What is the derivative of 4x^2+x+1 ?

The derivative of 4x^2+x+1 is 8x+1

What is the first derivative of 4x^2+x+1 ?

The first derivative of 4x^2+x+1 is 8x+1

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\frac{d}{d x} (4 x^{2} + x + 1)

8 x + 1

Solution steps

\frac{d}{d x} (4 x^{2} + x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{2}) + \frac{d x}{d x} + \frac{d}{d x} (1)

\frac{d}{d x} (4 x^{2}) = 8 x

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

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

= 8 x + 1 + 0

Simplify

= 8 x + 1

n = 1 \sum \infty \frac{( ln ( 3 n ) )}{n}

(2 x^{3} + 3)^{6}

(2 x - 3)^{6}

x \to 0 + lim (e^{c o t (x)})

What is the derivative of 4x^2+x+1 ?
The derivative of 4x^2+x+1 is 8x+1
What is the first derivative of 4x^2+x+1 ?
The first derivative of 4x^2+x+1 is 8x+1