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

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

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

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

y - 2 x

Solution steps

\frac{\partial}{\partial x} (3 - x^{2} + x y - 3 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} (3 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} (3 y^{2}) = 0

= 0 - 2 x + y - 0

Simplify

= y - 2 x

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

y = \frac{9}{x ^{2} + 2}, (1, 3)

\frac{d}{d y} (ln (\frac{( 5 y + 1 ) ^{2}}{y ^{2} + 1}))

ln (4 x^{3} + x)

\int 5 5 x d x

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