If y = tan^{-1}left(frac{12x - 64x^3}{1 - 48x | vedclass

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(C) Let $4x = 	an 	heta$,then $	heta = 	an^{-1}(4x)$.Substituting this into the expression for $y$:$y = 	an^{-1}\left(\frac{3(4x) - (4x)^3}{1 - 3

Vedclass · Accepted Answer

$\frac{12}{1 + 16x^2}$

Vedclass · Answer

(C) Let $4x = 	an 	heta$,then $	heta = 	an^{-1}(4x)$.Substituting this into the expression for $y$:$y = 	an^{-1}\left(\frac{3(4x) - (4x)^3}{1 - 3(4x)^2}ight)$Using the identity $	an(3	heta) = \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta}$,we get:$y = 	an^{-1}(	an(3	heta)) = 3	heta$Substituting $	heta = 	an^{-1}(4x)$ back:$y = 3 	an^{-1}(4x)$Differentiating with respect to $x$:$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + (4x)^2} \cdot \frac{d}{dx}(4x)$$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + 16x^2} \cdot 4 = \frac{12}{1 + 16x^2}$

If $y = \tan^{-1}\left(\frac{12x - 64x^3}{1 - 48x^2}\right)$,then $\frac{dy}{dx} = $

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