Question

If \( y = \tan^{-1} \left[ \frac{1}{1+x+x^2} \right] + \tan^{-1} \left[ \frac{1}{x^2+3x+3} \right] \), \( x > 0 \), then \(\frac{dy}{dx} =\)

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

Hint:

When you see \(\tan^{-1}\left(\frac{1}{1+x+x^2}\right)\), try to express it as \(\tan^{-1}(x+1) - \tan^{-1}(x)\). This trick often simplifies sums of such terms.

Answer 1

The Correct Option is B

Step 1: Understanding the Question: We are given \(y\) as sum of two inverse tangent terms and need to find \(\frac{dy}{dx}\) for \(x>0\).

Step 2: Key Formula or Approach: Use the identity \(\tan^{-1}\left(\frac{1}{t}\right) = \cot^{-1}(t) = \frac{\pi}{2} - \tan^{-1}(t)\) for \(t>0\). However, a better simplification is to recognise each term as the difference of two inverse tangents: \[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right). \] Set \(a = x+1\), \(b = x\): \(\tan^{-1}(x+1) - \tan^{-1}(x) = \tan^{-1}\left(\frac{1}{1+x(x+1)}\right) = \tan^{-1}\left(\frac{1}{1+x+x^2}\right)\). Similarly, \(\tan^{-1}(x+2) - \tan^{-1}(x+1) = \tan^{-1}\left(\frac{1}{1+(x+1)(x+2)}\right) = \tan^{-1}\left(\frac{1}{x^2+3x+3}\right)\).

Step 3: Detailed Explanation: Adding the two identities: \[ y = \bigl[\tan^{-1}(x+1) - \tan^{-1}(x)\bigr] + \bigl[\tan^{-1}(x+2) - \tan^{-1}(x+1)\bigr] = \tan^{-1}(x+2) - \tan^{-1}(x). \] Now differentiate: \[ \frac{dy}{dx} = \frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}. \] This is equivalent to \(-\frac{1}{1+x^2} + \frac{1}{1+(x+2)^2}\), which matches option (B).

Step 4: Final Answer: Option (B) is correct.