Let $z = x \cos y + y \sin x$ . Then the differential $d ^ { z}$ is

A) $( \cos y + y \cos x ) d x + ( - x \sin y - \sin x ) d y$
B) $( \cos y + y \cos x ) d x + ( - x \sin y + \sin x ) d y$
C) $( \cos y - y \cos x ) d x + ( x \sin y + \sin x ) d y$
D) $( \cos y - y \cos x ) d x - ( - x \sin y + \sin x ) d y$
E) $( \cos y - y \cos x ) d x + ( - x \sin y + \sin x ) d y$

Question

Let $z = x \cos y + y \sin x$ . Then the differential $d ^ { z}$ is

A) $( \cos y + y \cos x ) d x + ( - x \sin y - \sin x ) d y$
B) $( \cos y + y \cos x ) d x + ( - x \sin y + \sin x ) d y$
C) $( \cos y - y \cos x ) d x + ( x \sin y + \sin x ) d y$
D) $( \cos y - y \cos x ) d x - ( - x \sin y + \sin x ) d y$
E) $( \cos y - y \cos x ) d x + ( - x \sin y + \sin x ) d y$

Quizplus · Accepted Answer

To find the differential $dz$, we apply the differentiation rules, treating $x$ and $y$ as independent variables. The partial derivative of $z$ with respect to $x$ is $\frac{\partial z}{\partial x} = \cos y + y \cos x$, and the partial derivative of $z$ with respect to $y$ is $\frac{\partial z}{\partial y} = -x \sin y + \sin x$. Therefore, $dz = (\cos y + y \cos x) dx + (-x \sin y + \sin x) dy$.

Let z=xcos⁡y+ysin⁡xz = x \cos y + y \sin xz=xcosy+ysinx Then the Differential

Let $z = x \cos y + y \sin x$ Then the Differential