Patel Griffiths

03/26/2023 · Primary School

Find the partial derivative of the function with respect to each of the independent variables. \( f(x, y)=\sin ^{2}\left(-6 x y^{2}-y\right) \) \[ \begin{array}{l}\frac{\partial f}{\partial x}=-12 y^{2} \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) ; \\ \frac{\partial f}{\partial y}=(-24 x y-2) \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) \\ \frac{\partial f}{\partial x}=2 \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) ; \\ \frac{\partial f}{\partial y}=(-24 x-2) \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) \\ \text { C. } \frac{\partial f}{\partial x}=2 \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) ; \\ \frac{\partial f}{\partial y}=2 \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) \\ \text { D. } \frac{\partial f}{\partial x}=-12 y^{2} \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right) ; \frac{\partial f}{\partial y}=2 \sin \left(-6 x y^{2}-y\right) \cos \left(-6 x y^{2}-y\right)\end{array} \]