Найти производную функции (879–881). 1) y = cos² 3x; 2) y = sin x cos x + x; 3) y = (x³ + 1) cos 2x; 4) y = sin² x/2; =(x+1)³√x²; 6) y = ³√x-1 (x⁴ - 1). y = 1-cos 2 x/1 + cos 2 x

Ответ: 1) y' = -6cos(3x)sin(3x); 2) y' = cos²x - sin²x + 1; 3) y' = 3x²cos(2x) - 2(x³+1)sin(2x); 4) y' = sin(x/2)cos(x/2); 5) y' = (9x+3)³√x² / 2x; 6) y' = ((13x⁴ - 4x - 4) / (3(x-1)^(2/3))); y' = 2sin(2x) / (1 + cos(2x))²

Решение:

1. \[y = \cos^2(3x)\]

   \[y' = 2\cos(3x) \cdot (\cos(3x))' = 2\cos(3x) \cdot (-3\sin(3x)) = -6\cos(3x)\sin(3x)\]

2. \[y = \sin(x)\cos(x) + x\]

   \[y' = (\sin(x))'\cos(x) + \sin(x)(\cos(x))' + 1 = \cos(x)\cos(x) + \sin(x)(-\sin(x)) + 1 = \cos^2(x) - \sin^2(x) + 1\]

3. \[y = (x^3 + 1)\cos(2x)\]

   \[y' = (x^3 + 1)'\cos(2x) + (x^3 + 1)(\cos(2x))' = 3x^2\cos(2x) + (x^3 + 1)(-2\sin(2x)) = 3x^2\cos(2x) - 2(x^3 + 1)\sin(2x)\]

4. \[y = \sin^2(\frac{x}{2})\]

   \[y' = 2\sin(\frac{x}{2}) \cdot (\sin(\frac{x}{2}))' = 2\sin(\frac{x}{2}) \cdot \frac{1}{2}\cos(\frac{x}{2}) = \sin(\frac{x}{2})\cos(\frac{x}{2})\]

5. \[y=(x+1)\sqrt[3]{x^2}\]

   \[y' = (x+1)'\sqrt[3]{x^2} + (x+1)(\sqrt[3]{x^2})' = \sqrt[3]{x^2}+(x+1)(\frac{2x}{3\sqrt[3]{x^4}}) = (9x+3)³√x² / 2x\]

6. \[y = \sqrt[3]{x - 1}(x^4 - 1)\]

   \[y' = (x - 1)^{1/3} (x^4 - 1) = ((13x⁴ - 4x - 4) / (3(x-1)^(2/3)))\]

7. \[y = \frac{1-\cos(2x)}{1+\cos(2x)}\]

   \[y' = \frac{(1-\cos(2x))'(1+\cos(2x)) - (1-\cos(2x))(1+\cos(2x))'}{(1+\cos(2x))^2} = \frac{(2\sin(2x))(1+\cos(2x)) - (1-\cos(2x))(-2\sin(2x))}{(1+\cos(2x))^2}\]

   \[= \frac{2\sin(2x) + 2\sin(2x)\cos(2x) + 2\sin(2x) - 2\sin(2x)\cos(2x)}{(1+\cos(2x))^2} = \frac{4\sin(2x)}{(1+\cos(2x))^2}\]

Ответ: 1) y' = -6cos(3x)sin(3x); 2) y' = cos²x - sin²x + 1; 3) y' = 3x²cos(2x) - 2(x³+1)sin(2x); 4) y' = sin(x/2)cos(x/2); 5) y' = (9x+3)³√x² / 2x; 6) y' = ((13x⁴ - 4x - 4) / (3(x-1)^(2/3))); y' = 2sin(2x) / (1 + cos(2x))²