What is (\partial)/(\partial y)(x^2y-2x^2-4y^2) ?

The answer to (\partial)/(\partial y)(x^2y-2x^2-4y^2) is x^2-8y

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(\partial)/(\partial y)(x^2y-2x^2-4y^2)

\frac{\partial}{\partial y} (x^{2} y - 2 x^{2} - 4 y^{2})

x^{2} - 8 y

Solution steps

\frac{\partial}{\partial y} (x^{2} y - 2 x^{2} - 4 y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{2} y) - \frac{\partial}{\partial y} (2 x^{2}) - \frac{\partial}{\partial y} (4 y^{2})

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

\frac{\partial}{\partial y} (2 x^{2}) = 0

\frac{\partial}{\partial y} (4 y^{2}) = 8 y

= x^{2} - 0 - 8 y

Simplify

= x^{2} - 8 y

\frac{4 x - 5}{3 x + 2}

\frac{d}{d x} (\frac{11 x}{( 2 + x ^{2} )})

x = 64 - y^{2}, x = y^{2} - 64

\frac{d}{d x} (cos (5 x - 2))

\frac{\partial}{\partial y} (\frac{2 x y}{x ^{2} + y ^{2}})

What is (\partial)/(\partial y)(x^2y-2x^2-4y^2) ?
The answer to (\partial)/(\partial y)(x^2y-2x^2-4y^2) is x^2-8y