1. **State the problem:** Find $\frac{dy}{dx}$ when $y = (x^2 - 2x)^{\tan x}$ for $x > \sqrt{2}$. 2. **Use logarithmic differentiation:** Take natural logarithm on both sides: $$\ln y = \ln \left((x^2 - 2x)^{\tan x}\right)$$ 3. Apply log power rule: $$\ln y = \tan x \cdot \ln (x^2 - 2x)$$ 4. Differentiate both sides with respect to $x$: $$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left(\tan x \cdot \ln (x^2 - 2x)\right)$$ 5. Use product rule on right side: $$\frac{d}{dx} (u v) = u' v + u v'$$ where $u = \tan x$, $v = \ln (x^2 - 2x)$. 6. Compute derivatives: $$u' = \sec^2 x$$ $$v' = \frac{1}{x^2 - 2x} \cdot (2x - 2) = \frac{2(x - 1)}{x^2 - 2x}$$ 7. Substitute back: $$\frac{1}{y} \frac{dy}{dx} = \sec^2 x \cdot \ln (x^2 - 2x) + \tan x \cdot \frac{2(x - 1)}{x^2 - 2x}$$ 8. Multiply both sides by $y$: $$\frac{dy}{dx} = y \left(\sec^2 x \cdot \ln (x^2 - 2x) + \tan x \cdot \frac{2(x - 1)}{x^2 - 2x}\right)$$ 9. Recall $y = (x^2 - 2x)^{\tan x}$, so final answer is: $$\boxed{\frac{dy}{dx} = (x^2 - 2x)^{\tan x} \left(\sec^2 x \cdot \ln (x^2 - 2x) + \tan x \cdot \frac{2(x - 1)}{x^2 - 2x}\right)}$$