Question

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

• A. $\sin \left(\log x\right).\frac{1}{x\log x}$
• B. $cos\left(log\left(\frac{2x+3}{3-2x}\right)\right)\left(\frac{12}{9-4x^2}\right)$
• C. $\sin \left[\log \left(\frac{2x+3}{3-2x}\right)\right]$
• D. $\sin \left(\log x\right)$

Answer 1

The correct option is B $cos\left(log\left(\frac{2x+3}{3-2x}\right)\right)\left(\frac{12}{9-4x^2}\right)$

given, $f\left(x\right)=sin\left(logx\right)$

$\Rightarrow y=sin(log(2x+3)-log(3-2x))$

$\Rightarrow \frac{dy}{dx}=cos(log(2x+3)-log(3-2x))\left[\frac{dy}{dx}\left(log\right(2x+3)-log(3-2x\left)\right)\right]$

$\Rightarrow \frac{dy}{dx}=cos\left(log\frac{2x+3}{3-2x}\right)(\frac{2}{2x+3}+\frac{2}{3-2x})$

$\Rightarrow \frac{dy}{dx}=cos\left(log\left(\frac{2x+3}{3-2x}\right)\right)\left(\frac{12}{9-4x^2}\right)$