What is the d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) ?

The d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) is {z}(3{y}^2+6{x}{z})

What is the first d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) ?

The first d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) is {z}(3{y}^2+6{x}{z})

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\frac{d}{d y} ((z) (x^{3} + y^{3} + z^{3} + 6 x y z - 1))

z (3 y^{2} + 6 x z)

Solution steps

\frac{d}{d y} ((z) (x^{3} + y^{3} + z^{3} + 6 x y z - 1))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= z \frac{d}{d y} (x^{3} + y^{3} + z^{3} + 6 x y z - 1)

Apply the Sum/Difference Rule:

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

= z (\frac{d}{d y} (x^{3}) + \frac{d}{d y} (y^{3}) + \frac{d}{d y} (z^{3}) + \frac{d}{d y} (6 x y z) - \frac{d}{d y} (1))

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

\frac{d}{d y} (y^{3}) = 3 y^{2}

\frac{d}{d y} (z^{3}) = 0

\frac{d}{d y} (6 x y z) = 6 x z

\frac{d}{d y} (1) = 0

= z (0 + 3 y^{2} + 0 + 6 x z - 0)

Simplify

= z (3 y^{2} + 6 x z)

\frac{6 ( s + 1 )}{s ^{4}}

\frac{5 p ^{2} + 6 p - 6}{8 ( p + 1 ) ( p - 1 )}

\frac{x ^{2} + 2 a x}{( x + a ) ^{2}}

(3 - i)^{27}

\frac{( s + 4 s ^{2} + 1 )}{( s ^{2} ( s - 1 ) ^{2} + 4 )}

What is the d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) ?
The d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) is {z}(3{y}^2+6{x}{z})
What is the first d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) ?
The first d/(d{y)}(({z})({x}^3+{y}^3+{z}^3+6{x}{y}{z}-1)) is {z}(3{y}^2+6{x}{z})