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Popular >

d/(d{x)}({x}^4+{y}^4+{z}^4-4{x}{y}{z})

Solution

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

Solution

4 x^{3} - 4 y z

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) + \frac{d}{d x} (y^{4}) + \frac{d}{d x} (z^{4}) - \frac{d}{d x} (4 x y z)

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

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

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

\frac{d}{d x} (4 x y z) = 4 y z

= 4 x^{3} + 0 + 0 - 4 y z

Simplify

= 4 x^{3} - 4 y z

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\int \frac{x}{x - 3} d x

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\frac{d}{d x} (- x^{3} + x)

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\frac{( s + 1 )}{s ^{2} + 1}

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Frequently Asked Questions (FAQ)

What is the d/(d{x)}({x}^4+{y}^4+{z}^4-4{x}{y}{z}) ?
The d/(d{x)}({x}^4+{y}^4+{z}^4-4{x}{y}{z}) is 4{x}^3-4{y}{z}
What is the first d/(d{x)}({x}^4+{y}^4+{z}^4-4{x}{y}{z}) ?
The first d/(d{x)}({x}^4+{y}^4+{z}^4-4{x}{y}{z}) is 4{x}^3-4{y}{z}