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

The answer to (\partial)/(\partial x)(x^4+y^4-4xy+2) is 4x^3-4y

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

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

4 x^{3} - 4 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x^{4}) = 4 x^{3}

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

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

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

= 4 x^{3} + 0 - 4 y + 0

Simplify

= 4 x^{3} - 4 y

\int_{0}^{2} - 7 x^{3} x^{2} + 4 d x

\frac{d}{d x} ((1 + 4 x)^{\frac{3}{2}})

\int_{0}^{\infty} 2 e^{- 2 x} d x

\frac{d y}{d x}, y = 7 x^{8}

What is (\partial)/(\partial x)(x^4+y^4-4xy+2) ?
The answer to (\partial)/(\partial x)(x^4+y^4-4xy+2) is 4x^3-4y