What is the derivative of y=x^2+x+1 ?

The derivative of y=x^2+x+1 is 2x+1

What is the first derivative of y=x^2+x+1 ?

The first derivative of y=x^2+x+1 is 2x+1

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derivative of

y = 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

\frac{\partial}{\partial y} (x + \frac{1}{2} ln (x^{2} + y^{2}))

\frac{1}{y}

x \to 1 lim (\frac{e ^{x}}{2})

f (x) = cos (x^{5})

\frac{\partial}{\partial x} (p (x)^{2})

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