1. **Stating the problem:** We need to analyze the given expression involving $W$, $\lambda$, and $\alpha$ and prove the stated relationship or property related to the function involving $\sec(\lambda - \alpha)$ and the logarithmic term $2 \ln(\lambda - \alpha)$. 2. **Understanding the given expression:** The problem mentions a function of the form: $$ W \sec(\lambda - \alpha) $$ and a term involving: $$ 2 \ln(\lambda - \alpha) $$ with $\lambda$ and $\alpha$ as variables or parameters. 3. **Key formulas and rules:** - The derivative of $\sec x$ is $\sec x \tan x$. - The derivative of $\ln x$ is $\frac{1}{x}$. - The chain rule applies when differentiating composite functions. 4. **Differentiating the function:** Assuming we want to differentiate the function: $$ f(\lambda) = W \sec(\lambda - \alpha) $$ with respect to $\lambda$: $$ \frac{d}{d\lambda} f(\lambda) = W \frac{d}{d\lambda} \sec(\lambda - \alpha) = W \sec(\lambda - \alpha) \tan(\lambda - \alpha) \cdot \frac{d}{d\lambda}(\lambda - \alpha) $$ Since $\alpha$ is constant with respect to $\lambda$: $$ \frac{d}{d\lambda}(\lambda - \alpha) = 1 $$ So, $$ \frac{d}{d\lambda} f(\lambda) = W \sec(\lambda - \alpha) \tan(\lambda - \alpha) $$ 5. **Differentiating the logarithmic term:** $$ g(\lambda) = 2 \ln(\lambda - \alpha) $$ Differentiating: $$ \frac{d}{d\lambda} g(\lambda) = 2 \cdot \frac{1}{\lambda - \alpha} \cdot \frac{d}{d\lambda}(\lambda - \alpha) = \frac{2}{\lambda - \alpha} $$ 6. **Combining the results:** If the problem involves showing a relationship between these derivatives or proving a limit or identity, the above derivatives are the key components. 7. **Summary:** - The derivative of $W \sec(\lambda - \alpha)$ is $W \sec(\lambda - \alpha) \tan(\lambda - \alpha)$. - The derivative of $2 \ln(\lambda - \alpha)$ is $\frac{2}{\lambda - \alpha}$. These results can be used to prove further properties or limits as required. **Final answer:** $$ \frac{d}{d\lambda} \left( W \sec(\lambda - \alpha) \right) = W \sec(\lambda - \alpha) \tan(\lambda - \alpha) $$ $$ \frac{d}{d\lambda} \left( 2 \ln(\lambda - \alpha) \right) = \frac{2}{\lambda - \alpha} $$