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: Understanding the Question:
We need to find the derivative of one function with respect to another, which can be done by expressing $u$ in terms of $v$ or using parametric differentiation.

Step 2: Key Formula or Approach:
Notice that $(\sqrt{x+1} - \sqrt{x-1})(\sqrt{x+1} + \sqrt{x-1}) = (x+1) - (x-1) = 2$.
Let $t = \sqrt{x+1} - \sqrt{x-1}$. Then $t \cdot v = 2 \implies t = \frac{2}{v}$.

Step 3: Detailed Explanation:
Given $u = \log(\sqrt{x+1} - \sqrt{x-1})$, we can substitute $\sqrt{x+1} - \sqrt{x-1} = \frac{2}{v}$:
$u = \log(\frac{2}{v}) = \log 2 - \log v$.
Now, differentiate $u$ with respect to $v$:
$\frac{du}{dv} = \frac{d}{dv} (\log 2 - \log v)$
$\frac{du}{dv} = 0 - \frac{1}{v} = -\frac{1}{v}$.

Step 4: Final Answer:
The value of $\frac{du}{dv}$ is $-\frac{1}{v}$.