If y = tan^{-1}left(frac{4x}{1+5x^2}right) + | vedclass

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(A) Given $y = 	an^{-1}\left(\frac{5x-x}{1+5x \cdot x}ight) + \cot^{-1}\left(\frac{\frac{3}{2}-x}{1+\frac{3}{2}x}ight)$.Using the formula $	an^{

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$\frac{5}{1+25x^2}$

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(A) Given $y = 	an^{-1}\left(\frac{5x-x}{1+5x \cdot x}ight) + \cot^{-1}\left(\frac{\frac{3}{2}-x}{1+\frac{3}{2}x}ight)$.Using the formula $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we have:$y = (	an^{-1}(5x) - 	an^{-1}(x)) + (\cot^{-1}(x) - \cot^{-1}(\frac{3}{2}))$.Since $\cot^{-1}(x) = \frac{\pi}{2} - 	an^{-1}(x)$,we substitute:$y = 	an^{-1}(5x) - 	an^{-1}(x) + \frac{\pi}{2} - 	an^{-1}(x) - \cot^{-1}(\frac{3}{2})$.$y = 	an^{-1}(5x) - 2	an^{-1}(x) + \frac{\pi}{2} - \cot^{-1}(\frac{3}{2})$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an^{-1}(5x)) - 2\frac{d}{dx}(	an^{-1}(x)) + 0 - 0$.$\frac{dy}{dx} = \frac{5}{1+(5x)^2} - \frac{2}{1+x^2} = \frac{5}{1+25x^2} - \frac{2}{1+x^2}$.Note: Re-evaluating the expression $\cot^{-1}\left(\frac{3-2x}{2+3x}ight) = 	an^{-1}\left(\frac{2+3x}{3-2x}ight) = 	an^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3}x}ight) = 	an^{-1}(\frac{2}{3}) + 	an^{-1}(x)$.Thus,$y = 	an^{-1}(5x) - 	an^{-1}(x) + 	an^{-1}(\frac{2}{3}) + 	an^{-1}(x) = 	an^{-1}(5x) + 	an^{-1}(\frac{2}{3})$.$\frac{dy}{dx} = \frac{5}{1+25x^2}$.

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \cot^{-1}\left(\frac{3-2x}{2+3x}\right)$,then $\frac{dy}{dx}$ is equal to

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