Give an appropriate answer.
-$$f ( x , y ) = \sin ^ { 2 } \left( - 2 x y ^ { 2 } - y \right)$$
A) $\frac { \partial f } { \partial x } = 2 \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right) ; \frac { \partial f } { \partial y } = ( - 8 x - 2 ) \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right)$
B) $\frac { \partial f } { \partial x } = 2 \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right) ; \frac { \partial f } { \partial y } = 2 \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right)$
C) $\frac { \partial f } { \partial x } = - 4 y ^ { 2 } \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right) ; \frac { \partial f } { \partial y } = ( - 8 x y - 2 ) \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right)$
D) $\frac { \partial f } { \partial x } = - 4 y ^ { 2 } \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right) ; \frac { \partial f } { \partial y } = 2 \sin \left( - 2 x y ^ { 2 } - y \right) \cos \left( - 2 x y ^ { 2 } - y \right)$

Question

Quizplus · Accepted Answer

The correct partial derivatives are obtained by applying the chain rule to the function $f(x, y) = \sin^2(-2xy^2 - y)$. For $\frac{\partial f}{\partial x}$, the inner function is $-2xy^2 - y$, and its derivative with respect to $x$ is $-4y^2$. For $\frac{\partial f}{\partial y}$, the derivative of the inner function with respect to $y$ is $-4xy - 1$. Applying the chain rule and the derivative of $\sin^2(u)$ with respect to $u$, which is $2\sin(u)\cos(u)$, gives the results in option C.

Give an Appropriate Answer f(x,y)=sin⁡2(−2xy2−y)f ( x , y ) = \sin ^ { 2 } \left( - 2 x y ^ { 2 } - y \right)f(x,y)=sin2(−2xy2−y)

Give an Appropriate Answer $f ( x , y ) = \sin ^ { 2 } \left( - 2 x y ^ { 2 } - y \right)$