Question

The derivative of $\tan^{-1} \left(\sqrt{1+x^2}-1\right)$ is

• A. $\frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{x+1}+1)}$
• B. $\frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{1+x^2}+3)}$
• C. $\frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{x^2+1}+2)}$
• D. $\frac{x}{\sqrt{1+x^2}(x^2+2\sqrt{1+x^2}-3)}$

Hint:

Use chain rule carefully with inverse trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Let function. \[ y = \tan^{-1}(\sqrt{1+x^2}-1) \]
Step 2: Differentiate. \[ \frac{dy}{dx} = \frac{1}{1+(\sqrt{1+x^2}-1)^2} \cdot \frac{d}{dx}(\sqrt{1+x^2}-1) \] \[ = \frac{1}{1+(\sqrt{1+x^2}-1)^2} \cdot \frac{x}{\sqrt{1+x^2}} \]
Step 3: Simplify denominator. \[ 1+(\sqrt{1+x^2}-1)^2 = x^2 - 2\sqrt{1+x^2} + 2 \]
Step 4: Final answer. \[ \frac{x}{\sqrt{1+x^2}(x^2 - 2\sqrt{1+x^2} + 2)} \]
Step 5: Conclusion. \[ {\text{Option (C)}} \]