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

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

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

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

4 x^{3} - 4 y

Solution steps

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

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} (x^{4}) = 4 x^{3}

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

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

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

Simplify

= 4 x^{3} - 4 y

\frac{\partial}{\partial x} (y^{2} e^{x yz})

x \to - \infty lim (- 2)

\frac{\partial}{\partial y} (3 y cos (x))

(2 x y + x^{3}) y^{^{'}} = \frac{1}{2} x^{2} y + y^{2}

x \to - 4 lim (9 x + 11)

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