Решение:

1) $$y = 3x^2 - \frac{1}{x^3} = 3x^2 - x^{-3}$$ $$y' = (3x^2)' - (x^{-3})' = 3 \cdot 2x - (-3)x^{-4} = 6x + \frac{3}{x^4}$$ Ответ: $$y' = 6x + \frac{3}{x^4}$$ 2) $$y = \left(\frac{x}{3} + 7\right)^6$$ $$y' = 6 \cdot \left(\frac{x}{3} + 7\right)^5 \cdot \left(\frac{x}{3} + 7\right)' = 6 \cdot \left(\frac{x}{3} + 7\right)^5 \cdot \frac{1}{3} = 2\left(\frac{x}{3} + 7\right)^5$$ Ответ: $$y' = 2\left(\frac{x}{3} + 7\right)^5$$ 3) $$y = e^x \cos x$$ $$y' = (e^x)' \cdot \cos x + e^x \cdot (\cos x)' = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x)$$ Ответ: $$y' = e^x(\cos x - \sin x)$$ 4) $$y = \frac{2^x}{\sin x}$$ $$y' = \frac{(2^x)' \cdot \sin x - 2^x \cdot (\sin x)'}{\sin^2 x} = \frac{2^x \ln 2 \sin x - 2^x \cos x}{\sin^2 x} = \frac{2^x(\ln 2 \sin x - \cos x)}{\sin^2 x}$$ Ответ: $$y' = \frac{2^x(\ln 2 \sin x - \cos x)}{\sin^2 x}$$