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If $f^{\prime}(x)=\sin (\log x)$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$, then $\frac{d y}{d x}$ equals
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Verified Answer
The correct answer is:
$\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$
$\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$
Let $f^{\prime}(x)=\sin [\log x]$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$
Now, $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$
$$
\begin{aligned}
& =\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \frac{[(6-4 x-)-4 x-6]}{\left(3-2 x^2\right)} \\
& =\frac{12}{\left(3-2 x^2\right.} \cdot \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]
\end{aligned}
$$
Now, $\frac{d y}{d x}=f^{\prime}\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$
$$
\begin{aligned}
& =\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \frac{[(6-4 x-)-4 x-6]}{\left(3-2 x^2\right)} \\
& =\frac{12}{\left(3-2 x^2\right.} \cdot \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]
\end{aligned}
$$
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