Question

If $f^{\prime \prime }\left(x\right)=\frac{\cos \left(\log x\right)}{x},f^{\prime }\left(1\right)=0$ and $y=f\left(\frac{2x+3}{3-2x}\right)$ then $\frac{dy}{dx}$ is equal to

• A. $\left(\sin \left(\log x\right)\right)\frac{1}{\cos x}$
• 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. None of these

Answer 1

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

Given

$f^{\prime \prime }\left(x\right)=\frac{\cos \left(\log x\right)}{x}$, $f^{\prime }\left(1\right)=0$, & $y=f\left(\frac{2x+3}{3-2x}\right)$

Solution

$f^{\prime \prime }\left(x\right)=\cos \left(\log x\right)\times \frac{1}{x}$

$\Rightarrow \frac{d}{dx}\left[f^{\prime }\right(x\left)\right]=\frac{1}{x}\cos \left(\log x\right)$

$\Rightarrow \int d\left[f^{\prime }\right(x\left)\right]=\int \cos \left(\log x\right)\times \frac{1}{x}\times dx$

Let $\log x=t$

$\Rightarrow \frac{1}{x}=\frac{dt}{dx}$

$\Rightarrow dx=xdt$

$\Rightarrow f^{\prime }\left(x\right)=\int \cos \left(t\right)\times dt$

$\Rightarrow f^{\prime }\left(x\right)=\sin t+c$

$\Rightarrow f^{\prime }\left(x\right)=\sin \left(\log x\right)+c$

$\because f^{\prime }\left(1\right)=0$

$\therefore c=0$

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

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

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

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

$f^{\prime }\left(\frac{2x+3}{3-2x}\right)\times \frac{6-4x+4x+6}{(2x-3{)}^2}$

$=f^{\prime }\left(\frac{2x+3}{3-2x}\right)\times \frac{12}{(2x-3{)}^2}$

$=f^{\prime }\left(\frac{2x+3}{3-2x}\right)\times \frac{12}{(3-2x{)}^2}$

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

option c