Question

If \( y = \sinh^{-1} \left(\frac{1 - x}{1 + x} \right) \), then \( \frac{dy}{dx} \) is given by:

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

Hint:

For inverse hyperbolic differentiation, use the formula \( \frac{d}{dx} \sinh^{-1} u = \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{dx} \).

Answer 1

The Correct Option is A

Step 1: Differentiate both sides Using the derivative formula for inverse hyperbolic functions: \[ \frac{d}{dx} \sinh^{
-1} u = \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{dx}. \] Let \( u = \frac{1
- x}{1 + x} \), then differentiate: \[ \frac{du}{dx} = \frac{(1 + x)(
-1)
- (1
- x)(1)}{(1 + x)^2} = \frac{
- (1 + x)
- (1
- x)}{(1 + x)^2}. \] \[ = \frac{
-1
- x
- 1 + x}{(1 + x)^2} = \frac{
-2}{(1 + x)^2}. \] Step 2: Compute \( \frac{dy}{dx} \) \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{dx}. \] Using \( u = \frac{1
- x}{1 + x} \), \[ 1 + u^2 = 1 + \left(\frac{1
- x}{1 + x}\right)^2. \] Expanding and simplifying, we get: \[ \frac{dy}{dx} = \frac{
-\sqrt{2}}{|1 + x| \sqrt{1 + x^2}}. \]