Find frac{dx}{dy} if y = tan^{-1}left(frac{3x | vedclass

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(D) Given $y = 	an^{-1}\left(\frac{3x - x^3}{1 - 3x^2}ight)$.Let $x = 	an 	heta$. Since $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$,we have $-

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$\frac{1}{3(1+x^2)}$

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(D) Given $y = 	an^{-1}\left(\frac{3x - x^3}{1 - 3x^2}ight)$.Let $x = 	an 	heta$. Since $-\frac{1}{\sqrt{3}} Then $y = 	an^{-1}\left(\frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta}ight) = 	an^{-1}(	an 3	heta) = 3	heta$.Since $x = 	an 	heta$,we have $	heta = 	an^{-1} x$.Thus,$y = 3	an^{-1} x$.Now,differentiate with respect to $y$:$\frac{dx}{dy} = \frac{d}{dy}(	an 	heta) = \sec^2 	heta \cdot \frac{d	heta}{dy}$.Since $y = 3	heta$,we have $	heta = \frac{y}{3}$,so $\frac{d	heta}{dy} = \frac{1}{3}$.$\frac{dx}{dy} = \sec^2 	heta \cdot \frac{1}{3} = \frac{1 + 	an^2 	heta}{3} = \frac{1 + x^2}{3}$.

Find $\frac{dx}{dy}$ if $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ for $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

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