If y = tan^{-1} frac{x}{1+2x^2} + tan^{-1} fr | vedclass

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(C) We can rewrite each term using the formula $	an^{-1} \frac{a-b}{1+ab} = 	an^{-1} a - 	an^{-1} b$. $y = 	an^{-1} \frac{2x-x}{1+(2x)(x)} + 	an^

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(C) We can rewrite each term using the formula $	an^{-1} \frac{a-b}{1+ab} = 	an^{-1} a - 	an^{-1} b$. $y = 	an^{-1} \frac{2x-x}{1+(2x)(x)} + 	an^{-1} \frac{3x-2x}{1+(3x)(2x)} + 	an^{-1} \frac{4x-3x}{1+(4x)(3x)}$ $y = (	an^{-1} 2x - 	an^{-1} x) + (	an^{-1} 3x - 	an^{-1} 2x) + (	an^{-1} 4x - 	an^{-1} 3x)$ $y = 	an^{-1} 4x - 	an^{-1} x$ Now,differentiate with respect to $x$: $\frac{dy}{dx} = \frac{4}{1+(4x)^2} - \frac{1}{1+x^2} = \frac{4}{1+16x^2} - \frac{1}{1+x^2}$ Substitute $x = \frac{1}{2}$: $\left(\frac{dy}{dx}ight)_{x=\frac{1}{2}} = \frac{4}{1+16(\frac{1}{4})} - \frac{1}{1+(\frac{1}{4})} = \frac{4}{1+4} - \frac{1}{\frac{5}{4}} = \frac{4}{5} - \frac{4}{5} = 0$

If $y = \tan^{-1} \frac{x}{1+2x^2} + \tan^{-1} \frac{x}{1+6x^2} + \tan^{-1} \frac{x}{1+12x^2}$,then $\left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = $

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