To find the derivative of $f(x) = 3\sqrt{x^2 - 1}$ using the chain rule, let's go step by step.

Step 1: Rewrite the function

First, express the square root in exponential form: $f(x) = 3(x^2 - 1)^{1/2}$

Step 2: Apply the chain rule

The chain rule states that if you have a composition of functions, say $f(g(x))$, then the derivative is given by: $f'(x) = f'(g(x)) \cdot g'(x)$

Here, $f(x) = 3(u^{1/2})$ and $u = x^2 - 1$. So, we'll differentiate the outer function first and then the inner function.

Step 3: Differentiate the outer function

The outer function is $3u^{1/2}$, where $u = x^2 - 1$. The derivative of $u^{1/2}$ with respect to $u$ is: $\frac{d}{du} \left( u^{1/2} \right) = \frac{1}{2} u^{-1/2}$ So the derivative of the outer function is: $\frac{d}{dx} \left[ 3(x^2 - 1)^{1/2} \right] = 3 \cdot \frac{1}{2} (x^2 - 1)^{-1/2}$

Step 4: Differentiate the inner function

Now, differentiate the inner function $u = x^2 - 1$: $\frac{d}{dx}(x^2 - 1) = 2x$

Step 5: Multiply them together

Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function: $f'(x) = 3 \cdot \frac{1}{2}(x^2 - 1)^{-1/2} \cdot 2x$

Step 6: Simplify the expression

Now simplify the terms: $f'(x) = 3x(x^2 - 1)^{-1/2}$

Thus, the derivative of $f(x) = 3\sqrt{x^2 - 1}$ is: $f'(x) = \frac{3x}{\sqrt{x^2 - 1}}$

Final Answer:

$f'(x) = \frac{3x}{\sqrt{x^2 - 1}}$

Would you like further details on any part of this solution?

Here are 5 related questions to explore:

1. How would you find the second derivative of $f(x) = 3\sqrt{x^2 - 1}$?
2. Can you apply the product rule to functions involving square roots and polynomials?
3. What happens if $f(x) = \sqrt{x^2 + 1}$, and how does the derivative change?
4. How does the chain rule work when there are multiple layers of functions?
5. How does the chain rule apply in implicit differentiation?

Tip: When dealing with square roots, rewriting them as exponents (like $u^{1/2}$) often simplifies the process of taking derivatives.