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

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