Question

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

• A. 6 sin log (5)
• B. 5 sin log (6)
• C. 12 sin log (5)
• D. 5 sin log (12)
• E. 12 sin log (6)

Answer 1

Answer: (C)

$\frac{dy}{dx}=f\left(\frac{2x+3}{3-2x}\right)\frac{d}{dx}\left(\frac{2x+3}{3-2x}\right)$
$=\sin \log \left(\frac{2x+3}{3-2x}\right)\left(\frac{(3-2x)(2)-(2x+3)(-2)}{{(3-2x)}^2}\right)$ $=\frac{12}{{(3-2x)}^2}\sin \left\{\log \left(\frac{2x+3}{3-2x}\right)\right\}$
$\therefore$ ${\left(\frac{dy}{dx}\right)}_{x=1}=\frac{12}{{(3-2)}^2}.\sin \log 5$
= 12 sin log 5