Use the Chain Rule to find $\frac { \partial z } { \partial s }$ . $z = e ^ { y } \cos ( \theta ) , r = 8 s t , \theta = \sqrt { s ^ { 2 } + t ^ { 2 } }$
A) $\frac { \partial z } { \partial s } = e ^ { \gamma } \left( 8 t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)$
B) $\frac { \partial z } { \partial s } = e ^ { \gamma } \left( 8 t \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)$
C) $\frac { \partial z } { \partial s } = \left( 8 t \cos ( \theta ) + \frac { s e ^ { \gamma } \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)$
D) $\frac { \partial z } { \partial s } = e ^ { \gamma } \left( \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)$
E) $\frac { \partial z } { \partial s } = e ^ { \gamma } \left( t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t } } \right)$

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Explanation:Using the Chain Rule, we have:$\frac { \partial z } { \partial s } = \frac { \partial z } { \partial r } \frac { \partial r } { \partial s } + \frac { \partial z } { \partial \theta } \frac { \partial \theta } { \partial s }$First, we find the partial derivatives of $z$ with respect to $r$ and $\theta$:$\frac { \partial z } { \partial r } = e ^ { y } \cos ( \theta )$$\frac { \partial z } { \partial \theta } = - e ^ { y } \sin ( \theta )$Next, we find the partial derivatives of $r$ and $\theta$ with respect to $s$:$\frac { \partial r } { \partial s } = 8 t$$\frac { \partial \theta } { \partial s } = \frac { s } { \sqrt { s ^ { 2 } + t ^ { 2 } } }$Substituting these partial derivatives into the Chain Rule formula, we get:$\frac { \partial z } { \partial s } = e ^ { y } \cos ( \theta ) ( 8 t ) - e ^ { y } \sin ( \theta ) \left( \frac { s } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)$Simplifying, we get:$\frac { \partial z } { \partial s } = e ^ { \gamma } \left( 8 t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)$

Use the Chain Rule to Find ∂z∂s\frac { \partial z } { \partial s }∂s∂z​

Use the Chain Rule to Find $\frac { \partial z } { \partial s }$