Explanation

(c)
Let $y = \sqrt{x^{2} + 16}\ and\mathcal{\ z =}\frac{x}{x - 1}$
On differentiating w.r.t., $x$, we get
$\frac{dy}{dx} = \frac{1}{2}\left( x^{2} + 16 \right)^{- 1/2}(2x)$
and $\frac{d\mathcal{z}}{dx} = \frac{x - 1 - x}{(x - 1)^{2}} = \frac{- 1}{(x - 1)^{2}}$
$\therefore\frac{dy}{d\mathcal{z}} = \frac{dy/dx}{d\mathcal{z/}dx} = \frac{- x}{\sqrt{x^{2} + 16}}.\frac{1}{\frac{1}{(x - 1)^{2}}}$
$\left( \frac{dy}{d\mathcal{z}} \right)_{x = 3} = \frac{- 3(2)^{2}}{\sqrt{25}} = \frac{- 12}{5}$