If y=tan left(3 tan ^{-1} xright),then left(1 | vedclass

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(B) $y=	an \left(3 	an ^{-1} xight)$Let $	an ^{-1} x=	heta$,so $	an 	heta=x$.Using the formula $	an 3	heta = \frac{3	an 	heta - 	an^3 	h

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$6(y-x)$

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(B) $y=	an \left(3 	an ^{-1} xight)$Let $	an ^{-1} x=	heta$,so $	an 	heta=x$.Using the formula $	an 3	heta = \frac{3	an 	heta - 	an^3 	heta}{1-3	an^2 	heta}$,we get $y = \frac{3x-x^3}{1-3x^2}$.Thus,$(1-3x^2)y = 3x-x^3$.Differentiating both sides with respect to $x$:$(1-3x^2)\frac{dy}{dx} + y(-6x) = 3-3x^2$.Differentiating again with respect to $x$:$(1-3x^2)\frac{d^2y}{dx^2} + (-6x)\frac{dy}{dx} - 6(x\frac{dy}{dx} + y) = -6x$.$(1-3x^2)\frac{d^2y}{dx^2} - 6x\frac{dy}{dx} - 6x\frac{dy}{dx} - 6y = -6x$.$(1-3x^2)\frac{d^2y}{dx^2} - 12x\frac{dy}{dx} = 6y - 6x = 6(y-x)$.

If $y=\tan \left(3 \tan ^{-1} x\right)$,then $\left(1-3 x^2\right) \frac{d^2 y}{d x^2}-12 x \frac{d y}{d x}=$

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