Let f (x) = (px + q) $\left(\frac{r}{x}+s\right)$ ... (i) Differentiating (i) with respect to x, we get $\frac{d}{dx}$ (f (x)) = (px + q)' $\left(\frac{r}{x}+s\right)$ + (px + q) $\left(\frac{r}{x}+s\right)'$ = (p + 0) $\left(\frac{r}{x}+s\right)$ + (px + q) $\left(\frac{x r' - r x'}{x^2}+0\right)$ = p $\left(\frac{r}{x}+s\right)$ + (px + q) $\left(\frac{0-r}{x^2}\right)$ = p $\left(\frac{r}{x}+s\right)$ - $\frac{(px+q)r}{x^2}$ = $\frac{pr}{x} + ps - \frac{pr}{x} - \frac{qr}{x^2}$ = ps - $\frac{qr}{x^2}$