If y=tan ^{-1}left(frac{3 x-x^3}{1-3 x^2}righ | vedclass

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(D) Given,$y=	an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}ight)+	an ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}ight)$.Substitute $x=	an 	heta$,then $	h

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$\frac{7}{1+x^2}$

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(D) Given,$y=	an ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}ight)+	an ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}ight)$.Substitute $x=	an 	heta$,then $	heta = 	an^{-1} x$.The expression becomes:$y = 	an^{-1}(	an 3	heta) + 	an^{-1}(	an 4	heta)$.This simplifies to $y = 3	heta + 4	heta = 7	heta$.Substituting back,$y = 7 	an^{-1} x$.Differentiating with respect to $x$:$\frac{d y}{d x} = 7 	imes \frac{1}{1+x^2} = \frac{7}{1+x^2}$.

If $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$,then $\frac{d y}{d x}$ is equal to

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