To differentiate the expression:

$f(x) = \tan^{-1}(\sqrt{x^2 - 1}) + \csc^{-1}(x)$

we need to differentiate each term individually.

Step 1: Differentiate $\tan^{-1}(\sqrt{x^2 - 1})$

Let $u = \sqrt{x^2 - 1}$, so $\tan^{-1}(u)$ is a composite function. Using the chain rule: $\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$ Now, differentiate $u = \sqrt{x^2 - 1}$: $\frac{du}{dx} = \frac{d}{dx} (x^2 - 1)^{1/2} = \frac{1}{2}(x^2 - 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}}$ Substituting $u = \sqrt{x^2 - 1}$ and $\frac{du}{dx}$ back: $\frac{d}{dx} \left( \tan^{-1}(\sqrt{x^2 - 1}) \right) = \frac{1}{1 + (\sqrt{x^2 - 1})^2} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{x}{(x^2)\sqrt{x^2 - 1}} = \frac{1}{x\sqrt{x^2 - 1}}$

Step 2: Differentiate $\csc^{-1}(x)$

The derivative of $\csc^{-1}(x)$ is given by: $\frac{d}{dx} \csc^{-1}(x) = -\frac{1}{|x|\sqrt{x^2 - 1}}$

Step 3: Combine the results

Now, we combine the derivatives of both terms: $f'(x) = \frac{1}{x\sqrt{x^2 - 1}} - \frac{1}{|x|\sqrt{x^2 - 1}}$

For $x > 1$, where $|x| = x$, the expression simplifies to: $f'(x) = \frac{1}{x\sqrt{x^2 - 1}} - \frac{1}{x\sqrt{x^2 - 1}} = 0$

Thus, the derivative of the given expression is: $f'(x) = 0 \text{ for } x > 1$

Would you like more details on any part of the calculation?

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Here are 5 questions that extend this discussion:

1. How do we handle derivatives of inverse trigonometric functions in general?
2. Can the derivative change if $x$ is negative or less than 1?
3. How can we verify the behavior of $f'(x)$ for different intervals of $x$?
4. Why does the expression simplify to zero in this particular case?
5. Can we apply similar techniques to differentiate products or quotients of trigonometric functions?

Tip: Always be cautious with the domain restrictions for inverse trigonometric functions to avoid errors in the final expression.