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

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

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(\partial)/(\partial x)(x^4+y^4+z^4-4xyz)

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

4 x^{3} - 4 yz

Solution steps

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

Treat

y, z

as constants

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} (z^{4}) - \frac{\partial}{\partial x} (4 x yz)

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

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

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

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

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

Simplify

= 4 x^{3} - 4 yz

\int_{0}^{\frac{π}{4}} cos^{7} (x) d x

\frac{\partial}{\partial x} (y^{2} - 4 y + x y)

\frac{x ^{3}}{8}

\int cos^{4} (6 x) d x

\frac{d}{d x} (\frac{1}{e} e^{x})

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