Question

The derivative of \(tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})\) with respect to \(tan^{-1}(\frac{x}{\sqrt{1-x^2}})\) at \(x = \frac{1}{2}\) is

• A. \(\frac{\sqrt{3}}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{\sqrt{3}}{5}\)
• D. \(\frac{1}{4}\)

Hint:

The limit at 0 is of the form 0/0, so expand using the binomial approximation.

Answer 1

The Correct Option is C

Step 1: Use derivatives (L'Hopital):
Differentiate top and bottom: top $\frac14(256 - 8x)^{-3/4}(-8)$, bottom $-4\cdot\frac13(64 + 3x)^{-2/3}(3)$.

Step 2: Evaluate at 0:
Top at 0: $-2\times 256^{-3/4} = -2\times\frac{1}{64} = -\frac{1}{32}$. Bottom at 0: $-4\times 64^{-2/3} = -4\times\frac{1}{16} = -\frac14$.
Ratio $= \frac{-1/32}{-1/4} = \frac18$. Option (B).

Final Answer:
$\frac18$. \[ \boxed{\frac{1}{8}} \]