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

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

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

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

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derivative of

\frac{x ^{3} + x - 1}{2}

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

Solution steps

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

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (1) = 0

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

Simplify

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

e^{x} + 2

ln (x^{- 1})

(x^{2} + 3 x)^{4}

\frac{cos ( x ) + 1}{x}

\frac{x ^{2}}{( e ^{x} - 1 )}

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