What is the derivative of f(x)=x^2+x+1 ?

The derivative of f(x)=x^2+x+1 is 2x+1

What is the first derivative of f(x)=x^2+x+1 ?

The first derivative of f(x)=x^2+x+1 is 2x+1

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derivative of

f (x) = x^{2} + x + 1

2 x + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 x + 1 + 0

Simplify

= 2 x + 1

x \to 1 lim (\frac{3}{x ^{3} - 1})

\frac{\partial}{\partial x} (2 (x - a))

y^{^{'}} + 7 y = 4 t

\frac{d}{d x} (e^{4 x^{3}})

x \to 0 + lim (ln (x))

What is the derivative of f(x)=x^2+x+1 ?
The derivative of f(x)=x^2+x+1 is 2x+1
What is the first derivative of f(x)=x^2+x+1 ?
The first derivative of f(x)=x^2+x+1 is 2x+1