Question

Найдите диффернциал (Г,Д,Е)

Answer 1

г)

$z = {x}^{y}$

$dz = Z'xdx + Z'ydy$

$Z'x = y {x}^{y - 1} \\$

$Z'y = ln(x) \times {x}^{y}$

$dz = y {x}^{y - 1} dx + ln(x) \times {x}^{y} dy \\ dz = {x}^{y} ( \frac{y}{x} dx + ln(x) dy)$

д)

$z = \frac{ {y}^{4} }{x} \\$

$Z'x = {y}^{4} \times ( {x}^{ - 1} )' = \\ = - {y}^{4} {x}^{ - 2} = - \frac{ {y}^{4} }{ {x}^{2} }$

$Z'y = \frac{1}{x} \times 4 {y}^{3} = \frac{4 {y}^{3} }{x} \\$

$dz = - \frac{ {y}^{4} }{ {x}^{2} } dx + \frac{4 {y}^{3} }{x} dy \\ dz = \frac{ {y}^{3} }{x} ( - \frac{ y}{x} dx + 4dy)$

е)

$z = {e}^{ \sin( \frac{x}{y} ) } \\$

$Z'x = {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} ) \times \frac{1}{y} \\$

$Z'y = {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} ) \times x \times ( {y}^{ - 1} )' = \\ = {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} ) \times ( - x {y}^{ - 2} ) = \\ = - \frac{x}{ {y}^{2} } {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} )$

$dz = {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} ) \times \frac{1}{y} dx - \frac{x}{ {y}^{2} } {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} ) dy \\ dz = {e}^{ \sin( \frac{x}{y} ) } \cos( \frac{x}{y} ) \times ( \frac{1}{y}dx - \frac{ x}{ {y}^{2} } dy)$