Question

If $u=\log(\sqrt{x+1}-\sqrt{x-1})$ and $v=\sqrt{x+1}+\sqrt{x-1}$ then $\frac{du}{dv}=...$

• A. u
• B. v
• C. $-1/u$
• D. $-1/v$

Hint:

Identifying conjugates in radical expressions often leads to simple logarithmic relations.

Answer 1

The Correct Option is D

Step 1: Concept
Note the relationship between the two expressions: $(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1}) = (x+1) - (x-1) = 2$.

Step 2: Analysis
So, $\sqrt{x+1}-\sqrt{x-1} = 2/v$.
Substitute in $u$: $u = \log(2/v) = \log 2 - \log v$.

Step 3: Calculation
Differentiate $u$ with respect to $v$:
$\frac{du}{dv} = \frac{d}{dv}(\log 2 - \log v) = 0 - \frac{1}{v}$.

Step 4: Conclusion
Hence, $\frac{du}{dv} = -1/v$. Final Answer: (D)