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

The answer to (\partial)/(\partial x)(4x^2-2xy+y^2) is 8x-2y

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

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

8 x - 2 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (4 x^{2}) = 8 x

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

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

= 8 x - 2 y + 0

Simplify

= 8 x - 2 y

x \to 4 lim (x - \frac{4}{x ^{2} - 16})

x^{3} - 3, [- 2, 0]

e^{3 x + 1}

\frac{d}{d x} (\frac{sin ( 5 x )}{5})

\int (2 x^{2} - 3 x + 2) d x

What is (\partial)/(\partial x)(4x^2-2xy+y^2) ?
The answer to (\partial)/(\partial x)(4x^2-2xy+y^2) is 8x-2y