1. We are asked to find the derivative $\frac{dy}{dx}$ when $y = \cosh^{-1}(\sqrt{5x} + 5)$.\n\n2. Recall the formula for the derivative of the inverse hyperbolic cosine function: $$\frac{d}{dx} \cosh^{-1}(u) = \frac{1}{\sqrt{u^2 - 1}} \cdot \frac{du}{dx}.$$\n\n3. Let $u = \sqrt{5x} + 5$. We need to find $\frac{du}{dx}$.\n\n4. Compute $\frac{du}{dx}$:\n$$u = (5x)^{1/2} + 5$$\n$$\frac{du}{dx} = \frac{1}{2}(5x)^{-1/2} \cdot 5 + 0 = \frac{5}{2\sqrt{5x}}.$$\n\n5. Substitute $u$ and $\frac{du}{dx}$ into the derivative formula:\n$$\frac{dy}{dx} = \frac{1}{\sqrt{(\sqrt{5x} + 5)^2 - 1}} \cdot \frac{5}{2\sqrt{5x}}.$$\n\n6. Simplify the expression inside the square root:\n$$(\sqrt{5x} + 5)^2 - 1 = (\sqrt{5x})^2 + 2 \cdot \sqrt{5x} \cdot 5 + 25 - 1 = 5x + 10\sqrt{5x} + 24.$$\n\n7. Therefore, the derivative is:\n$$\frac{dy}{dx} = \frac{5}{2\sqrt{5x} \sqrt{5x + 10\sqrt{5x} + 24}}.$$\n\nThis is the final expression for $\frac{dy}{dx}$.