Найти частные производные по каждой из независимых переменных. 3. z = u * sin^2(v), где u = xy, v = tg(y/x)

Решение:

Дано: \( z = u \cdot \sin^2(v) \), где \( u = xy \) и \( v = \tan(\frac{y}{x}) \).

Сначала найдем частные производные \(\frac{\partial z}{\partial x}\) и \(\frac{\partial z}{\partial y}\). Воспользуемся правилом дифференцирования сложной функции:

\[\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}\] \[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}\]

Найдем необходимые производные:

\[\frac{\partial z}{\partial u} = \sin^2(v)\] \[\frac{\partial z}{\partial v} = u \cdot 2 \sin(v) \cos(v) = u \sin(2v)\] \[\frac{\partial u}{\partial x} = y\] \[\frac{\partial u}{\partial y} = x\] \[\frac{\partial v}{\partial x} = \frac{1}{\cos^2(\frac{y}{x})} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 \cos^2(\frac{y}{x})}\] \[\frac{\partial v}{\partial y} = \frac{1}{\cos^2(\frac{y}{x})} \cdot \frac{1}{x} = \frac{1}{x \cos^2(\frac{y}{x})}\]

Подставим найденные производные в формулы для \(\frac{\partial z}{\partial x}\) и \(\frac{\partial z}{\partial y}\):

\[\frac{\partial z}{\partial x} = \sin^2(v) \cdot y + u \sin(2v) \cdot \left(-\frac{y}{x^2 \cos^2(\frac{y}{x})}\right)\] \[\frac{\partial z}{\partial y} = \sin^2(v) \cdot x + u \sin(2v) \cdot \frac{1}{x \cos^2(\frac{y}{x})}\]

Теперь подставим \( u = xy \) и \( v = \tan(\frac{y}{x}) \):

\[\frac{\partial z}{\partial x} = y \sin^2\left(\tan\left(\frac{y}{x}\right)\right) - xy \sin\left(2 \tan\left(\frac{y}{x}\right)\right) \cdot \frac{y}{x^2 \cos^2(\frac{y}{x})}\] \[\frac{\partial z}{\partial y} = x \sin^2\left(\tan\left(\frac{y}{x}\right)\right) + xy \sin\left(2 \tan\left(\frac{y}{x}\right)\right) \cdot \frac{1}{x \cos^2(\frac{y}{x})}\]

Упростим:

\[\frac{\partial z}{\partial x} = y \sin^2\left(\tan\left(\frac{y}{x}\right)\right) - \frac{y^2}{x} \cdot \frac{\sin\left(2 \tan\left(\frac{y}{x}\right)\right)}{\cos^2(\frac{y}{x})}\] \[\frac{\partial z}{\partial y} = x \sin^2\left(\tan\left(\frac{y}{x}\right)\right) + y \cdot \frac{\sin\left(2 \tan\left(\frac{y}{x}\right)\right)}{\cos^2(\frac{y}{x})}\]

Ответ:

\[\frac{\partial z}{\partial x} = y \sin^2\left(\tan\left(\frac{y}{x}\right)\right) - \frac{y^2}{x} \cdot \frac{\sin\left(2 \tan\left(\frac{y}{x}\right)}{\cos^2(\frac{y}{x})}\] \[\frac{\partial z}{\partial y} = x \sin^2\left(\tan\left(\frac{y}{x}\right)\right) + y \cdot \frac{\sin\left(2 \tan\left(\frac{y}{x}\right)}{\cos^2(\frac{y}{x})}\]