Question

If f(x) = sin (log x) 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. $\frac{12}{{\left(3-2x\right)}^2}$
• C. $\frac{12}{{\left(3-2x\right)}^2}sin\left[log\left(\frac{2x+3}{3-2x}\right)\right]$
• D. $\frac{12}{{\left(3-2x\right)}^2}cos\left[log\left(\frac{2x+3}{3-2x}\right)\right]$

Answer 1

Answer: (C)

Let $f^{\prime }\left(x\right)=sin\left[logx\right]$ and $y=f\left(\frac{2x+3}{3-2x}\right)$
Now, $\frac{dy}{dx}=f^{\prime }\left(\frac{2x+3}{3-2x}\right).\frac{d}{dx}\left(\frac{2x+3}{3-2x}\right)$
$=sin\left[log\left(\frac{2x+3}{3-2x}\right)\right]\frac{\left[\left(6-4x-\right)-4x\left(-6\right)\right]}{\left(3-2x^2\right)}$
$=\frac{12}{\left(3-2x^2\right)}sin\left[log\left(\frac{2x+3}{3-2x}\right)\right]$