What is the d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) ?

The d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) is 8{x}^3-4{y}^2{x}{z}

What is the first d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) ?

The first d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) is 8{x}^3-4{y}^2{x}{z}

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Popular >

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

8 x^{3} - 4 y^{2} x z

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{4}) + \frac{d}{d x} (2 y^{4}) + \frac{d}{d x} (z^{4}) - \frac{d}{d x} (2 x^{2} y^{2} z)

\frac{d}{d x} (2 x^{4}) = 8 x^{3}

\frac{d}{d x} (2 y^{4}) = 0

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

\frac{d}{d x} (2 x^{2} y^{2} z) = 4 x y^{2} z

= 8 x^{3} + 0 + 0 - 4 x y^{2} z

Simplify

= 8 x^{3} - 4 y^{2} x z

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

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

\frac{3}{( s + 1 ) ^{2} + 9}

\frac{d}{d x} (tan (13 x))

2 e^{4 x} cos (5 x)

What is the d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) ?
The d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) is 8{x}^3-4{y}^2{x}{z}
What is the first d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) ?
The first d/(d{x)}(2{x}^4+2{y}^4+{z}^4-2{x}^2{y}^2{z}) is 8{x}^3-4{y}^2{x}{z}