What is the first (ln(x+2))^' ?

The first (ln(x+2))^' is 1/(x+2)

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(ln(x+2))^'

(ln (x + 2))^{^{'}}

\frac{1}{x + 2}

Solution steps

(ln (x + 2))^{'}

Apply the chain rule:

\frac{1}{x + 2} (x + 2)^{'}

= \frac{1}{x + 2} (x + 2)^{'}

(x + 2)^{'} = 1

= \frac{1}{x + 2} \cdot 1

Simplify

= \frac{1}{x + 2}

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