentiation rules such as the product rule, quotient rule, and chain rule. 9. For \( y=\left(3 x^{2}+6\right)(2 x-1) \), we apply the product rule: \[ y' = (3x^2 + 6)'(2x - 1) + (3x^2 + 6)(2x - 1)' = (6x)(2x - 1) + (3x^2 + 6)(2) \] Simplifying, we find \( y' = 12x^2 - 6x + 6x^2 + 12 = 18x^2 + 12 \). 10. For \( f(x)=e^{x}(2 x+\sin x) \), we again use the product rule: \[ f'(x) = (e^x)'(2x + \sin x) + e^x(2x + \sin x)' = e^x(2x + \sin x) + e^x(2 + \cos x) \] Hence, \( f'(x) = e^x(2x + \sin x + 2 + \cos x) \). 11. For \( y=\frac{4 x}{x^{2}+1} \), we use the quotient rule: \[ y' = \frac{(4)(x^2 + 1) - (4x)(2x)}{(x^2 + 1)^2} = \frac{4(x^2 + 1) - 8x^2}{(x^2 + 1)^2} \] This simplifies to \( y' = \frac{-4x^2 + 4}{(x^2 + 1)^2} \) or \( y' = \frac{4(1 - x^2)}{(x^2 + 1)^2} \). 12. For \( y=\frac{x-2}{2 x^{2}-\cos x} \), the quotient rule gives us: \[ y' = \frac{(1)(2x^2 - \cos x) - (x-2)(4x + \sin x)}{(2x^2 - \cos x)^2} \] After some simplification, this results in: \[ y' = \frac{(2x^2 - \cos x) - (4x^2 - 8 + (x-2)\sin x)}{(2x^2 - \cos x)^2} \]. 13. For \( f(x)=\sin 5x \), we use the chain rule: \[ f'(x) = \cos(5x) \cdot (5) = 5\cos(5x) \]. 14. For \( f(x)=\log 2x \), recall that \( \log 2x = \log 2 + \log x \), so: \[ f'(x) = 0 + \frac{1}{x} = \frac{1}{x} \]. 15. For \( f(x)=\left(x^{3}+2 x\right)^{27} \), we again utilize the chain rule: \[ f'(x) = 27\left(x^3 + 2x\right)^{26}(3x^2 + 2) \]. And there you go! Each derivative is calculated with precision!