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: Understanding the Concept:
For continuity at \(x = 0\), \(f(0)\) must equal \(\lim_{x\to0}f(x)\). The limit is of the form \(\frac00\).

Step 2: Expand the numerator:
\((256 - 8x)^{1/4} = 4\left(1 - \frac{x}{32}\right)^{1/4} \approx 4\left(1 - \frac{x}{128}\right)\). So the numerator is about \(-\frac{x}{32}\).

Step 3: Expand the denominator:
\((64 + 3x)^{1/3} = 4\left(1 + \frac{3x}{64}\right)^{1/3} \approx 4\left(1 + \frac{x}{64}\right)\). So \(16 - 4(64 + 3x)^{1/3} \approx 16 - 16 - \frac{x}{4} = -\frac{x}{4}\).

Step 4: Limit:
\[ \lim_{x\to0}f(x) = \frac{-x/32}{-x/4} = \frac{4}{32} = \frac18 \]

Final Answer:
\(f(0) = \frac18\), option (B). \[ \boxed{\frac{1}{8}} \]