We are tasked with differentiating $\tan^{-1}(3\tan x)$ with respect to $x$.

Let $y = \tan^{-1}(3\tan x)$.

Step 1: Use Chain Rule

The chain rule tells us: $\frac{dy}{dx} = \frac{d}{d(\theta)}\left( \tan^{-1}(\theta) \right) \cdot \frac{d}{dx}\left( 3\tan x \right)$ where $\theta = 3\tan x$.

Step 2: Differentiate the outer function

We know that the derivative of $\tan^{-1}(u)$ with respect to $u$ is: $\frac{d}{du} \tan^{-1}(u) = \frac{1}{1 + u^2}$ So, $\frac{d}{dx} \tan^{-1}(3\tan x) = \frac{1}{1 + (3\tan x)^2} \cdot \frac{d}{dx}(3\tan x)$

Step 3: Differentiate the inner function

The derivative of $3\tan x$ is: $\frac{d}{dx}(3\tan x) = 3 \sec^2 x$

Step 4: Combine the results

Now, combine the derivatives: $\frac{dy}{dx} = \frac{1}{1 + 9\tan^2 x} \cdot 3 \sec^2 x$

Final Answer:

$\frac{d}{dx} \left( \tan^{-1}(3\tan x) \right) = \frac{3 \sec^2 x}{1 + 9\tan^2 x}$

Would you like more details or have any questions about this?

Here are some related questions for further exploration:

1. What is the derivative of $\tan^{-1}(x)$ in general?
2. How do you differentiate inverse trigonometric functions?
3. What is the derivative of $\tan^{-1}(\sin x)$?
4. What is the derivative of $\sec^{-1}(3x)$?
5. How does the chain rule work with trigonometric functions?

Tip: Always remember to apply the chain rule when you have a composition of functions!