$\text { Find } \mathrm { f } _ { \mathbf { x } } , \mathrm { f } _ { \mathbf { y } } \text {, and } \mathrm { f } _ { \mathbf { Z } }$
-$f ( x , y , z ) = e^{ ( y z + \sin x )}$
A) $f _ { X } = ( \cos x ) e^{ ( y z + \sin x }) ; f _ { y } = e ^ { ( y z + \sin x ) } ; f _ { z } = e ^{( y z + \sin x )}$
B) $f _ { X } = ( \cos x ) e^{ ( y z + \sin x )} ; f _ { y } = z e^{ ( y z + \sin x )} ; f _ { z } = y e^{ ( y z + \sin x )}$
C) $f _ { x } = e ^{( y z + \sin x ) }; f _ { y } = z e ^{( y z + \sin x )} ; f _ { z } = y e ^{( y z + \sin x )}$
D) $f _ { X } = ( \cos x + y z ) e ^{( y z + \sin x )} ; f _ { y } = z e^{ ( y z + \sin x ) }; f _ { z } = y e ^ { ( y z + \sin x ) }$

Question

Quizplus · Accepted Answer

The partial derivatives of $f(x, y, z) = e^{(yz + \sin x)}$ with respect to $x$, $y$, and $z$ are calculated as follows: $f_x = \cos(x)e^{(yz + \sin x)}$ because the derivative of $\sin(x)$ with respect to $x$ is $\cos(x)$, and the chain rule is applied. $f_y = ze^{(yz + \sin x)}$ because the derivative of $yz$ with respect to $y$ is $z$, applying the chain rule. $f_z = ye^{(yz + \sin x)}$ because the derivative of $yz$ with respect to $z$ is $y$, again applying the chain rule.

Find fx,fy, and fZ\text { Find } \mathrm { f } _ { \mathbf { x } } , \mathrm { f } _ { \mathbf { y } } \text {, and } \mathrm { f } _ { \mathbf { Z } } Find fx​,fy​, and fZ​

$\text { Find } \mathrm { f } _ { \mathbf { x } } , \mathrm { f } _ { \mathbf { y } } \text {, and } \mathrm { f } _ { \mathbf { Z } }$