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

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Aicharya · Accepted Answer

$\begin{aligned} & y=\mathrm{f}\left(\frac{2 x+3}{3-2 x}ight) \ & 	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{f}^{\prime}\left(\frac{2 x+3}{3-2 x}ight) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 x+3}{3-2 x}ight) \ & =\mathrm{f}^{\prime}\left(\frac{2 x+3}{3-2 x}ight) \cdot \frac{(3-2 x) \cdot 2-(2 x+3)(-2)}{(3-2 x)^2} \ & =\sin \left[\log \left(\frac{2 x+3}{3-2 x}ight)ight] \cdot \frac{12}{(3-2 x)^2} \ & 	herefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{x=1}=\sin (\log 5) \cdot 12=12 \sin (\log 5) \ & \end{aligned}$

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

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