What is (\partial)/(\partial y)(cos(2y)-3x^2y^2) ?

The answer to (\partial)/(\partial y)(cos(2y)-3x^2y^2) is -sin(2y)*2-6x^2y

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(\partial)/(\partial y)(cos(2y)-3x^2y^2)

\frac{\partial}{\partial y} (cos (2 y) - 3 x^{2} y^{2})

- sin (2 y) \cdot 2 - 6 x^{2} y

Solution steps

\frac{\partial}{\partial y} (cos (2 y) - 3 x^{2} y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (cos (2 y)) - \frac{\partial}{\partial y} (3 x^{2} y^{2})

\frac{\partial}{\partial y} (cos (2 y)) = - sin (2 y) \cdot 2

\frac{\partial}{\partial y} (3 x^{2} y^{2}) = 6 x^{2} y

= - sin (2 y) \cdot 2 - 6 x^{2} y

\frac{d y}{d x} = (64 x y)^{\frac{1}{3}}

x \to 2 - lim (\frac{1}{( 1.9 ) - 2})

\frac{\partial}{\partial y} (\frac{y}{x + 6 x})

(cos (x))^{5 l n (x)}

\int_{0}^{1} (e^{x})^{2} d x

What is (\partial)/(\partial y)(cos(2y)-3x^2y^2) ?
The answer to (\partial)/(\partial y)(cos(2y)-3x^2y^2) is -sin(2y)*2-6x^2y