Question

If \(u=\sin^{-1}(\frac{2x}{1+x^2})\) and \(v=\tan^{-1}(\frac{2x}{1-x^2})\) then \(\frac{du}{dv}\) is

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

Hint:

When solving inverse trigonometric functions and their derivatives, remember to use the chain rule. For inverse sine, cosine, and tangent functions, use the standard derivatives and simplify the expressions step by step. The goal is to simplify complex expressions and identify common terms to reach the final answer.

Answer 1

The Correct Option is C

The correct answer is (C) : 1.

Answer 2

The correct answer is: (C) 1.

We are given the following expressions for \( u \) and \( v \):

\( u = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)

\( v = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \)

We need to find \( \frac{du}{dv} \).

Step 1: Differentiate \( u \) with respect to \( x \)

We start by differentiating \( u = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) using the chain rule. The derivative of \( \sin^{-1}(y) \) with respect to \( y \) is \( \frac{1}{\sqrt{1 - y^2}} \). Therefore: \[ \frac{du}{dx} = \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) \] First, simplify \( 1 - \left( \frac{2x}{1+x^2} \right)^2 \): \[ 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} \] This simplifies to \( \frac{(1 - x^2)^2}{(1+x^2)^2} \), so: \[ \frac{du}{dx} = \frac{(1 + x^2)}{\sqrt{(1 + x^2)^2 - 4x^2}} \cdot \frac{2}{(1 + x^2)^2} \] Step 2: Differentiate \( v \) with respect to \( x \)

Now, differentiate \( v = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \): \[ \frac{dv}{dx} = \frac{1}{1 + \left(\frac{2x}{1-x^2}\right)^2} \cdot \frac{d}{dx} \left( \frac{2x}{1-x^2} \right) \] Differentiating the expression inside the parentheses: \[ \frac{d}{dx} \left( \frac{2x}{1 - x^2} \right) = \frac{(1 - x^2)(2) - 2x(-2x)}{(1 - x^2)^2} = \frac{2 - 2x^2 + 4x^2}{(1 - x^2)^2} = \frac{2 + 2x^2}{(1 - x^2)^2} \] Now substitute into the formula: \[ \frac{dv}{dx} = \frac{1}{1 + \frac{4x^2}{(1 - x^2)^2}} \cdot \frac{2 + 2x^2}{(1 - x^2)^2} \] Simplifying the denominator: \[ \frac{dv}{dx} = \frac{1}{\frac{(1 - x^2)^2 + 4x^2}{(1 - x^2)^2}} \cdot \frac{2 + 2x^2}{(1 - x^2)^2} \] Which simplifies further to: \[ \frac{dv}{dx} = \frac{2 + 2x^2}{(1 - x^2) \left( (1 - x^2)^2 + 4x^2 \right)} \] Step 3: Find \( \frac{du}{dv} \)

Now that we have the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), the value of \( \frac{du}{dv} \) is: \[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = 1 \] Therefore, the correct answer is (C) 1.