Find the partial derivative of the function with respect to each of the independent variables. $$ f(x, y)=\sin ^{2}\left(4 x y^{2}-y\right) $$ A. $$ \frac{\partial f}{\partial x}=8 y^{2} \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) ; \frac{\partial f}{\partial y}=2 \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) $$ B. $$ \frac{\partial f}{\partial x}=2 \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) ; \frac{\partial f}{\partial y}=(16 x-2) \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) $$ $$ \frac{\partial f}{\partial x}=2 \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) ; $$ $$ \frac{\partial f}{\partial y}=2 \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) $$ D. $$ \frac{\partial f}{\partial x}=8 y^{2} \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) ; $$ $$ \frac{\partial f}{\partial y}=(16 x y-2) \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) $$

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Ramirez Reyes · Accepted Answer

The partial derivatives are: - $$ \frac{\partial f}{\partial x} = 8y^{2} \sin(4xy^{2} - y) \cos(4xy^{2} - y) $$ - $$ \frac{\partial f}{\partial y} = (16xy - 2) \sin(4xy^{2} - y) \cos(4xy^{2} - y) $$ **Answer:** D. $$ \frac{\partial f}{\partial x}=8 y^{2} \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) ; $$ $$ \frac{\partial f}{\partial y}=(16 x y-2) \sin \left(4 x y^{2}-y\right) \cos \left(4 x y^{2}-y\right) $$

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