What is the first derivative of y=3x+8 ?

The first derivative of y=3x+8 is 3

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derivative of y=3x+8

derivative of

y = 3 x + 8

3

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3 x) = 3

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

= 3 + 0

Simplify

= 3

\int 2^{x} 5^{x} d x

\int (e^{- x^{3}} x^{2}) d x

y^{^{'}} - 2 y = 1 - 3 x

\frac{\partial}{\partial v} (\frac{u}{v})

y \cdot ln (x) - x^{2} y = x y^{^{'}}