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
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
-
$\sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$
-
$\frac{12}{\left(3-2 x^2\right)}$
-
$\frac{12}{\left(3-2 x^2\right)} \sin \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$
-
$\frac{12}{\left(3-2 x^2\right.} \cos \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$
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}
$
Asked in: JEE Main 2012 (12 May Online)
Practice more Differentiation questions on Aicharya