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

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

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

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

y - 2 x

Solution steps

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

\frac{\partial}{\partial x} (3) = 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

\int 7 x d x

\frac{7 ^{x}}{x ^{3}}

\int (\frac{3}{x} - 9 e^{5 x} + x^{9}) d x

y = m x + b

\int_{4}^{6} (x^{2} + 2 x - 2) d x

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