If y = f left( frac{3 + 2x}{3 - 2x} right), w | vedclass

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(D) Given $y = 	an \left( \log \left( \frac{3 + 2x}{3 - 2x} ight) ight)$.Let $u = \log \left( \frac{3 + 2x}{3 - 2x} ight) = \log(3 + 2x) - \log

Vedclass · Accepted Answer

$12, 9, -4$

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(D) Given $y = 	an \left( \log \left( \frac{3 + 2x}{3 - 2x} ight) ight)$.Let $u = \log \left( \frac{3 + 2x}{3 - 2x} ight) = \log(3 + 2x) - \log(3 - 2x)$.Then $y = 	an(u)$, so $\frac{dy}{dx} = \sec^2(u) \cdot \frac{du}{dx}$.Calculate $\frac{du}{dx} = \frac{d}{dx} [\log(3 + 2x) - \log(3 - 2x)] = \frac{2}{3 + 2x} - \frac{-2}{3 - 2x} = \frac{2}{3 + 2x} + \frac{2}{3 - 2x}$.Simplify $\frac{du}{dx} = \frac{2(3 - 2x) + 2(3 + 2x)}{(3 + 2x)(3 - 2x)} = \frac{6 - 4x + 6 + 4x}{9 - 4x^2} = \frac{12}{9 - 4x^2}$.Thus, $\frac{dy}{dx} = \sec^2 \left( \log \left( \frac{3 + 2x}{3 - 2x} ight) ight) \cdot \left( \frac{12}{9 - 4x^2} ight)$.Comparing with $\frac{dy}{dx} = \left( \frac{A}{B + Cx^2} ight) \cdot \sec^2 \left( \log \left( \frac{3 + 2x}{3 - 2x} ight) ight)$, we get $A = 12$, $B = 9$, and $C = -4$.

If $y = f \left( \frac{3 + 2x}{3 - 2x} \right)$, where $f(x) = \tan(\log x)$ and $\frac{dy}{dx} = \left( \frac{A}{B + Cx^2} \right) \cdot \sec^2 \left( \log \left( \frac{3 + 2x}{3 - 2x} \right) \right)$, then the respective values of $A, B$ and $C$ are ......

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