frac{d}{d x}left(sqrt{frac{1-tan x}{1+tan x}} | vedclass

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(D) Let $y = \sqrt{\frac{1-	an x}{1+	an x}}$. Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1-	an x}{1+	an x}}} \cdot \frac{d}{dx}\l

Vedclass · Accepted Answer

$\frac{-\sec ^2 x}{(1+	an x)^{3 / 2}(1-	an x)^{1 / 2}}$

Vedclass · Answer

(D) Let $y = \sqrt{\frac{1-	an x}{1+	an x}}$. Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1-	an x}{1+	an x}}} \cdot \frac{d}{dx}\left(\frac{1-	an x}{1+	an x}ight)$. Applying the quotient rule to the inner term: $\frac{d}{dx}\left(\frac{1-	an x}{1+	an x}ight) = \frac{(1+	an x)(-\sec^2 x) - (1-	an x)(\sec^2 x)}{(1+	an x)^2} = \frac{-\sec^2 x - 	an x \sec^2 x - \sec^2 x + 	an x \sec^2 x}{(1+	an x)^2} = \frac{-2\sec^2 x}{(1+	an x)^2}$. Now,substitute this back: $\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1+	an x}{1-	an x}} \cdot \left(\frac{-2\sec^2 x}{(1+	an x)^2}ight) = \frac{-\sec^2 x}{\sqrt{1-	an x} \cdot (1+	an x)^{3/2}}$.

$\frac{d}{d x}\left(\sqrt{\frac{1-\tan x}{1+\tan x}}\right) = $

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