Use the Chain Rule to find $\frac { \partial z } { \partial s }$.
$$z = e ^ { \gamma } \cos ( \theta ) , r = 6 s t , \theta = \sqrt { s ^ { 2 } + t ^ { 2 } }$$
A)
$$\frac { \partial z } { \partial s } = e ^ { y } \left( 6 t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)$$
B)
$\frac { \partial z } { \partial s } = \left( 6 t \cos ( \theta ) + \frac { \operatorname { se } ^ { \gamma } \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)$
C)
$\frac { \partial z } { \partial s } = e ^ { y } \left( t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t } } \right)$
D)
$\frac { \partial z } { \partial s } = e ^ { \gamma } \left( 6 t \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)$
E)
$\frac { \partial z } { \partial s } = e ^ { \gamma } \left( \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)$

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Using the Chain Rule, we have:$\frac { \partial z } { \partial s } = \frac { \partial z } { \partial r } \frac { \partial r } { \partial s } + \frac { \partial z } { \partial \theta } \frac { \partial \theta } { \partial s }$where$\frac { \partial z } { \partial r } = e ^ { \gamma } \cos ( \theta )$$\frac { \partial r } { \partial s } = 6 t$$\frac { \partial z } { \partial \theta } = - e ^ { \gamma } \sin ( \theta )$$\frac { \partial \theta } { \partial s } = \frac { s } { \sqrt { s ^ { 2 } + t ^ { 2 } } }$Substituting these partial derivatives into the Chain Rule equation, we get:$\frac { \partial z } { \partial s } = e ^ { \gamma } \cos ( \theta ) ( 6 t ) - e ^ { \gamma } \sin ( \theta ) \frac { s } { \sqrt { s ^ { 2 } + t ^ { 2 } } }$Simplifying, we get:$\frac { \partial z } { \partial s } = e ^ { \gamma } \left( 6 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 $\frac { \partial z } { \partial s }$