If $5 f(x)+3 f\left(\frac{1}{x}\right)=x+2$ and $y=x f(x)$, then $\frac{d y}{d x}$ at x = 1 is equal to

$\frac{7}{8}$

We have,

$5 f(x)+3 f\left(\frac{1}{x}\right)=x+2$  for all  $x \neq 0$              ......(i)

Replacing x by $\frac{1}{x}$, we get

$5 f\left(\frac{1}{x}\right)+3 f(x)=\frac{1}{x}+2$                ......(ii)

Eliminating $f\left(\frac{1}{x}\right)$ between (i) and (ii), we get

$16 f(x)=5 x-\frac{3}{x}+4$

$\Rightarrow f(x)=\frac{5}{16} x-\frac{3}{16 x}+\frac{1}{4} \Rightarrow f'(x)=\frac{5}{16}+\frac{3}{16 x^2}$

Now,

$y=x f(x)$

$\Rightarrow \frac{d y}{d x}=f(x)+x f'(x)$

$\Rightarrow \left(\frac{d y}{d x}\right)_{x=1}=f(1)+f'(1)=\frac{5}{16}-\frac{3}{16}+\frac{1}{4}+\frac{5}{16}+\frac{3}{16}=\frac{7}{8}$