What is the first (ln(3x+1))^' ?

The first (ln(3x+1))^' is 3/(3x+1)

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(ln(3x+1))^'

(ln (3 x + 1))^{^{'}}

\frac{3}{3 x + 1}

Solution steps

(ln (3 x + 1))^{'}

Apply the chain rule:

\frac{1}{3 x + 1} (3 x + 1)^{'}

= \frac{1}{3 x + 1} (3 x + 1)^{'}

(3 x + 1)^{'} = 3

= \frac{1}{3 x + 1} \cdot 3

Simplify

\frac{1}{3 x + 1} \cdot 3 : \frac{3}{3 x + 1}

= \frac{3}{3 x + 1}

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