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

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

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

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

2 x - 3 yz

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

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

= 2 x + 0 + 0 - 3 yz

Simplify

= 2 x - 3 yz

(x + 1) \frac{d y}{d x} + (x + 2) y = 8 x e^{- x}

\frac{\partial}{\partial x} (5 x^{3} y z^{2})

\int_{- 1}^{2} x^{3} - 1 d x

x \to 0 lim (x^{4})

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