1. Найти производную функции: 1) y = 4x³+x⁶-3x²+x+7 2) y = 3x³-4x x² 3) y = (3x²+5x)(x+1) 4) y = cos² x 5) y = 2sin 4x

Решение:

Используем правила дифференцирования.

1. \( y = 4x^3 + x^6 - 3x^2 + x + 7 \)
   \( y' = (4x^3)' + (x^6)' - (3x^2)' + (x)' + (7)' \)
   \( y' = 12x^2 + 6x^5 - 6x + 1 \)
2. \( y = \frac{3x^3 - 4x}{x^2} = 3x - \frac{4}{x} = 3x - 4x^{-1} \)
   \( y' = (3x)' - (4x^{-1})' \)
   \( y' = 3 - 4(-1)x^{-2} = 3 + \frac{4}{x^2} \)
3. \( y = (3x^2 + 5x)(x+1) \)
   \( y' = (3x^2 + 5x)'(x+1) + (3x^2 + 5x)(x+1)' \)
   \( y' = (6x + 5)(x+1) + (3x^2 + 5x)(1) \)
   \( y' = 6x^2 + 6x + 5x + 5 + 3x^2 + 5x \)
   \( y' = 9x^2 + 16x + 5 \)
4. \( y = cos^2 x = (cos x)^2 \)
   \( y' = 2(cos x) \cdot (cos x)' \)
   \( y' = 2(cos x)(-\sin x) = -2\sin x \cos x = -\sin(2x) \)
5. \( y = 2\sin 4x \)
   \( y' = 2(\sin 4x)' \)
   \( y' = 2(\cos 4x) \cdot (4x)' \)
   \( y' = 2(\cos 4x) \cdot 4 = 8\cos 4x \)

Ответ: 1) \( 12x^2 + 6x^5 - 6x + 1 \); 2) \( 3 + \frac{4}{x^2} \); 3) \( 9x^2 + 16x + 5 \); 4) \( -\sin(2x) \); 5) \( 8\cos 4x \).