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

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

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

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

4 x - 4 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (4 x y) = 4 y

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

= 4 x - 4 y + 0

Simplify

= 4 x - 4 y

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