Question

If $f(x)=\sin (\log x), y=f\left(\frac{2 x+3}{3-2 x}\right)$ and $\left(\frac{d y}{d x}\right)_{x=1}=\frac{a}{b} \cos (\log c)$, then find $a+b+c .$

Answer 1

$f(x)=\sin (\log x)$
$\Rightarrow f'(x)=\cos (\log x) \cdot \frac{1}{x}$ ...(i)
$y=f\left(\frac{2 x+3}{3-2 x}\right)$
$\Rightarrow \frac{d y}{d x}=f'\left(\frac{2 x+3}{3-2 x}\right) \cdot \frac{d}{d x}\left(\frac{2 x+3}{3-2 x}\right)$
$=f'\left(\frac{2 x+3}{3-2 x}\right) \cdot\left[\frac{12}{(3-2 x)^{2}}\right]$
$=\cos \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right] \cdot \frac{1}{\frac{2 x+3}{3-2 x}} \cdot \frac{12}{(3-2 x)^{2}}$... [From (i)]
$\Rightarrow \frac{d y}{d x}=\frac{12}{(2 x+3)(3-2 x)} \cos \left[\log \left(\frac{2 x+3}{3-2 x}\right)\right]$
$\Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=\frac{12}{5} \cos (\log 5)$
$\Rightarrow a=12, b=5, c=5$
$\Rightarrow a+ b+ c=12+5+5=22$