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

Question 92

Multiple Choice

Use the Chain Rule to find zs\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)
zs=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)
zs=(6tcos(θ) +seγsin(θ) s2+t2) \frac { \partial z } { \partial s } = \left( 6 t \cos ( \theta ) + \frac { \operatorname { se } ^ { \gamma } \sin ( \theta ) } { \sqrt { s ^ { 2 } + t ^ { 2 } } } \right)
C)
zs=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)
zs=eγ(6tcos(θ) +ssin(θ) s2t2) \frac { \partial z } { \partial s } = e ^ { \gamma } \left( 6 t \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)
E)
zs=eγ(cos(θ) +ssin(θ) s2t2) \frac { \partial z } { \partial s } = e ^ { \gamma } \left( \cos ( \theta ) + \frac { s \sin ( \theta ) } { \sqrt { s ^ { 2 } - t ^ { 2 } } } \right)

Correct Answer:

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