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

Question

Quizplus · Accepted Answer

We use the chain rule as follows:
$$\frac {\partial z} {\partial s} = \frac {\partial z} {\partial \gamma} \cdot \frac {\partial \gamma} {\partial s} + \frac {\partial z} {\partial \theta} \cdot \frac {\partial \theta} {\partial s}$$
Using the given expressions for $z$ and $\theta$, we have:
$$\frac {\partial z} {\partial \gamma} = e^\gamma \cos \theta \qquad \frac {\partial \gamma} {\partial s} = 0 \qquad \frac {\partial z} {\partial \theta} = - e^\gamma \sin\theta \qquad \frac {\partial \theta} {\partial s} = \frac {s} {\sqrt{s^2+t^2}}$$
Substituting these values and simplifying, we get:
$$\frac {\partial z} {\partial s} = e^\gamma \cos \theta \cdot 0 - e^\gamma \sin\theta \cdot \frac {s} {\sqrt{s^2+t^2}} = e^y \left(6t\cos\theta - \frac {s\sin\theta} {\sqrt{s^2+t^2}}\right)$$

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

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