С-2. Вычислите производную функций по правилу дифференцирования: если y = C·u, то y' = C·u'

Для решения данных задач используем правило вынесения константы: если $y = C \cdot u(x)$, то $y' = C \cdot u'(x)$. 1) $y = -3x^2 \implies y' = -3 \cdot 2x = -6x$ 2) $y = -3x^{1/6} \implies y' = -3 \cdot \frac{1}{6}x^{-5/6} = -\frac{1}{2}x^{-5/6} = -\frac{1}{2\sqrt[6]{x^5}}$ 3) $y = -\frac{1}{8}x^6 \implies y' = -\frac{1}{8} \cdot 6x^5 = -\frac{3}{4}x^5$ 4) $y = \frac{1}{6}\operatorname{tg}x \implies y' = \frac{1}{6} \cdot \frac{1}{\cos^2 x} = \frac{1}{6\cos^2 x}$ 5) $y = 4 \cdot 2^x \implies y' = 4 \cdot 2^x \ln 2$ 6) $y = -\frac{1}{x} = -x^{-1} \implies y' = -(-1)x^{-2} = \frac{1}{x^2}$ 7) $y = -\frac{1}{2}\operatorname{ctg}x \implies y' = -\frac{1}{2} \cdot \left(-\frac{1}{\sin^2 x}\right) = \frac{1}{2\sin^2 x}$ 8) $y = 4e^x \implies y' = 4e^x$ 9) $y = \frac{4}{x} = 4x^{-1} \implies y' = 4(-1)x^{-2} = -\frac{4}{x^2}$ 10) $y = -4\cos x \implies y' = -4(-\sin x) = 4\sin x$ 11) $y = 2 \cdot (1/3)^x \implies y' = 2 \cdot (1/3)^x \ln(1/3) = -2(1/3)^x \ln 3$ 12) $y = 5\log_2 x \implies y' = 5 \cdot \frac{1}{x \ln 2} = \frac{5}{x \ln 2}$ 13) $y = \frac{1}{2x^4} = \frac{1}{2}x^{-4} \implies y' = \frac{1}{2}(-4)x^{-5} = -\frac{2}{x^5}$ 14) $y = 3x^6 \implies y' = 3 \cdot 6x^5 = 18x^5$ 15) $y = \frac{1}{3}x^6 \implies y' = \frac{1}{3} \cdot 6x^5 = 2x^5$ 16) $y = 3\operatorname{ctg}x \implies y' = 3(-\frac{1}{\sin^2 x}) = -\frac{3}{\sin^2 x}$ 17) $y = \frac{\cos x}{5} = \frac{1}{5}\cos x \implies y' = \frac{1}{5}(-\sin x) = -\frac{1}{5}\sin x$ 18) $y = -\frac{1}{7}x^7 \implies y' = -\frac{1}{7} \cdot 7x^6 = -x^6$ 19) $y = -\frac{2}{5}x^5 \implies y' = -\frac{2}{5} \cdot 5x^4 = -2x^4$ 20) $y = -\operatorname{ctg}x \implies y' = -(-\frac{1}{\sin^2 x}) = \frac{1}{\sin^2 x}$ 21) $y = -2\sin x \implies y' = -2\cos x$ 22) $y = \frac{1}{2} \cdot 6^x \implies y' = \frac{1}{2} \cdot 6^x \ln 6$ 23) $y = 3\ln x \implies y' = 3 \cdot \frac{1}{x} = \frac{3}{x}$ 24) $y = -4x^{1/4} \implies y' = -4 \cdot \frac{1}{4}x^{-3/4} = -x^{-3/4} = -\frac{1}{\sqrt[4]{x^3}}$ 25) $y = -2\log_{1/4} x \implies y' = -2 \cdot \frac{1}{x \ln(1/4)} = -2 \cdot \frac{1}{x(- \ln 4)} = \frac{2}{x \ln 4}$ 26) $y = \frac{2}{x^5} = 2x^{-5} \implies y' = 2(-5)x^{-6} = -\frac{10}{x^6}$ 27) $y = -\frac{1}{2}\ln x \implies y' = -\frac{1}{2x}$ 28) $y = -4\log_{1/5} x \implies y' = -4 \cdot \frac{1}{x \ln(1/5)} = -4 \cdot \frac{1}{x(-\ln 5)} = \frac{4}{x \ln 5}$ 29) $y = \frac{1}{5}x^{10} \implies y' = \frac{1}{5} \cdot 10x^9 = 2x^9$ 30) $y = -7\operatorname{ctg}x \implies y' = -7(-\frac{1}{\sin^2 x}) = \frac{7}{\sin^2 x}$