What is the first derivative of 9x+8 ?

The first derivative of 9x+8 is 9

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derivative of 9x+8

\frac{d}{d x} (9 x + 8)

9

Solution steps

\frac{d}{d x} (9 x + 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (9 x) + \frac{d}{d x} (8)

\frac{d}{d x} (9 x) = 9

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

= 9 + 0

Simplify

= 9

4 x^{6} (x^{4} - 3 x)

x \to - 4 lim (\frac{x ^{2} - 8}{4 - x})

\frac{d}{d x} \frac{( 3 - e ^{\frac{1}{x}} )}{3 + e ^{\frac{1}{x}}}

\frac{\partial}{\partial x} (3 x^{2} y^{2} + 4)

f (x) = \frac{2}{x - 4}, (6, 1)