876. Найти значение производной функции f(x) в точке x0, если:

Решение:

1) \( f(x) = \cos x \sin x, x_0 = \frac{\pi}{6} \)

• \( f'(x) = (\cos x \sin x)' = -\sin^2 x + \cos^2 x = \cos 2x \)
• \( f'(\frac{\pi}{6}) = \cos (2 \cdot \frac{\pi}{6}) = \cos (\frac{\pi}{3}) = \frac{1}{2} \)

2) \( f(x) = e^x \ln x, x_0 = 1 \)

• \( f'(x) = (e^x \ln x)' = e^x \ln x + e^x \cdot \frac{1}{x} \)
• \( f'(1) = e^1 \ln 1 + e^1 \cdot \frac{1}{1} = e \cdot 0 + e = e \)

3) \( f(x) = \frac{2 \cos x}{\sin x} = 2 \cot x, x_0 = \frac{\pi}{4} \)

• \( f'(x) = (2 \cot x)' = -2 \csc^2 x \)
• \( f'(\frac{\pi}{4}) = -2 \csc^2 (\frac{\pi}{4}) = -2 (\sqrt{2})^2 = -2 \cdot 2 = -4 \)

4) \( f(x) = \ln (x^3) = 3\ln x, x_0 = \frac{1}{3} \)

• \( f'(x) = (3\ln x)' = \frac{3}{x} \)
• \( f'(\frac{1}{3}) = \frac{3}{\frac{1}{3}} = 3 \cdot 3 = 9 \)

Ответ: 1) \( \frac{1}{2} \); 2) \( e \); 3) \( -4 \); 4) \( 9 \).