1. **State the problem:** We want to find the derivative $\frac{dy}{d\theta}$ of the function $$y = \frac{40^{8} \sin \theta}{7 \sqrt{\sec \theta}}.$$\n\n2. **Rewrite the function for easier differentiation:** Recall that $\sqrt{\sec \theta} = (\sec \theta)^{1/2}$. So,\n$$y = \frac{40^{8} \sin \theta}{7 (\sec \theta)^{1/2}} = \frac{40^{8}}{7} \sin \theta (\sec \theta)^{-1/2}.$$\n\n3. **Apply logarithmic differentiation:** Take natural logarithm on both sides:\n$$\ln y = \ln \left( \frac{40^{8}}{7} \right) + \ln (\sin \theta) + \ln \left( (\sec \theta)^{-1/2} \right).$$\nSimplify:\n$$\ln y = \ln \left( \frac{40^{8}}{7} \right) + \ln (\sin \theta) - \frac{1}{2} \ln (\sec \theta).$$\n\n4. **Differentiate both sides with respect to $\theta$:**\n$$\frac{1}{y} \frac{dy}{d\theta} = \frac{d}{d\theta} \ln (\sin \theta) - \frac{1}{2} \frac{d}{d\theta} \ln (\sec \theta).$$\nRecall derivatives:\n$$\frac{d}{d\theta} \ln (\sin \theta) = \cot \theta,$$\n$$\frac{d}{d\theta} \ln (\sec \theta) = \tan \theta.$$\nSo,\n$$\frac{1}{y} \frac{dy}{d\theta} = \cot \theta - \frac{1}{2} \tan \theta.$$\n\n5. **Solve for $\frac{dy}{d\theta}$:**\n$$\frac{dy}{d\theta} = y \left( \cot \theta - \frac{1}{2} \tan \theta \right).$$\nSubstitute back $y$:\n$$\frac{dy}{d\theta} = \frac{40^{8} \sin \theta}{7 \sqrt{\sec \theta}} \left( \cot \theta - \frac{1}{2} \tan \theta \right).$$\n\n6. **Final answer:**\n$$\boxed{\frac{dy}{d\theta} = \frac{40^{8} \sin \theta}{7 \sqrt{\sec \theta}} \left( \cot \theta - \frac{1}{2} \tan \theta \right)}.$$