Conundrum #2

We have $y = f(g(x))\ \ \ \ \ \ \ \text{with}\ \ g(x) = \frac{2x+3}{3-2x}.$ It is easy to see that $g(1) = 5$ . The derivative of $g$ is $g'(x) = \frac{2(3-2x)-(-2)(2x+3)}{(3-2x)^2};\ \ \ \ g'(1) = \frac{2\cdot 1 + 2 \cdot 5}{1^2} = 12.$ According to the chain rule, $\left.\frac{dy}{dx}\right|_{x=1} = \left.f'(g(x))\cdot g'(x)\right|_{x=1} = \sin(\log g(1))\cdot g'(1) = \sin(\log 5)\cdot 12.$ The correct answer is therefore $\boxed{\mathbf C}$ .

$\begin{aligned} y & = f \left(\frac {2x+3}{3-2x} \right) \\ \frac {dy}{dx}\bigg|_{x=1} & = f '\left(\frac {2x+3}{3-2x} \right)\cdot \frac d{dx}\left(\frac {2x+3}{3-2x} \right)\bigg|_{x=1} \\ & = \sin \left(\log \frac {2x+3}{3-2x} \right) \left(\frac {2(3-2x) + 2(2x+3)}{(3-2x)^2} \right)\bigg|_{x=1} \\ & = \sin \left(\log \frac 51 \right) \left(\frac {2 + 2(5)}{1^2} \right) \\ & = \boxed{12 \sin (\log 5)} \end{aligned}$