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

Question

Quizplus · Accepted Answer

The partial derivatives of $f(x, y, z) = \sin(xy)\cos(yz^2)$ are calculated using the product rule and chain rule. For $f_x$, differentiate with respect to $x$ treating $y$ and $z$ as constants, which gives $y\cos(xy)\cos(yz^2)$. For $f_y$, differentiate with respect to $y$, applying the product rule and chain rule, resulting in $x\cos(xy)\cos(yz^2) - z^2\sin(xy)\sin(yz^2)$. For $f_z$, differentiate with respect to $z$, treating $x$ and $y$ as constants, which gives $-2yz\sin(xy)\sin(yz^2)$.

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​

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