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

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

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

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

y - 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (4) = 0

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

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

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

= 0 - 2 x + y - 0

Simplify

= y - 2 x

x^{2} + y^{2} = 29, (- 5, 2)

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{6}} d x

\int e^{- 2 l n (x)} d x

\frac{\partial}{\partial x} (e^{x z} + x y)

(x^{2} + 4) \frac{d y}{d x} = x y

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