What is (\partial)/(\partial x)(6-x^2+xy-2y^2) ?

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

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(\partial)/(\partial x)(6-x^2+xy-2y^2)

\frac{\partial}{\partial x} (6 - x^{2} + x y - 2 y^{2})

y - 2 x

Solution steps

\frac{\partial}{\partial x} (6 - 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} (6) - \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial x} (2 y^{2})

\frac{\partial}{\partial x} (6) = 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

\frac{( s + 2 ) ^{2}}{( s + 1 ) ^{3}}

\int 10 x - x e^{x} + e d x

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

\int b x^{3} d x

\frac{x}{1 + x}

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