Найти производную функции: 1) 3/x + 2√x - e^x; 2) (3x-5)⁴; 3) 3 sin 2x cos x; 4) x³/(x²+5)

Ответ: 1) f'(x) = -3/x² + 1/√x - e^x; 2) f'(x) = 12(3x-5)³; 3) f'(x) = 6cos(2x)cos(x) - 3sin(2x)sin(x); 4) f'(x) = (x⁴ + 15x²)/(x² + 5)²

Решение:

1. \[f(x) = \frac{3}{x} + 2\sqrt{x} - e^x\]

   \[f'(x) = (\frac{3}{x})' + (2\sqrt{x})' - (e^x)' = -\frac{3}{x^2} + \frac{1}{\sqrt{x}} - e^x\]

2. \[f(x) = (3x - 5)^4\]

   \[f'(x) = 4(3x - 5)^3 \cdot (3x - 5)' = 4(3x - 5)^3 \cdot 3 = 12(3x - 5)^3\]

3. \[f(x) = 3 \sin(2x) \cos(x)\]

   \[f'(x) = 3(\sin(2x))'\cos(x) + 3\sin(2x)(\cos(x))' = 3(2\cos(2x))\cos(x) + 3\sin(2x)(-\sin(x)) = 6\cos(2x)\cos(x) - 3\sin(2x)\sin(x)\]

4. \[f(x) = \frac{x^3}{x^2 + 5}\]

   \[f'(x) = \frac{(x^3)'(x^2 + 5) - x^3(x^2 + 5)'}{(x^2 + 5)^2} = \frac{3x^2(x^2 + 5) - x^3(2x)}{(x^2 + 5)^2} = \frac{3x^4 + 15x^2 - 2x^4}{(x^2 + 5)^2} = \frac{x^4 + 15x^2}{(x^2 + 5)^2}\]

Ответ: 1) f'(x) = -3/x² + 1/√x - e^x; 2) f'(x) = 12(3x-5)³; 3) f'(x) = 6cos(2x)cos(x) - 3sin(2x)sin(x); 4) f'(x) = (x⁴ + 15x²)/(x² + 5)²