What is the d/(dy)(4*x^2*y^3-4) ?

The d/(dy)(4*x^2*y^3-4) is 12x^2y^2

What is the first d/(dy)(4*x^2*y^3-4) ?

The first d/(dy)(4*x^2*y^3-4) is 12x^2y^2

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\frac{d}{d y} (4 \cdot x^{2} \cdot y^{3} - 4)

12 x^{2} y^{2}

Solution steps

\frac{d}{d y} (4 x^{2} y^{3} - 4)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{d}{d y} (4 x^{2} y^{3}) = 12 x^{2} y^{2}

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

= 12 x^{2} y^{2} - 0

Simplify

= 12 x^{2} y^{2}

d er i v a t i v e y = (t^{2} - 7)^{2}

\frac{\partial}{\partial x} (4 cos (x))

\frac{d}{d x} (8 arcsin (x^{4}))

y^{^{'}} + 5 y = x

\frac{d}{d x} (x^{3} - 3 x^{2} + 3)

What is the d/(dy)(4*x^2*y^3-4) ?
The d/(dy)(4*x^2*y^3-4) is 12x^2y^2
What is the first d/(dy)(4*x^2*y^3-4) ?
The first d/(dy)(4*x^2*y^3-4) is 12x^2y^2