Question

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

• A. \( \frac{4}{16x^2+1} - \frac{3}{9x^2+1} \)
• B. \( \frac{3}{9x^2+1} - \frac{1}{x^2+1} \)
• C. \( \frac{3}{9x^2+1} - \frac{2}{4x^2+1} \)
• D. \( \frac{1}{9x^2+1} - \frac{1}{x^2+1} \)

Hint:

Use inverse trigonometric identities to simplify expressions before differentiation. Key identities: \( \tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \) \( \tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \) Try to write the argument of \( \tan^{-1}\left(\frac{\text{expression}}{1+\text{product}}\right) \) in the form \( \frac{A-B}{1+AB} \) or \( \frac{A+B}{1-AB} \). Derivative of \( \tan^{-1}u \) is \( \frac{1}{1+u^2} \frac{du}{dx} \).

Answer 1

The Correct Option is B

We use the formula \( \tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \).
Term 1: \( \tan^{-1}\left(\frac{x}{1+2x^2}\right) = \tan^{-1}\left(\frac{2x-x}{1+(2x)(x)}\right) = \tan^{-1}(2x) - \tan^{-1}(x) \).
Term 2: \( \tan^{-1}\left(\frac{x}{1+6x^2}\right) = \tan^{-1}\left(\frac{3x-2x}{1+(3x)(2x)}\right) = \tan^{-1}(3x) - \tan^{-1}(2x) \).
So, \( y = (\tan^{-1}(2x) - \tan^{-1}(x)) + (\tan^{-1}(3x) - \tan^{-1}(2x)) \).
\[ y = \tan^{-1}(3x) - \tan^{-1}(x) \] Now differentiate \(y\) with respect to \(x\).
Recall \( \frac{d}{du}(\tan^{-1}u) = \frac{1}{1+u^2} \).
\[ \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(3x)) - \frac{d}{dx}(\tan^{-1}(x)) \] \[ = \frac{1}{1+(3x)^2} \cdot \frac{d}{dx}(3x) - \frac{1}{1+x^2} \cdot \frac{d}{dx}(x) \] \[ = \frac{1}{1+9x^2} \cdot 3 - \frac{1}{1+x^2} \cdot 1 \] \[ = \frac{3}{1+9x^2} - \frac{1}{1+x^2} \] This can be rewritten as \( \frac{3}{9x^2+1} - \frac{1}{x^2+1} \).
This matches option (2).