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