y = operatorname{Tan}^{-1}left(frac{x}{1+2x^2 | vedclass

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(B) We know that $\operatorname{Tan}^{-1}(a) - \operatorname{Tan}^{-1}(b) = \operatorname{Tan}^{-1}\left(\frac{a-b}{1+ab}\right)$.We can rewrite the t

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$\frac{3}{9x^2+1} - \frac{1}{x^2+1}$

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(B) We know that $\operatorname{Tan}^{-1}(a) - \operatorname{Tan}^{-1}(b) = \operatorname{Tan}^{-1}\left(\frac{a-b}{1+ab}\right)$.We can rewrite the terms as:$\operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) = \operatorname{Tan}^{-1}\left(\frac{2x-x}{1+(2x)(x)}\right) = \operatorname{Tan}^{-1}(2x) - \operatorname{Tan}^{-1}(x)$.$\operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right) = \operatorname{Tan}^{-1}\left(\frac{3x-2x}{1+(3x)(2x)}\right) = \operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(2x)$.Substituting these into the expression for $y$:$y = (\operatorname{Tan}^{-1}(2x) - \operatorname{Tan}^{-1}(x)) + (\operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(2x)) = \operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\operatorname{Tan}^{-1}(3x)) - \frac{d}{dx}(\operatorname{Tan}^{-1}(x)) = \frac{3}{1+(3x)^2} - \frac{1}{1+x^2} = \frac{3}{9x^2+1} - \frac{1}{x^2+1}$.

$y = \operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right)$,then $\frac{dy}{dx} = $

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