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

question 97

Multiple Choice

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)
• \(\frac { \partial z } { \partial s } = \left( 6 t \cos ( \theta ) + \frac { \operatorname { se } ^ { y } \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)\)
• 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)
• \(\frac { \partial z } { \partial s } = e ^ { y } \left( \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)\)

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