Question

If \[ y = \tan^{-1} \frac{x}{1+2x^2} + \tan^{-1} \frac{x}{1+6x^2} + \tan^{-1} \frac{x}{1+12x^2}, \] then \( \left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} \) =

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

Hint:

To differentiate \( y = \tan^{-1} f(x) \), use: \[ \frac{d}{dx} \tan^{-1} f(x) = \frac{f'(x)}{1+f(x)^2}. \]

Answer 1

The Correct Option is C

Step 1: Differentiate Each Term
The derivative of \( \tan^{-1} f(x) \) is given by: \[ \frac{d}{dx} \tan^{-1} f(x) = \frac{f'(x)}{1 + f(x)^2}. \] For each term in \( y \): 1. \( y_1 = \tan^{-1} \frac{x}{1+2x^2} \), let \( f(x) = \frac{x}{1+2x^2} \), 2. \( y_2 = \tan^{-1} \frac{x}{1+6x^2} \), let \( g(x) = \frac{x}{1+6x^2} \), 3. \( y_3 = \tan^{-1} \frac{x}{1+12x^2} \), let \( h(x) = \frac{x}{1+12x^2} \).
Step 2: Compute Derivatives
Using quotient rule: \[ f'(x) = \frac{(1+2x^2)(1) - x(4x)}{(1+2x^2)^2} = \frac{1+2x^2 - 4x^2}{(1+2x^2)^2} = \frac{1-2x^2}{(1+2x^2)^2}. \] Similarly, \[ g'(x) = \frac{1 - 6x^2}{(1+6x^2)^2}, \quad h'(x) = \frac{1 - 12x^2}{(1+12x^2)^2}. \] Step 3: Compute \( \frac{dy}{dx} \) \[ \frac{dy}{dx} = \frac{f'(x)}{1+f(x)^2} + \frac{g'(x)}{1+g(x)^2} + \frac{h'(x)}{1+h(x)^2}. \] Substituting \( x = \frac{1}{2} \): \[ \left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = 0. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{0}. \]