Question:

Step 1: Decompose the second inverse tangent term using the addition formula. We recognize that the second term can be rewritten using the addition formula for inverse tangent: $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$. Rewrite $\frac{2+3x}{3-2x}$ by factoring: $$\frac{2+3x}{3-2x} = \frac{\frac{2}{3}+x}{1-\frac{2}{3}x}$$ This matches the form $\frac{A+B}{1-AB}$ with $A = \frac{2}{3}$ and $B = x$. Therefore: $$\tan^{-1}\left(\frac{2+3x}{3-2x}\right) = \tan^{-1}\left(\frac{2}{3}\right) + \tan^{-1}(x)$$ This is valid for $x \in \left(0, \frac{2}{3}\right)$. Step 2: Decompose the first inverse tangent term using the subtraction formula. We use the subtraction formula for inverse tangent: $\tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$. Rewrite the numerator and denominator of the first term: $$\frac{4x}{1+5x^2} = \frac{5x - x}{1 + (5x)(x)}$$ This matches the form $\frac{A-B}{1+AB}$ with $A = 5x$ and $B = x$. Therefore: $$\tan^{-1}\left(\frac{4x}{1+5x^2}\right) = \tan^{-1}(5x) - \tan^{-1}(x)$$ Step 3: Combine all terms to simplify $y$. Substitute both decompositions into the original expression: $$y = \left[\tan^{-1}(5x) - \tan^{-1}(x)\right] + \left[\tan^{-1}\left(\frac{2}{3}\right) + \tan^{-1}(x)\right]$$ The $\tan^{-1}(x)$ terms cancel: $$y = \tan^{-1}(5x) + \tan^{-1}\left(\frac{2}{3}\right)$$ Step 4: Differentiate both sides with respect to $x$. Since $\tan^{-1}\left(\frac{2}{3}\right)$ is a constant: $$\frac{dy}{dx} = \frac{d}{dx}\left[\tan^{-1}(5x)\right] + \frac{d}{dx}\left[\tan^{-1}\left(\frac{2}{3}\right)\right]$$ Using the chain rule and the derivative formula $\frac{d}{dx}\tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$: $$\frac{dy}{dx} = \frac{1}{1+(5x)^2} \cdot 5 + 0$$ $$\frac{dy}{dx} = \frac{5}{1+25x^2}$$ Step 5: Compare with the given form and find $\alpha$. We are given that $\frac{dy}{dx} = \frac{\alpha}{1+25x^2}$. Comparing our result with this form: $$\frac{5}{1+25x^2} = \frac{\alpha}{1+25x^2}$$ Therefore: $\alpha = 5$ The answer is **Option 3: 5**