frac{d}{dx} left[ tan^{-1} left( frac{sqrt{x} | vedclass

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(D) Let $y = 	an^{-1} \left( \frac{\sqrt{x}(3 - x)}{1 - 3x} ight)$.Substitute $\sqrt{x} = 	an 	heta$,which implies $	heta = 	an^{-1}(\sqrt{x})$

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

Vedclass · Answer

(D) Let $y = 	an^{-1} \left( \frac{\sqrt{x}(3 - x)}{1 - 3x} ight)$.Substitute $\sqrt{x} = 	an 	heta$,which implies $	heta = 	an^{-1}(\sqrt{x})$.Then $x = 	an^2 	heta$,so the expression becomes $	an^{-1} \left( \frac{	an 	heta (3 - 	an^2 	heta)}{1 - 3	an^2 	heta} ight)$.Using the trigonometric identity $	an(3	heta) = \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta}$,the expression simplifies to $	an^{-1}(	an 3	heta) = 3	heta$.Substituting back,$y = 3	an^{-1}(\sqrt{x})$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(	an^{-1}(\sqrt{x}))$$= 3 \cdot \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x})$$= 3 \cdot \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}$$= \frac{3}{2(1 + x)\sqrt{x}}$.

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{x}(3 - x)}{1 - 3x} \right) \right] =$

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