What is the derivative of 2x^3y ?

The derivative of 2x^3y is 2(3x^2y+(dy)/(dx)x^3)

What is the first derivative of 2x^3y ?

The first derivative of 2x^3y is 2(3x^2y+(dy)/(dx)x^3)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3}, g = y

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

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

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

y^{^{'}} = (x + y + 3)^{2}

\int \frac{1}{x - 3} d x

y = 12 - x, y = x, y = 1

\frac{\partial}{\partial x} (ln (x + 2 y))

e^{t} cos (6 t)

What is the derivative of 2x^3y ?
The derivative of 2x^3y is 2(3x^2y+(dy)/(dx)x^3)
What is the first derivative of 2x^3y ?
The first derivative of 2x^3y is 2(3x^2y+(dy)/(dx)x^3)