Question

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

Answer 1

Answer: 22

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