Find the derivative of the following functions:

Okay, let's find the derivatives of these functions. I'll provide the solution step-by-step. 1) $$y = (5x + 6)^4$$ $$y' = 4(5x + 6)^3 cdot 5 = 20(5x + 6)^3$$ 2) $$y = (2 - 7x^2 + 3x)^3$$ $$y' = 3(2 - 7x^2 + 3x)^2 cdot (-14x + 3) = (9 - 42x)(2 - 7x^2 + 3x)^2$$ 3) $$y = 4(2x - 9)^2$$ $$y' = 4 cdot 2(2x - 9) cdot 2 = 16(2x - 9)$$ 4) $$y = \frac{1}{(3x + 5)^3} = (3x + 5)^{-3}$$ $$y' = -3(3x + 5)^{-4} cdot 3 = -\frac{9}{(3x + 5)^4}$$ 5) $$y = \frac{5}{(6 - 4x)^5} = 5(6 - 4x)^{-5}$$ $$y' = 5 cdot (-5)(6 - 4x)^{-6} cdot (-4) = \frac{100}{(6 - 4x)^6}$$ 6) $$y = 2\sqrt{6x + 2} = 2(6x + 2)^{\frac{1}{2}}$$ $$y' = 2 cdot \frac{1}{2} (6x + 2)^{-\frac{1}{2}} cdot 6 = \frac{6}{\sqrt{6x + 2}}$$ 7) $$y = \sqrt[4]{\frac{x}{4} - 12} = (\frac{x}{4} - 12)^{\frac{1}{4}}$$ $$y' = \frac{1}{4}(\frac{x}{4} - 12)^{-\frac{3}{4}} cdot \frac{1}{4} = \frac{1}{16(\frac{x}{4} - 12)^{\frac{3}{4}}}$$ 8) $$y = \sin(6x - \frac{\pi}{3})$$ $$y' = \cos(6x - \frac{\pi}{3}) cdot 6 = 6\cos(6x - \frac{\pi}{3})$$ 9) $$y = 4\cos(2x + \pi)$$ $$y' = 4(-\sin(2x + \pi)) cdot 2 = -8\sin(2x + \pi)$$ 10) $$y = tg(3x - \frac{\pi}{4})$$ $$y' = \frac{1}{\cos^2(3x - \frac{\pi}{4})} cdot 3 = \frac{3}{\cos^2(3x - \frac{\pi}{4})}$$ 11) $$y = 4ctg(\frac{x}{2} + \frac{\pi}{6})$$ $$y' = 4(-\frac{1}{\sin^2(\frac{x}{2} + \frac{\pi}{6})}) cdot \frac{1}{2} = -\frac{2}{\sin^2(\frac{x}{2} + \frac{\pi}{6})}$$ 12) $$y = 5\sin^3(3x + \frac{\pi}{2})$$ $$y' = 5 cdot 3\sin^2(3x + \frac{\pi}{2}) cdot \cos(3x + \frac{\pi}{2}) cdot 3 = 45\sin^2(3x + \frac{\pi}{2})\cos(3x + \frac{\pi}{2})$$