Explanation

(a)
$x^{2x} - 2x^{x}\cot{y - 1 = 0}$ …(i)
At $x = 1,$
$1 - 2\cot{y - 1 = 0}$
$\Rightarrow \ \cot{y = 0 \Rightarrow y = \frac{\pi}{2}}$
On differentiating Eq. (i), w.r.t.$'x'$, we get
$2x^{2x}(1 + \log{x) - 2\left\lbrack x^{x}\left( - {cosec}^{2}\ y \right)\frac{dy}{dx} + \cot{y\ x^{x}(1 + \log{x)}} \right\rbrack} = 0$
At $\left( 1,\frac{\pi}{2} \right),$
$2(1 + \log{1) - 2\left( 1( - 1)\left( \frac{dy}{dx} \right)_{\left( 1,\frac{\pi}{2} \right)} + 0 \right)\ } = 0$
$\Rightarrow \$2+2$\left( \frac{dy}{dx} \right)_{(1,\frac{\pi}{2})} = 0 \Rightarrow \left( \frac{dy}{dx} \right)_{\left( 1,\frac{\pi}{2} \right)} = \ - 1$