Question

найти частные производные второго порядка функции z=z(x,y). Функция: z=xy+y+ctg(xy)

Answer 1

Ответ:

$z = xy + y + ctg(xy)$

$z'_x = y - \frac{y}{ \sin {}^{2} (xy) }$

$z'_y = x + 1 - \frac{x}{ \sin {}^{2} (xy) }$

$z''_{xx} = ( - y( \sin(xy)) {}^{ - 2} )'_x = 2y( \sin(xy)) {}^{ - 3} \times ( \sin(xy))'_x \times (xy)'_x= \\ = \frac{2y}{ \sin {}^{3} (xy) } \times \cos(xy) \times y = \frac{2 {y}^{2} \cos(xy) }{ \sin {}^{3} (xy) }$

$z''_{yy} = ( - x \sin {}^{ - 2} (xy))'_y = 2x \sin {}^{ - 3} (xy) \times \cos(xy) \times x = \\ = \frac{2 {x}^{2} \cos(xy) }{ \sin {}^{3} (xy) }$

$z''_{xy} = 1 - \frac{y' \sin {}^{2} (xy) - (\sin {}^{2} (xy)) '\times y}{ \sin {}^{4} (xy) } = \\ = 1 - \frac{ \sin {}^{2} (xy) - 2 \sin(xy) \times \cos(xy) \times x \times y}{ \sin {}^{4} (xy) } = \\ = 1 - \frac{ \sin {}^{2} (xy) - xy \sin(2xy) }{ \sin {}^{4} (xy) }$

Answer 2

Ответ:

Пошаговое объяснение:

$z'_{x}=y-\frac{y}{sin^2xy}\\z'_{y}=x+1-\frac{x}{sin^2xy}$

$z''_{x}=-\frac{y*(-2)*y*cosxy}{sin^3xy}=\frac{2y^2}{sin^3xy} \\\\z''_{y}=-\frac{x}{sin^2xy}=2x^2*sin^{-3}xy=\frac{2x^2}{sin^3xy}$