What is the derivative of x/3 (2x+1^3) ?

The derivative of x/3 (2x+1^3) is ((2x+1)^3+6x(2x+1)^2)/3

What is the first derivative of x/3 (2x+1^3) ?

The first derivative of x/3 (2x+1^3) is ((2x+1)^3+6x(2x+1)^2)/3

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\frac{d}{d x} (\frac{x}{3} (2 x + 1)^{3})

\frac{( 2 x + 1 ) ^{3} + 6 x ( 2 x + 1 ) ^{2}}{3}

Solution steps

\frac{d}{d x} (\frac{x}{3} (2 x + 1)^{3})

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (x (2 x + 1)^{3})

Apply the Product Rule:

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

f = x, g = (2 x + 1)^{3}

= \frac{1}{3} (\frac{d x}{d x} (2 x + 1)^{3} + \frac{d}{d x} ((2 x + 1)^{3}) x)

\frac{d x}{d x} = 1

\frac{d}{d x} ((2 x + 1)^{3}) = 6 (2 x + 1)^{2}

= \frac{1}{3} (1 \cdot (2 x + 1)^{3} + 6 (2 x + 1)^{2} x)

Simplify

\frac{1}{3} (1 \cdot (2 x + 1)^{3} + 6 (2 x + 1)^{2} x) : \frac{( 2 x + 1 ) ^{3} + 6 x ( 2 x + 1 ) ^{2}}{3}

= \frac{( 2 x + 1 ) ^{3} + 6 x ( 2 x + 1 ) ^{2}}{3}

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

x \to 0 lim (\frac{1}{e ^{\frac{1}{x}}})

\frac{\partial}{\partial z} (\frac{x}{z})

g (x) = ln (x e^{- 7 x})

\frac{\partial}{\partial y} (x^{2} y^{\frac{1}{3}})

What is the derivative of x/3 (2x+1^3) ?
The derivative of x/3 (2x+1^3) is ((2x+1)^3+6x(2x+1)^2)/3
What is the first derivative of x/3 (2x+1^3) ?
The first derivative of x/3 (2x+1^3) is ((2x+1)^3+6x(2x+1)^2)/3