$\text { Find } \mathrm { f } _ { \mathbf { x } } , \mathrm { f } _ { \mathbf { y } } \text {, and } \mathrm { f } _ { \mathbf { Z } }$
-$f ( x , y , z ) = \frac { \cos y } { x z ^ { 2 } }$
A) $f _ { x } = - \frac { \cos y } { x ^ { 2 } z ^ { 2 } } ; f _ { y } = - \frac { \sin y } { x z _ { 2 } ^ { 2 } } ; f _ { z } = - \frac { 2 \cos y } { x z ^ { 3 } }$
B) $\mathrm { f } _ { \mathrm { x } } = \frac { \cos \mathrm { y } } { \mathrm { z } ^ { 2 } } ; \mathrm { f } _ { \mathrm { y } } = \frac { \sin \mathrm { y } } { \mathrm { xz } ^ { 2 } } ; \mathrm { f } _ { \mathrm { z } } = \frac { 2 \cos \mathrm { y } } { \mathrm { xz } }$
C) $f _ { x } = - \frac { \cos y } { z ^ { 2 } } ; f _ { y } = - \frac { \sin y } { x z ^ { 2 } } ; f _ { z } = - \frac { 2 \cos y } { x z }$

Question

Quizplus · Accepted Answer

The partial derivatives of $f(x, y, z) = \frac{\cos y}{xz^2}$ with respect to $x$, $y$, and $z$ are calculated using the rules of differentiation. For $f_x$, the derivative with respect to $x$ treats $y$ and $z$ as constants, resulting in a negative sign due to the power rule and the $x$ in the denominator. For $f_y$, the derivative of $\cos y$ is $-\sin y$, and since $x$ and $z$ are treated as constants, the derivative is $-\frac{\sin y}{xz^2}$. For $f_z$, the derivative with respect to $z$ also introduces a negative sign and increases the power of $z$ in the denominator due to the power rule, resulting in $-\frac{2\cos y}{xz^3}$.

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​

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