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

Question 27

Multiple Choice

Use the Chain Rule to find ∂z∂s\frac { \partial z } { \partial s } .
z=eγcos⁡(θ) ,r=6st,θ=s2+t2z = e ^ { \gamma } \cos ( \theta ) , r = 6 s t , \theta = \sqrt { s ^ { 2 } + t ^ { 2 } }


A)
∂z∂s=ey(6tcos⁡(θ) −ssin⁡(θ) s2+t2) \frac { \partial z } { \partial s } = e ^ { y } \left( 6 t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)
B)
∂z∂s=(6tcos⁡(θ) +seγsin⁡(θ) s2+t2) \frac { \partial z } { \partial s } = \left( 6 t \cos ( \theta ) + \frac { s e ^ { \gamma } \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)
C)
∂z∂s=ey(tcos⁡(θ) −ssin⁡(θ) s2+t) \frac { \partial z } { \partial s } = e ^ { y } \left( t \cos ( \theta ) - \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } + t } } \right)
D)
∂z∂s=eγ(6tcos⁡(θ) +ssin⁡(θ) s2−t2) \frac { \partial z } { \partial s } = e ^ { \gamma } \left( 6 t \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)
E)
∂z∂s=ey(cos⁡(θ) +ssin⁡(θ) s2−t2) \frac { \partial z } { \partial s } = e ^ { y } \left( \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)

Correct Answer:

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