What is the first d/(dk)(ln(3+k)) ?

The first d/(dk)(ln(3+k)) is 1/(3+k)

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d/(dk)(ln(3+k))

\frac{d}{d k} (ln (3 + k))

\frac{1}{3 + k}

Solution steps

\frac{d}{d k} (ln (3 + k))

Apply the chain rule:

\frac{1}{3 + k} \frac{d}{d k} (3 + k)

= \frac{1}{3 + k} \frac{d}{d k} (3 + k)

\frac{d}{d k} (3 + k) = 1

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

Simplify

= \frac{1}{3 + k}