Question

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

• A. $\frac{\sin \left(\log x\right)}{x\log x}$
• B. $\left(\frac{12}{(3-2x{)}^2}\right)\sin \log \frac{2x+3}{3-2x}$
• C. $\frac{\cos \left(\log x\right)}{x\log x}$
• D. none of these

Answer 1

The correct option is B

$\left(\frac{12}{(3-2x{)}^2}\right)\sin \log \frac{2x+3}{3-2x}$

Explanation for the correct option.

Given that, $y=f\left(\frac{(2x+3)}{(3-2x)}\right)$ and $f^{\text{'}}\left(x\right)=\sin \left(\log x\right)$

Differentiate $y$ with respect to $x$using chain rule.

$\frac{dy}{dx}={\mathrm{f}}^{\text{'}}\left(\frac{(2\mathrm{x}+3)}{(3-2\mathrm{x})}\right)\left(\frac{d{\displaystyle \frac{(2\mathrm{x}+3)}{(3-2\mathrm{x})}}}{dx}\right)$

Use quotient rule i.e; $\frac{d\left({\displaystyle \frac{u}{v}}\right)}{dx}=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}$.

$\begin{array}{rcl}\frac{dy}{dx}& =& {\mathrm{f}}^{\text{'}}\left(\frac{(2\mathrm{x}+3)}{(3-2\mathrm{x})}\right)\times \frac{[{(3-2\mathrm{x})\times 2}^{}-(2\mathrm{x}+3)\times -2]}{(3-2\mathrm{x}{)}^2}\\ & =& \sin \log \frac{2x+3}{3-2x}\times \frac{12}{{\left(3-2x\right)}^2}\end{array}$

Hence, option B is correct.