Question

If $f^1\left(x\right)=sin\left(logx\right)andy=f\left(\frac{2x+3}{3-2x}\right),$ then $\frac{dy}{dx}$ equals

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

Answer 1

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

Given $f^{{}^{\prime }}\left(x\right)=\sin \log x$

and $y=f\left(\frac{2x+3}{3-2x}\right)$

$\frac{dy}{dx}=f^{\prime }\left(\frac{2x+3}{3-2x}\right)\ast \left(\frac{2(3-2x)-(-2)(2x+3)}{{(3-2x)}^2}\right)=\sin \log \left(\frac{2x+3}{3-2x}\right)\ast \frac{12}{{(3-2x)}^2}$

so answer is option C