What is the derivative of x^3+x+1 ?

The derivative of x^3+x+1 is 3x^2+1

What is the first derivative of x^3+x+1 ?

The first derivative of x^3+x+1 is 3x^2+1

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

3 x^{2} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

= \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

= 3 x^{2} + 1 + 0

Simplify

= 3 x^{2} + 1

n = 1 \sum \infty n^{2} \cdot e^{- n}

\int_{1}^{2} 7 4 x^{2} - 3 d x

\int x^{- 10} d x

x \to 2 lim (\frac{- 1}{x - 2})

3 t sin (3 t)

What is the derivative of x^3+x+1 ?
The derivative of x^3+x+1 is 3x^2+1
What is the first derivative of x^3+x+1 ?
The first derivative of x^3+x+1 is 3x^2+1