What is the derivative of 3x^2+8x ?

The derivative of 3x^2+8x is 6x+8

What is the first derivative of 3x^2+8x ?

The first derivative of 3x^2+8x is 6x+8

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\frac{d}{d x} (3 x^{2} + 8 x)

6 x + 8

Solution steps

\frac{d}{d x} (3 x^{2} + 8 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (8 x)

\frac{d}{d x} (3 x^{2}) = 6 x

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

= 6 x + 8

x \to 1 lim (\frac{x + 6}{x + 1})

x \to 1 lim (arctan (ln (x)))

\frac{d}{d x} (4 x^{3} - 5 x^{2} + ln (x))

y^{2} - 2 x = 1, x - y = 1

cos (t) + t^{2} sin (t)

What is the derivative of 3x^2+8x ?
The derivative of 3x^2+8x is 6x+8
What is the first derivative of 3x^2+8x ?
The first derivative of 3x^2+8x is 6x+8