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

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(C) Given $y = 	an^{-1} \left( \frac{4x}{1 + 5x^2} ight) + 	an^{-1} \left( \frac{2 + 3x}{3 - 2x} ight)$.First,simplify the first term: $	an^{-1

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

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(C) Given $y = 	an^{-1} \left( \frac{4x}{1 + 5x^2} ight) + 	an^{-1} \left( \frac{2 + 3x}{3 - 2x} ight)$.First,simplify the first term: $	an^{-1} \left( \frac{5x - x}{1 + 5x \cdot x} ight) = 	an^{-1}(5x) - 	an^{-1}(x)$.Next,simplify the second term: $	an^{-1} \left( \frac{\frac{2}{3} + x}{1 - \frac{2}{3} \cdot x} ight) = 	an^{-1} \left( \frac{2}{3} ight) + 	an^{-1}(x)$.Adding these together,$y = 	an^{-1}(5x) - 	an^{-1}(x) + 	an^{-1} \left( \frac{2}{3} ight) + 	an^{-1}(x) = 	an^{-1}(5x) + 	an^{-1} \left( \frac{2}{3} ight)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [	an^{-1}(5x)] + \frac{d}{dx} [	an^{-1} \left( \frac{2}{3} ight)]$.Since $	an^{-1} \left( \frac{2}{3} ight)$ is a constant,its derivative is $0$.Therefore,$\frac{dy}{dx} = \frac{1}{1 + (5x)^2} \cdot \frac{d}{dx}(5x) = \frac{5}{1 + 25x^2}$.

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

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