Question

Найти dz, если z=yx^y

Answer 1

Ответ:

$z = y {x}^{y}$

$Z'x = y \times y {x}^{y - 1} = {y}^{2} {x}^{y - 1}$

$Z'y = (y)' {x}^{y} + ( {x}^{y} )'y = {x}^{y} + y \times ln(x) \times {x}^{y} = {x}^{y} (1 + y ln(x) )$

$dz = {y}^{2} {x}^{y - 1} dx + {x}^{y} (1 + y ln(x)) dy$