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

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

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

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

2 x - 2 y

Solution steps

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

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

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

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

= 2 x - 2 y + 0

Simplify

= 2 x - 2 y

x \to 6 lim (6 \cdot \frac{5}{6} \cdot (- \frac{2}{3}))

g (x) = x^{2} (1 - 9 x)

\int \frac{1}{3 - 2 y - y ^{2}} d y

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

\int \frac{e ^{x}}{5 - e ^{2 x}} d x

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