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

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

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(\partial)/(\partial x)(x^3+y^3-2xy)

\frac{\partial}{\partial x} (x^{3} + y^{3} - 2 x y)

3 x^{2} - 2 y

Solution steps

\frac{\partial}{\partial x} (x^{3} + y^{3} - 2 x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (2 x y)

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

\frac{\partial}{\partial x} (y^{3}) = 0

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

= 3 x^{2} + 0 - 2 y

Simplify

= 3 x^{2} - 2 y

x \to 0 lim (- 4)

g (t) = t^{2} - \frac{4}{t ^{3}}

\int (5 - 2^{- 1}) d x

\frac{d}{d x} (x + cos (x))

\frac{\partial}{\partial x} (8 x^{3} y^{2} + 6 x y^{4})

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