Question

If $f^{\prime \prime }\left(x\right)=\frac{\cos \left(\log x\right)}{x}$, $f^{\prime }\left(1\right)=0$ and $y=f\left(\frac{2x+3}{3-2x}\right)$ then $\frac{dy}{dx}$ is equal to

• A. $\left(\sin \left(\log x\right)\right)\frac{1}{\cos x}$
• B. $\sin \left(\log \left(\frac{2x+3}{3-2x}\right)\right)$
• C. $\frac{12}{{(3-2x)}^2}\sin \left(\log \left(\frac{2x+3}{3-2x}\right)\right)$
• D. none of these

Answer 1

Answer: (C)

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

$f^{\prime \prime }\left(x\right)=\frac{\cos \left(\log x\right)}{x}=\frac{d}{dx}(\sin \left(\log x\right)+C)$
So $f^{\prime }\left(x\right)=\sin \left(\log x\right)+C$ but $f^{\prime }\left(1\right)=0$ so $C=0.$
Thus $f^{\prime }\left(x\right)=\sin \left(\log x\right)$.
Now
$\frac{dy}{dx}=f^{\prime }\left(\frac{2x+3}{3-2x}\right)\frac{d}{dx}\left(\frac{2x+3}{3-2x}\right)$

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