What is the derivative of (x^3+6x)/3 ?

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

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

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

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derivative of

\frac{x ^{3} + 6 x}{3}

x^{2} + 2

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (6 x) = 6

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

Simplify

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

= x^{2} + 2

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

\int \frac{x}{2 - x - x ^{2}} d x

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

\frac{1}{s ( s ^{2} + 5 )}

x \to 0 lim (sin (3 x))

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