If f'(x) = sin(log x) and y = fleft(frac{2x + | vedclass

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(D) Let $t = \frac{2x + 3}{3 - 2x}$. Then $y = f(t)$.By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}$.Given $f'(x) = \sin(\log x)$,so $f'

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None of these

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(D) Let $t = \frac{2x + 3}{3 - 2x}$. Then $y = f(t)$.By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}$.Given $f'(x) = \sin(\log x)$,so $f'(t) = \sin(\log t) = \sin\left(\log \frac{2x + 3}{3 - 2x}\right)$.Now,calculate $\frac{dt}{dx}$ using the quotient rule:$\frac{dt}{dx} = \frac{(3 - 2x)(2) - (2x + 3)(-2)}{(3 - 2x)^2} = \frac{6 - 4x + 4x + 6}{(3 - 2x)^2} = \frac{12}{(3 - 2x)^2}$.Therefore,$\frac{dy}{dx} = \sin\left(\log \frac{2x + 3}{3 - 2x}\right) \cdot \frac{12}{(3 - 2x)^2}$.Comparing this with the given options,none of them match the result. Thus,the correct option is $(d)$.

If $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$,then $\frac{dy}{dx} = $

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