Q46 / 90 · JEE Main 2012 (12 May Online) · Mathematics

If $f^{\prime}(x)=\sin (\log x)$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$, then $\frac{d y}{d x}$ equals

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

Solution

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

Hence, the correct answer is option: 'C'.

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