1. **Problem:** Find $\frac{dy}{dx}$ using implicit differentiation for the equation: $$x + \sec(y) = \ln(y)$$ 2. **Formula and rules:** - Differentiate both sides with respect to $x$. - Use the chain rule for $\sec(y)$ and $\ln(y)$ since $y$ is a function of $x$. - Recall: $\frac{d}{dx} \sec(y) = \sec(y) \tan(y) \frac{dy}{dx}$ and $\frac{d}{dx} \ln(y) = \frac{1}{y} \frac{dy}{dx}$. 3. **Differentiate both sides:** $$\frac{d}{dx} x + \frac{d}{dx} \sec(y) = \frac{d}{dx} \ln(y)$$ $$1 + \sec(y) \tan(y) \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$$ 4. **Group terms with $\frac{dy}{dx}$ on one side:** $$\sec(y) \tan(y) \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = -1$$ 5. **Factor out $\frac{dy}{dx}$:** $$\left(\sec(y) \tan(y) - \frac{1}{y}\right) \frac{dy}{dx} = -1$$ 6. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{-1}{\sec(y) \tan(y) - \frac{1}{y}}$$ 7. **Simplify the denominator if desired:** $$\frac{dy}{dx} = \frac{-1}{\sec(y) \tan(y) - \frac{1}{y}} = \frac{-1}{\sec(y) \tan(y) - \frac{1}{y}}$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{-1}{\sec(y) \tan(y) - \frac{1}{y}}}$$