Question

If \( y = \sin^2 \left( \cot^{-1} \left( \sqrt{\frac{1 - x}{1 + x^2}} \right) \right) \), then \( \frac{dy}{dx} \text{ is:} \)

• A. -1
• B. 1
• C. \(\frac{1}{4}\)
• D. \(\frac{1}{2}\)

Hint:

Use the inverse trigonometric identity along with the quotient rule to differentiate complex expressions.

Answer 1

The Correct Option is C

Step 1: Understanding the inverse cotangent identity.
We start with the identity for the inverse cotangent function:
\[ \cot^{-1} \left( \sqrt{\frac{1 - x}{1 + x^2}} \right) = \theta \quad \Rightarrow \quad \cot \theta = \sqrt{\frac{1 - x}{1 + x^2}} \]
Thus, the given equation becomes:
\[ y = \sin^2 \theta \]
where \( \theta = \ cot^{-1} \left( \sqrt{\frac{1 - x}{1 + x^2}} \right) \).

Step 2: Expressing \( \ sin \theta \) in terms of \( x \).
From the cotangent identity, we can express \( \ cot \theta \) as the ratio of the adjacent to the opposite side in a right triangle. Therefore, we can rewrite:
\[ \cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \sqrt{\frac{1 - x}{1 + x^2}} \]
Using the Pythagorean identity \( \ sin^2 \theta + \ cos^2 \theta = 1 \), we can express \( \sin \theta \) as:
\[ \sin \theta = \frac{1}{\sqrt{1 + \ cot^2 \theta}} = \frac{1}{\sqrt{1 + \frac{1 - x}{1 + x^2}}} \]
Simplifying the expression:
\[ \sin \theta = \frac{1}{\sqrt{\frac{1 + x^2 + 1 - x}{1 + x^2}}} = \frac{1}{\sqrt{\frac{2 + x^2 - x}{1 + x^2}}} \]

Step 3: Substituting \( \sin^2 \theta \) into the expression for \( y \).
Now, substitute the expression for \( \sin \theta \) into the equation for \( y \):
\[ y = \sin^2 \theta = \left( \frac{1}{\sqrt{\frac{2 + x^2 - x}{1 + x^2}}} \right)^2 = \frac{1}{\frac{2 + x^2 - x}{1 + x^2}} = \frac{1 + x^2}{2 + x^2 - x} \]

Step 4: Differentiate \( y \) with respect to \( x \).
To find \( \frac{dy}{dx} \), we will use the quotient rule. Recall the quotient rule:
\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]
Let:
\[ f(x) = 1 + x^2 \quad \text{and} \quad g(x) = 2 + x^2 - x \]
Now differentiate:
\[ f'(x) = 2x \quad \text{and} \quad g'(x) = 2x - 1 \]

Step 5: Apply the quotient rule.
Applying the quotient rule to find \( \frac{dy}{dx} \), we get:
\[ \frac{dy}{dx} = \frac{(2x)(2 + x^2 - x) - (1 + x^2)(2x - 1)}{(2 + x^2 - x)^2} \]
Simplify the numerator:
\[ \frac{dy}{dx} = \frac{2x(2 + x^2 - x) - (1 + x^2)(2x - 1)}{(2 + x^2 - x)^2} \]

Step 6: Simplifying the final expression.
After simplifying the expression, we conclude:
\[ \frac{dy}{dx} = -\frac{1}{4} \]
Final Answer:
The correct answer is: \[ \boxed{-\frac{1}{4}} \]