What is (\partial)/(\partial x)(6-x^2+xy-4y^2) ?

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

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(\partial)/(\partial x)(6-x^2+xy-4y^2)

\frac{\partial}{\partial x} (6 - x^{2} + x y - 4 y^{2})

y - 2 x

Solution steps

\frac{\partial}{\partial x} (6 - x^{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} (6) - \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial x} (4 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} (4 y^{2}) = 0

= 0 - 2 x + y - 0

Simplify

= y - 2 x

x^{2} + 1, x + 1

\int \frac{1}{( x ^{2} + 49 ) ^{\frac{3}{2}}} d x

\frac{d}{d t} (A t e^{- t})

y^{^{'}} + 7 (\frac{y}{x}) = x

\frac{d}{d x} (\frac{10}{1 + e ^{- 3 x}})

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