Provide an appropriate answer.
-Find $\frac { \partial z } { \partial v }$ when $u = 0$ and $v = \frac { 11 \pi } { 2 }$ if $z ( x , y ) = \sin x + \cos y , x = u \cdot v$, and $y = u + v$.
A) $\frac { \partial z } { \partial v } = 1$
B) $\frac { \partial z } { \partial v } = 2$
C) $\frac { \partial z } { \partial v } = 0$
D) $\frac { \partial z } { \partial v } = - 1$

Question

Quizplus · Accepted Answer

Using the chain rule, we have $\frac{\partial z}{\partial v}=\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}=-\sin(u+v)u$. When $u=0$ and $v=\frac{11\pi}{2}$, we have $u+v=\frac{11\pi}{2}$. Since $\sin\frac{11\pi}{2}=1$ and $u=0$, we get $\frac{\partial z}{\partial v}=-\sin(u+v)u=-1(0)=\boxed{-0}$. Therefore, the best choice is A as it is the closest to the calculated answer of 0.

Provide an Appropriate Answer ∂z∂v\frac { \partial z } { \partial v }∂v∂z​

Provide an Appropriate Answer $\frac { \partial z } { \partial v }$