What is the derivative of 2x^2y ?

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{2}, g = y

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

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

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

(x^{2} - x) \cdot e^{- x}

\int \frac{2 x + 4}{x ^{3} - 2 x ^{2}} d x

\int π cos (n x) d x

\int_{0}^{1} 6 e^{- 6 x} d x

\frac{\partial}{\partial x} (\frac{1}{2} ln (x))

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