What is (\partial)/(\partial x)(3x^2-2xy-4y^2) ?

The answer to (\partial)/(\partial x)(3x^2-2xy-4y^2) is 6x-2y

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(\partial)/(\partial x)(3x^2-2xy-4y^2)

\frac{\partial}{\partial x} (3 x^{2} - 2 x y - 4 y^{2})

6 x - 2 y

Solution steps

\frac{\partial}{\partial x} (3 x^{2} - 2 x y - 4 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (2 x y) - \frac{\partial}{\partial x} (4 y^{2})

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

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

\frac{\partial}{\partial x} (4 y^{2}) = 0

= 6 x - 2 y - 0

Simplify

= 6 x - 2 y

\frac{d y}{d t} \cdot \frac{1}{4} + y = cos (t)

\int - \frac{1}{3} e^{- 3 t} d t

\frac{d}{d x} (3 x + e^{x})

\frac{\partial}{\partial y} (4 - x^{2} - 7 y^{2})

2 y^{^{''}} + 3 y^{^{'}} + y = t^{2} + 2 sin (t)

What is (\partial)/(\partial x)(3x^2-2xy-4y^2) ?
The answer to (\partial)/(\partial x)(3x^2-2xy-4y^2) is 6x-2y