What is the d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) ?

The d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) is 2{z}{y}

What is the first d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) ?

The first d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) is 2{z}{y}

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\frac{d}{d y} ((z) (x^{2} + y^{2} - 2 z^{2} + 12 x - 8 z - 4))

2 z y

Solution steps

\frac{d}{d y} ((z) (x^{2} + y^{2} - 2 z^{2} + 12 x - 8 z - 4))

Take the constant out:

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

= z \frac{d}{d y} (x^{2} + y^{2} - 2 z^{2} + 12 x - 8 z - 4)

Apply the Sum/Difference Rule:

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

= z (\frac{d}{d y} (x^{2}) + \frac{d}{d y} (y^{2}) - \frac{d}{d y} (2 z^{2}) + \frac{d}{d y} (12 x) - \frac{d}{d y} (8 z) - \frac{d}{d y} (4))

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

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

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

\frac{d}{d y} (12 x) = 0

\frac{d}{d y} (8 z) = 0

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

= z (0 + 2 y - 0 + 0 - 0 - 0)

Simplify

= 2 z y

(e^{x} + k)^{4}

1 + \frac{k}{j w ^{2} ( j w + 3.6 )}

\frac{x - 5}{( x + 3 ) ( x ^{2} + 7 x + 6 )}

\frac{2 x ( 216 - 216 x + 72 x ^{2} - 8 x ^{3} )}{81}

\frac{( x ^{2} + 3 x - 5 ) z ^{2} ( z - 3 ) ^{2} + 3}{x ^{2} + 3 x + 4}

What is the d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) ?
The d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) is 2{z}{y}
What is the first d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) ?
The first d/(d{y)}(({z})({x}^2+{y}^2-2{z}^2+12{x}-8{z}-4)) is 2{z}{y}