Find an expression for $\frac{\text{d}y}{\text{d}x}$ in terms of $x$ if $y = \tan^{-1} \sqrt{x^2 - 1}$. It is given that $y = x^{\cos x}$ for $x > 0$. By taking logarithm first, find an expression for $\frac{\text{d}y}{\text{d}x}$ in terms of $x$.

1. Apply the chain rule to differentiate the inverse tangent function.

$\frac{\text{d}y}{\text{d}x} = \frac{1}{1 + (\sqrt{x^2 - 1})^2} \cdot \frac{\text{d}}{\text{d}x}(\sqrt{x^2 - 1})$

2. Differentiate the inner square root function.

$\frac{\text{d}}{\text{d}x}(\sqrt{x^2 - 1}) = \frac{1}{2}(x^2 - 1)^{-\frac{1}{2}}(2x) = \frac{x}{\sqrt{x^2 - 1}}$

3. Simplify the expression to obtain the final derivative.

$\frac{\text{d}y}{\text{d}x} = \frac{1}{1 + x^2 - 1} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{x^2} \cdot \frac{x}{\sqrt{x^2 - 1}} = \frac{1}{x\sqrt{x^2 - 1}}$

1. Take the natural logarithm of both sides and apply logarithm properties to simplify.

$\ln y = \ln(x^{\cos x}) = \cos x \ln x$

2. Differentiate both sides with respect to $x$ using implicit differentiation and the product rule.

$\frac{1}{y} \frac{\text{d}y}{\text{d}x} = \cos x \cdot \frac{1}{x} + (-\sin x) \ln x$
$\frac{1}{y} \frac{\text{d}y}{\text{d}x} = \frac{\cos x}{x} - \sin x \ln x$

3. Make $\frac{\text{d}y}{\text{d}x}$ the subject and express it purely in terms of $x$ by substituting $y = x^{\cos x}$.

$\frac{\text{d}y}{\text{d}x} = y \left( \frac{\cos x}{x} - \sin x \ln x \right)$
$\frac{\text{d}y}{\text{d}x} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \ln x \right)$