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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3) = 0

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

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

= \frac{1}{6} (0 + 2 - 3 x^{2})

Simplify

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

y^{^{'}} = t \cdot y

\int 6 sin (x) d x

f (x) = x^{2} - 6 x, at x = 7

\int - 2 x + 1 d x

f (x) = \frac{- 8 x + 2}{3 x + 5}

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