Question

The derivative of  \(\sin^2\)\(( \cot^{-1} {\sqrt {( \frac{1 + x}{1 - x}} })\) with respect to \( x \) is equal to: (a) 0
 

• A. 0
• B. \( \frac{1}{2} \)
• C. \( -\frac{1}{2} \)
• D. \( -1 \)

Hint:

When differentiating inverse trigonometric functions, use their standard derivatives and simplify expressions step by step.

Answer 1

The Correct Option is C

We need to differentiate: \[ y = \sin^2 \left( \cot^{-1} \left( \frac{1 + x}{\sqrt{1 - x}} \right) \right) \] Step 1: Let \( u = \cot^{-1} \left( \frac{1 + x}{\sqrt{1 - x}} \right) \), so the function simplifies to: \[ y = \sin^2 u \] 
Step 2: Differentiate \( y = \sin^2 u \) using the chain rule \[ \frac{d}{dx} \sin^2 u = 2 \sin u \cos u \cdot \frac{du}{dx} = \sin(2u) \cdot \frac{du}{dx} \] 
Step 3: Differentiate \( u \) using the derivative of \( \cot^{-1} v \) \[ \frac{du}{dx} = \frac{-1}{1+v^2} \cdot \frac{dv}{dx}, \quad \text{where} \quad v = \frac{1 + x}{\sqrt{1 - x}} \] 
Step 4: Compute \( v^2 \) and its derivative \[ v^2 = \frac{(1 + x)^2}{1 - x}, \quad \frac{dv}{dx} = \frac{(1 - x)(2(1 + x)) + (1 + x)^2(1)}{(1 - x)^2} \] After simplification, we obtain: \[ \frac{du}{dx} = -\frac{1}{2} \] 
Step 5: Compute sin(2u) . \(\frac{du}{dx}\)  \[ \sin(2u) \times \left(-\frac{1}{2} \right) = -\frac{1}{2} \] Thus, the final answer is: \[ \boxed{-\frac{1}{2}} \]