What is (\partial)/(\partial x)(x^2-xy+8y^2) ?

The answer to (\partial)/(\partial x)(x^2-xy+8y^2) is 2x-y

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(\partial)/(\partial x)(x^2-xy+8y^2)

\frac{\partial}{\partial x} (x^{2} - x y + 8 y^{2})

2 x - y

Solution steps

\frac{\partial}{\partial x} (x^{2} - x y + 8 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (x y) + \frac{\partial}{\partial x} (8 y^{2})

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

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

\frac{\partial}{\partial x} (8 y^{2}) = 0

= 2 x - y + 0

Simplify

= 2 x - y

\int_{0}^{1} 5 e^{- 2 x} d x

x^{3} - 4 x^{2} - 7 x + 8

\frac{d}{d x} (\frac{6 x}{x + 1})

\frac{\partial}{\partial y} (7 z e^{x yz})

g (x) = \frac{- 7 e ^{x}}{5 e ^{x} - 8}

What is (\partial)/(\partial x)(x^2-xy+8y^2) ?
The answer to (\partial)/(\partial x)(x^2-xy+8y^2) is 2x-y