What is (\partial)/(\partial x)(x^3+y^3+z^3+6xyz) ?

The answer to (\partial)/(\partial x)(x^3+y^3+z^3+6xyz) is 3x^2+6yz

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(\partial)/(\partial x)(x^3+y^3+z^3+6xyz)

\frac{\partial}{\partial x} (x^{3} + y^{3} + z^{3} + 6 x yz)

3 x^{2} + 6 yz

Solution steps

\frac{\partial}{\partial x} (x^{3} + y^{3} + z^{3} + 6 x yz)

Treat

y, z

as constants

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} (z^{3}) + \frac{\partial}{\partial x} (6 x yz)

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

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

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

\frac{\partial}{\partial x} (6 x yz) = 6 yz

= 3 x^{2} + 0 + 0 + 6 yz

Simplify

= 3 x^{2} + 6 yz

\frac{2}{s ^{3}} + \frac{6}{s ^{2}} - \frac{3}{s}

y 1 - x^{2} y^{^{'}} - x 1 - y^{2} = 0

\frac{d P}{d t} = P (9 - P) - \frac{81}{4}

\int \frac{3 - 4 x ^{3} + 6 x ^{6}}{x ^{6}} d x

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

What is (\partial)/(\partial x)(x^3+y^3+z^3+6xyz) ?
The answer to (\partial)/(\partial x)(x^3+y^3+z^3+6xyz) is 3x^2+6yz