Задача 78915 Надо решить задачу , номер 1 и 2 (...
Условие
Решение
[m]y' = \frac{3 \cdot (1/3)(x^5 + 5x^4 - 5)^{-2/3} \cdot (5x^4 + 20x^3) \cdot x - 3(x^5 + 5x^4 - 5)^{1/3}}{x^2} = \frac{x(5x^4 + 20x^3)(x^5 + 5x^4 - 5)^{-2/3} - 3(x^5 + 5x^4 - 5)^{1/3}}{x^2}[/m]
б) [m]y = \frac{1}{3} tg^3 x\ arcsin\ 4x[/m]
[m]y' = \frac{1}{3}(3tg^2 x \cdot \frac{1}{\cos^2 x} \cdot arcsin\ 4x + \frac{tg^3 x \cdot 4}{\sqrt{1 - 16x^2}}) = \frac{tg^2 x }{\cos^2 x} \cdot arcsin\ 4x + \frac{4tg^3 x}{3 \sqrt{1 - 16x^2}}[/m]
в) [m]y = \sqrt{2 + \ln(1 - ctg\ 2x)}[/m]
[m]y' = \frac{1}{2\sqrt{2 + \ln(1 - ctg\ 2x)}} \cdot \frac{1}{1 - ctg\ 2x} \cdot \frac{1}{\sin^2 x}[/m]
г) [m]y = (1 - \ln x)^{\sqrt{x}} = e^{\ln (1 - \ln x)^{\sqrt{x}}} = e^{\sqrt{x} \cdot \ln (1 - \ln x)}[/m]
[m]y' = e^{\sqrt{x} \cdot \ln (1 - \ln x)} \cdot (\frac{\ln (1 - \ln x)}{2\sqrt{x}} + \sqrt{x} \cdot \frac{1}{1 - \ln x} (-\frac{1}{x})) =[/m]
[m]= (1 - \ln x)^{\sqrt{x}} \cdot (\frac{\ln (1 - \ln x)}{2\sqrt{x}} - \frac{1}{\sqrt{x}(1 - \ln x)})[/m]
д) [m]\cos (x^3+y^4) = \frac{x^2}{y}[/m]
Неявно заданная функция. "y = " в начале, очевидно, ошибка.
Берем производные от левой и правой части:
[m]-\sin (x^3 + y^4) \cdot (3x^2 + 4y^3y') = \frac{2xy - x^2y'}{y^2}[/m]
[m]-3x^2 \sin (x^3 + y^4) - 4y^3y' \sin (x^3 + y^4) = \frac{2xy}{y^2} - \frac{x^2y'}{y^2}[/m]
[m]\frac{x^2}{y^2} \cdot y' - 4y^3\sin (x^3 + y^4) \cdot y' = \frac{2x}{y} + 3x^2 \sin (x^3 + y^4)[/m]
[m]y' = (\frac{2x}{y} + 3x^2 \sin (x^3 + y^4)) : (\frac{x^2}{y^2} - 4y^3\sin (x^3 + y^4))[/m]
2. а) [m]y = x^3(1 - \ln x)[/m]
[m]\frac{dy}{dx} = 3x^2(1 - \ln x) + x^3(-\frac{1}{x}) = 3x^2 - 3x^2\ln x - x^2 = 2x^2 - 3x^2\ln x [/m]
[m]\frac{d^2y}{dx^2} = 4x - 3(2x \cdot \ln x + \frac{x^2}{x}) = 4x - 6x\cdot \ln x - 3x = x - 6x\cdot \ln x[/m]
б) [m]\begin{cases}
x = t + \ln \cos t \\
y = t = \ln \sin t \\
\end{cases}[/m]
Находим производные по t:
[m]\frac{dx}{dt} = 1 + \frac{1}{\cos t} \cdot (-\sin t) = 1 - tg\ t[/m]
[m]\frac{dy}{dt} = 1 + \frac{1}{\sin t} \cdot \cos t = 1 + ctg\ t[/m]
Первая производная:
[m]\frac{dy}{dx} = \frac{y'_{t}}{x'_{t}} = \frac{dy}{dt} : \frac{dx}{dt} = \frac{1 + ctg\ t}{1 - tg\ t} = [/m]
[m] = \frac{1 + \cos t/\sin t}{1 - \sin t/\cos t} = \frac{\sin t + \cos t}{\sin t} : \frac{\cos t - \sin t}{\cos t} = [/m]
[m] = \frac{\cos t(\sin t + \cos t)}{\sin t(\cos t - \sin t)} = \frac{\sin t \cos t + \cos^2 t}{\sin t \cos t - \sin^2 t}[/m]
Вторая производная:
[m]\frac{d^2 y}{dx^2} = \frac{(y'_{x})'_{t}}{x'_{t}} = (\frac{\sin t \cos t + \cos^2 t}{\sin t \cos t - \sin^2 t})'_{t} : (1 - tg\ t) =[/m]
[m]= \frac{(\cos t \cos t + \sin t(-\sin t) + 2\cos t(-\sin t))(\sin t \cos t - \sin^2 t)}{(\sin t \cos t - \sin^2 t)^2(1 - tg\ t)} -[/m]
[m]- \frac{(\sin t \cos t + \cos^2 t)(\cos t \cos t + \sin t(-\sin t) - 2\sin t \cos t)}{(\sin t \cos t - \sin^2 t)^2(1 - tg\ t)}[/m]
[m]= \frac{(\cos^2 t - \sin^2 t - 2\sin t \cos t)(\sin t \cos t - \sin^2 t)}{(\sin t \cos t - \sin^2 t)^2(\cos t - \sin t)/\cos t} -[/m]
[m]- \frac{(\sin t \cos t + \cos^2 t)(\cos^2 t - \sin^2 t - 2\sin t \cos t)}{(\sin t \cos t - \sin^2 t)^2(\cos t - \sin t)/\cos t}[/m]
Закончи это упрощать сам, я задолбался