If f^{prime}(x)=sin (log x) and y=fleft(frac{ | vedclass

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(C) Given $y=f\left(\frac{2 x+3}{3-2 x}\right)$.Applying the chain rule,we have $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \fra

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$12 \sin (\log 5)$

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(C) Given $y=f\left(\frac{2 x+3}{3-2 x}\right)$.Applying the chain rule,we have $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$.Using the quotient rule for $\frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$:$\frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right) = \frac{(3-2 x)(2) - (2 x+3)(-2)}{(3-2 x)^2} = \frac{6-4 x+4 x+6}{(3-2 x)^2} = \frac{12}{(3-2 x)^2}$.Since $f^{\prime}(x)=\sin (\log x)$,then $f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) = \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)$.Thus,$\frac{d y}{d x} = \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right) \cdot \frac{12}{(3-2 x)^2}$.At $x=1$,the argument of the function is $\frac{2(1)+3}{3-2(1)} = \frac{5}{1} = 5$.The derivative at $x=1$ is $\sin (\log 5) \cdot \frac{12}{(3-2)^2} = 12 \sin (\log 5)$.

If $f^{\prime}(x)=\sin (\log x)$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$,then $\frac{d y}{d x}$ at $x=1$ is

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