Derivative of log(t+1)

1. Let $u = t + 1$.

2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$.

3. Then, apply the chain rule. Multiply by $\frac{d}{d t} \left(t + 1\right)$:

   1. Differentiate $t + 1$ term by term:

      1. Apply the power rule: $t$ goes to $1$

      2. The derivative of the constant $1$ is zero.

      The result is: $1$

   The result of the chain rule is:

   $$\frac{1}{t + 1}$$

4. Now simplify:

   $$\frac{1}{t + 1}$$

The answer is:

$$\frac{1}{t + 1}$$