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

x^{2} - 6 x

x \to - 2 - lim (- 1)

3 (x^{2} + 3 x + 1)^{2} (2 x + 3)

2 x^{7}

\frac{\partial}{\partial x} (sin (3 x) cos (4 y))