Limits, Continuity & Differentiability
Differentiation
nta_abhyas_2025
Grade 12
Question:
If $y = \tan^{-1}\left(\frac{1}{x^2+1}\right) + \tan^{-1}\left(\frac{1}{x^2+3}\right)$, find $\frac{dy}{dx}$
Step-by-Step Solution
Key Concept: Inverse tangent difference formula: $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$ and telescoping series simplification
Using the inverse tangent addition formula and telescoping properties, $y = \tan^{-1}(x+1) - \tan^{-1}(x) + \tan^{-1}(x+2) - \tan^{-1}(x+1)$. This simplifies to $y = \tan^{-1}(x+2) - \tan^{-1}(x)$. Therefore, $\frac{dy}{dx} = \frac{1}{1+(x+2)^2} - \frac{1}{1+x^2} = \frac{-2(x+1)}{(1+x^2)(1+(x+2)^2)}$. The value of $k$ where $\frac{dy}{dx} = -\frac{1}{k}$ gives $k = 16$.
Correct Answer: 16