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

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

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(\partial)/(\partial x)(4+x^3+y^3-3xy)

\frac{\partial}{\partial x} (4 + x^{3} + y^{3} - 3 x y)

3 x^{2} - 3 y

Solution steps

\frac{\partial}{\partial x} (4 + x^{3} + y^{3} - 3 x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

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

Simplify

= 3 x^{2} - 3 y

2 e^{- t} - \frac{3}{2} e^{- 2 t}

\frac{d}{d x} (sec (x) - 2 cos (x))

\int 8 sin (t) d t

\frac{\partial}{\partial x} (3 x^{4} y^{5})

\frac{d}{d x} ((x^{2} + 1)^{5})

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