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

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(C) Given $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$.Using the chain rule,we differentiate $y$ with respect to $x$:$\frac{dy

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$\frac{12}{(3 - 2x)^2} \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right]$

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(C) Given $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x + 3}{3 - 2x}\right)$.Using the chain rule,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = f'\left(\frac{2x + 3}{3 - 2x}\right) \cdot \frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right)$.First,calculate the derivative of the inner function using the quotient rule:$\frac{d}{dx}\left(\frac{2x + 3}{3 - 2x}\right) = \frac{(3 - 2x)(2) - (2x + 3)(-2)}{(3 - 2x)^2} = \frac{6 - 4x + 4x + 6}{(3 - 2x)^2} = \frac{12}{(3 - 2x)^2}$.Now,substitute this back into the expression for $\frac{dy}{dx}$:$\frac{dy}{dx} = \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right] \cdot \frac{12}{(3 - 2x)^2}$.Thus,$\frac{dy}{dx} = \frac{12}{(3 - 2x)^2} \sin\left[\log\left(\frac{2x + 3}{3 - 2x}\right)\right]$.

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

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