Step 1: Understanding the Concept:
We need to find the derivative of a composite inverse trigonometric function. We apply the chain rule iteratively: outer function ($\tan^{-1} u$), inner algebraic expression, and nested square root.
Step 2: Key Formula or Approach:
1. Derivative of inverse tangent: $\frac{d}{du} (\tan^{-1} u) = \frac{1}{1+u^2} \cdot u'$.
2. Chain rule and Power rule: $\frac{d}{dx} (\sqrt{1+x^2}) = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$.
Step 3: Detailed Explanation:
Let $y = \tan^{-1} \left(\sqrt{1+x^2}-1\right)$.
Let $u = \sqrt{1+x^2}-1$. Then $y = \tan^{-1} u$.
Apply the chain rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{d}{dx} (\sqrt{1+x^2} - 1) \]
Calculate the derivative of the inner function $u$:
\[ \frac{du}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot \frac{d}{dx}(1+x^2) - 0 = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}} \]
Now, substitute $u$ and $du/dx$ back into the main equation:
\[ \frac{dy}{dx} = \frac{1}{1 + (\sqrt{1+x^2} - 1)^2} \cdot \frac{x}{\sqrt{1+x^2}} \]
Expand the squared binomial in the denominator:
\[ (\sqrt{1+x^2} - 1)^2 = (\sqrt{1+x^2})^2 - 2(1)(\sqrt{1+x^2}) + 1^2 = (1+x^2) - 2\sqrt{1+x^2} + 1 \]
\[ (\sqrt{1+x^2} - 1)^2 = x^2 - 2\sqrt{1+x^2} + 2 \]
Substitute this expanded form back into the denominator:
\[ \frac{dy}{dx} = \frac{1}{1 + (x^2 - 2\sqrt{1+x^2} + 2)} \cdot \frac{x}{\sqrt{1+x^2}} \]
Simplify the constant terms in the denominator ($1 + 2 = 3$):
\[ \frac{dy}{dx} = \frac{1}{x^2 - 2\sqrt{1+x^2} + 3} \cdot \frac{x}{\sqrt{1+x^2}} \]
Combine into a single fraction to match the options:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{1+x^2}(x^2 - 2\sqrt{1+x^2} + 3)} \]
Step 4: Final Answer:
The derivative is $\frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{1+x^2}+3)}$.