Limits, Continuity & Differentiability
Differentiation
nta_abhyas_2025
Grade 12
Question:
If $y = \tan^{-1}\left(\frac{1}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ ($y > 0$), then $\frac{dy}{dx}$ is equal to
\frac{2}{1+x^2}
\frac{1}{1+x^2}
\frac{1}{1+x^2} + \frac{2}{1+x^4}
\frac{2}{1+x^4}
Step-by-Step Solution
Key Concept: Use the addition formula for inverse tangent: $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$ (when $ab < 1$)
$y = \tan^{-1}\left(\frac{1+x^2}{1-x^2}\right) + \tan^{-1}\left(\frac{2x+1}{1-2x}\right)$ can be simplified using $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$. This simplifies to $y = \tan^{-1}(3x - \frac{3}{3}) = \frac{dx}{dy} = \frac{3}{1+9x^2}$
Correct Answer: 1