The problem asks for the first derivative with respect to $x$ of the function:

$f(x) = \left( \ln(3 - x^2) \right)^4 \cdot \sqrt[4]{x^2 + 1}$

Step-by-Step Derivation:

This function is a product of two terms:

1. $\left( \ln(3 - x^2) \right)^4$
2. $\left( x^2 + 1 \right)^{1/4}$

We'll apply the product rule for differentiation, which states:

$(fg)' = f'g + fg'$

Here, $f(x) = \left( \ln(3 - x^2) \right)^4$ and $g(x) = \left( x^2 + 1 \right)^{1/4}$.

Step 1: Differentiate $f(x) = \left( \ln(3 - x^2) \right)^4$

We'll use the chain rule here: $f'(x) = 4 \left( \ln(3 - x^2) \right)^3 \cdot \frac{d}{dx} \left( \ln(3 - x^2) \right)$

Next, differentiate $\ln(3 - x^2)$: $\frac{d}{dx} \left( \ln(3 - x^2) \right) = \frac{-2x}{3 - x^2}$

Thus, the derivative of $f(x)$ is: $f'(x) = 4 \left( \ln(3 - x^2) \right)^3 \cdot \frac{-2x}{3 - x^2}$

Step 2: Differentiate $g(x) = \left( x^2 + 1 \right)^{1/4}$

We'll use the chain rule again: $g'(x) = \frac{1}{4} \left( x^2 + 1 \right)^{-3/4} \cdot \frac{d}{dx}(x^2 + 1)$

$g'(x) = \frac{1}{4} \left( x^2 + 1 \right)^{-3/4} \cdot 2x$

Thus, the derivative of $g(x)$ is: $g'(x) = \frac{1}{2} \cdot \frac{x}{\left( x^2 + 1 \right)^{3/4}}$

Step 3: Apply the Product Rule

Now, applying the product rule: $f'(x) = f'(x)g(x) + f(x)g'(x)$

Substitute the derivatives $f'(x)$ and $g'(x)$ we found:

[ f'(x) = \left[ 4 \left( \ln(3 - x^2) \right)^3 \cdot \frac{-2x}{3 - x^2} \cdot \left( x^2 + 1 \right)^{1/4} \right] + \left[ \left( \ln(3 - x^2) \right)^4 \cdot \frac{1}{2} \cdot \frac{x}{(x^2 + 1)^{3/4}} \right] ]

This is the derivative of the function $f(x)$.

Would you like further details or simplifications on any step? Let me know if you have any questions.

Related Questions:

1. How can the chain rule be applied in more complex functions like this one?
2. What is the importance of the product rule when dealing with products of functions?
3. How would you differentiate a function involving multiple logarithmic terms?
4. Can this derivative be simplified further using any mathematical techniques?
5. How does this derivative behave as $x$ approaches specific values (e.g., 0, infinity)?

Tip:

Always break down composite functions into smaller parts and apply the rules (like chain or product rule) step-by-step to avoid confusion.