Explanation

(a)
Since, $x^{y}.y^{x} = 16$
$\therefore\log_{e}{x^{y} + \log_{e}{y^{x} = \log_{e}16}}$
$\Rightarrow y\log_{e}{x + x\log_{e}{y = 4\log_{e}2}}$
Now, on differentiating both sides w.r.t. $x,$ we get
$\frac{y}{x} + \log_{e}x\frac{dy}{dx} + \frac{x}{y}\frac{dy}{dx} + \log_{e}{y.1 = 0}$
$\Rightarrow \frac{dy}{dx} = - \frac{\left( \log_{e}{y +}\frac{y}{x} \right)}{\left( \log_{e}x + \frac{x}{y} \right)}$
$\therefore\left. \ \frac{dy}{dx} \right|_{(2,2)} = - \frac{(\log_{e}{2 + 1)}}{(\log_{e}{2 + 1)}} = - 1$