What is the derivative of y=(x^2+x+8) ?

The derivative of y=(x^2+x+8) is 2x+1

What is the first derivative of y=(x^2+x+8) ?

The first derivative of y=(x^2+x+8) is 2x+1

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derivative of

y = (x^{2} + x + 8)

2 x + 1

Solution steps

\frac{d}{d x} ((x^{2} + x + 8))

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} (8)

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

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

\frac{d}{d x} (8) = 0

= 2 x + 1 + 0

Simplify

= 2 x + 1

\frac{2 ^{x}}{x}

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

\frac{tanh ( 4 x + 1 )}{5}

\frac{d}{d x} \frac{1}{( x + { n } ) \cdot x}

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

What is the derivative of y=(x^2+x+8) ?
The derivative of y=(x^2+x+8) is 2x+1
What is the first derivative of y=(x^2+x+8) ?
The first derivative of y=(x^2+x+8) is 2x+1