If y = frac{tan x cos^{-1} x}{sqrt{1-x^2}}, t | vedclass

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(B) Given $y = \frac{	an x \cos^{-1} x}{\sqrt{1-x^2}}$.Using the product rule for differentiation: $\frac{dy}{dx} = \frac{d}{dx} [(	an x) \cdot (\co

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$\frac{\pi}{2}$

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(B) Given $y = \frac{	an x \cos^{-1} x}{\sqrt{1-x^2}}$.Using the product rule for differentiation: $\frac{dy}{dx} = \frac{d}{dx} [(	an x) \cdot (\cos^{-1} x) \cdot (1-x^2)^{-1/2}]$.Let $u = 	an x$, $v = \cos^{-1} x$, and $w = (1-x^2)^{-1/2}$.Then $\frac{dy}{dx} = u'vw + uv'w + uvw'$.$u' = \sec^2 x$, $v' = -\frac{1}{\sqrt{1-x^2}}$, $w' = -\frac{1}{2}(1-x^2)^{-3/2}(-2x) = \frac{x}{(1-x^2)^{3/2}}$.At $x = 0$:$u = 	an(0) = 0$, $u' = \sec^2(0) = 1$.$v = \cos^{-1}(0) = \frac{\pi}{2}$, $v' = -\frac{1}{\sqrt{1-0}} = -1$.$w = (1-0)^{-1/2} = 1$, $w' = \frac{0}{(1-0)^{3/2}} = 0$.Substituting these values:$\frac{dy}{dx} = (1)(\frac{\pi}{2})(1) + (0)(-1)(1) + (0)(\frac{\pi}{2})(0) = \frac{\pi}{2} + 0 + 0 = \frac{\pi}{2}$.

If $y = \frac{\tan x \cos^{-1} x}{\sqrt{1-x^2}}$, then the value of $\frac{dy}{dx}$ when $x = 0$ is:

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