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

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

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

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

- 4 x - 4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (4 x y) = 4 x

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

= 0 - 4 x - 4 y

Simplify

= - 4 x - 4 y

\int (3 x^{2} - 6 x + 3) d x

\int_{0}^{\frac{1}{4}} 5 arcsin (4 x) d x

\int \frac{6 x ^{3} + 5 x ^{2} - 47 x - 36}{x ^{2} - 9} d x

10 - x, 1

arccot (\frac{x}{6})

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