Question

Найти дифференциал функции:

$u=\frac{tgx^{2} xy}{z}$

Answer 1

Пошаговое объяснение:

$u = \frac{ {tg}^{2}x \times xy }{z} \\$

$du = U'xdx + U'ydy + U'zdz$

$U'x = \frac{y}{ z } \times (x {tg}^{2} x) = \frac{y}{z} \times ( {tg}^{2} x + 2tgx \times \frac{1}{ { \cos }^{2} x} \times x) = \\ = \frac{ytgx}{z} (tgx + \frac{2x}{ { \cos }^{2} x} )$

$U'y = x {tg}^{2} x \times \frac{1}{z} \times (y) = \frac{x {tg}^{2}x }{z} \\$

$U'z = xy {tg}^{2} x \times ( {z}^{ - 1} ) = xy {tg}^{2} x( - {z}^{ - 2} ) = \\ = - \frac{xy {tg}^{2}x }{ {z}^{2} }$

$du = \frac{ytgx}{z}(tgx + \frac{2x}{ { \cos }^{2}x } )dx + \frac{x {tg}^{2} x}{z} dy - \frac{xy {tg}^{2}x }{ {z}^{2} } dz = \\ = \frac{tgx}{z} ((ytgx + \frac{2xy}{ { \cos}^{2} x} )dx + xtgxdy - \frac{xytgx}{z} dz)$