Question

If $f\left(x\right)=\sin \left(\log x\right)$ and $y=f\left(\frac{2x+3}{3-2x}\right),$ then $\frac{dy}{dx}$ equals

• A. $\sin \left[\log \left(\frac{(2x+3)}{3-2x}\right)\right]$
• B. $\frac{12}{(3-2x^2)}$
• C. $\frac{12}{(3-2x^2)}\sin \left[\log \left(\frac{2x+3}{3-2x}\right)\right]$
• D. $\frac{12}{(9-16x^2)}\cos \left[\log \left(\frac{2x+3}{3-2x}\right)\right]$

Answer 1

The correct option is C $\frac{12}{(3-2x^2)}\sin \left[\log \left(\frac{2x+3}{3-2x}\right)\right]$