The derivative of $\ln(x)$ with respect to $x$ is indeed $\frac{1}{x}$. Here’s an example illustrating how to differentiate functions involving $\ln(x)$ with different expressions inside.

Example:

1. Find the derivative of $f(x) = \ln(3x^2 + 1)$:

   To solve this, we apply the chain rule, where we differentiate the outer function and then multiply by the derivative of the inner function.

   $f(x) = \ln(3x^2 + 1)$

   • Step 1: Differentiate the outer function $\ln(u)$ with respect to $u$, which gives $\frac{1}{u}$.

   • Step 2: Substitute $u = 3x^2 + 1$, so the derivative becomes $\frac{1}{3x^2 + 1}$.

   • Step 3: Now multiply by the derivative of the inner function $3x^2 + 1$ with respect to $x$:

     $\frac{d}{dx}(3x^2 + 1) = 6x$

   • Combine steps: So the derivative of $f(x) = \ln(3x^2 + 1)$ is:

     $f'(x) = \frac{1}{3x^2 + 1} \cdot 6x = \frac{6x}{3x^2 + 1}$

Answer:

$f'(x) = \frac{6x}{3x^2 + 1}$

Would you like further details on the steps or more examples?

Related Questions:

1. What is the derivative of $\ln(x^2)$?
2. How would you differentiate $\ln(\sin(x))$?
3. What is the derivative of $\ln(ax + b)$ where $a$ and $b$ are constants?
4. How can we apply this to functions like $\ln(\cos(x) + x^2)$?
5. What are the steps to find the second derivative of $\ln(x)$?

Tip:

When differentiating $\ln(g(x))$, remember to use the chain rule: start with $\frac{1}{g(x)}$ and then multiply by $g'(x)$.