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

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

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

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

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

8 x + 1

Solution steps

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

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} (4 x^{2}) = 8 x

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

= 8 x + 1

d er i v a t i v e 2 x^{7}

s l o p e (- 3.1) (- 17.2)

\int_{0}^{2} x^{2} e^{x} d x

d er i v a t i v e e^{3}

\int 9 x 1 + 2 x^{2} d x