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

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

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

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

- 2 y - 4 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 - 2 y - 4 x

Simplify

= - 2 y - 4 x

f (x) = 5 x + 7

\frac{d}{d x} (\frac{1}{( 1 - 5 x ) ^{2}})

\frac{d}{d x} (30)

\frac{d}{d x} (2 sin (x) + 2 cos (x))

\frac{d}{d x} (lo g_{7} (x^{2} + 1))

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