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МатематикаВопрос:
Найти производные функций:
Ответ:
Нахождение производных функций:
- \(y = -8x + x^{12}\)
\(y' = -8 + 12x^{11}\) - \(y = 9x^2 + 5x^4 + 15\)
\(y' = 18x + 20x^3\) - \(y = 9x - 3\sqrt{x}\)
\(y' = 9 - 3 \cdot \frac{1}{2\sqrt{x}} = 9 - \frac{3}{2\sqrt{x}}\) - \(y = \frac{1}{x} - 7x^{-4} + 10\)
\(y' = -\frac{1}{x^2} - 7 \cdot (-4)x^{-5} = -\frac{1}{x^2} + \frac{28}{x^5}\) - \(y = -\frac{2}{x^4} - 3\sin x\)
\(y' = -2 \cdot (-4)x^{-5} - 3\cos x = \frac{8}{x^5} - 3\cos x\) - \(y = \operatorname{tg} x + \sqrt{x}\)
\(y' = \frac{1}{\cos^2 x} + \frac{1}{2\sqrt{x}}\) - \(y = \frac{5}{x^{-6}} + \operatorname{ctg} x\)
\(y = 5x^6 + \operatorname{ctg} x\)
\(y' = 30x^5 - \frac{1}{\sin^2 x}\) - \(y = (x^4 + 7)(1 + x^5)\)
\(y' = (4x^3)(1+x^5) + (x^4+7)(5x^4) = 4x^3 + 4x^8 + 5x^8 + 35x^4 = 9x^8 + 35x^4 + 4x^3\) - \(y = \sqrt{x}(3-4x)\)
\(y = 3\sqrt{x} - 4x\sqrt{x} = 3x^{1/2} - 4x^{3/2}\)
\(y' = 3 \cdot \frac{1}{2}x^{-1/2} - 4 \cdot \frac{3}{2}x^{1/2} = \frac{3}{2\sqrt{x}} - 6\sqrt{x}\) - \(y = x^8 \cos x\)
\(y' = 8x^7 \cos x + x^8 (-\sin x) = 8x^7 \cos x - x^8 \sin x\) - \(y = (\frac{6}{x} - 7)(x+2)\)
\(y = (6x^{-1} - 7)(x+2)\)
\(y' = (-6x^{-2})(x+2) + (6x^{-1}-7)(1) = -\frac{6}{x^2}(x+2) + \frac{6}{x} - 7 = -\frac{6x}{x^2} - \frac{12}{x^2} + \frac{6}{x} - 7 = -\frac{6}{x} - \frac{12}{x^2} + \frac{6}{x} - 7 = -\frac{12}{x^2} - 7\) - \(y = \frac{8x^3}{2x-9}\)
\(y' = \frac{(24x^2)(2x-9) - (8x^3)(2)}{(2x-9)^2} = \frac{48x^3 - 216x^2 - 16x^3}{(2x-9)^2} = \frac{32x^3 - 216x^2}{(2x-9)^2}\) - \(y = \frac{4\sqrt{x}}{x^3+5}\)
\(y = \frac{4x^{1/2}}{x^3+5}\)
\(y' = \frac{(4 \cdot \frac{1}{2}x^{-1/2})(x^3+5) - (4x^{1/2})(3x^2)}{(x^3+5)^2} = \frac{(2x^{-1/2})(x^3+5) - 12x^{5/2}}{(x^3+5)^2} = \frac{\frac{2x^3}{\sqrt{x}} + \frac{10}{\sqrt{x}} - 12x^{5/2}}{(x^3+5)^2} = \frac{2x^{5/2} + 10x^{-1/2} - 12x^{5/2}}{(x^3+5)^2} = \frac{10x^{-1/2} - 10x^{5/2}}{(x^3+5)^2} = \frac{\frac{10}{\sqrt{x}} - 10x^2\sqrt{x}}{(x^3+5)^2}\) - \(y = \frac{\sin x}{4x^3}\)
\(y' = \frac{(\cos x)(4x^3) - (\sin x)(12x^2)}{(4x^3)^2} = \frac{4x^3 \cos x - 12x^2 \sin x}{16x^6} = \frac{x^2(4x \cos x - 12 \sin x)}{16x^6} = \frac{4x \cos x - 12 \sin x}{16x^4} = \frac{x \cos x - 3 \sin x}{4x^4}\)
