Explanation

(a)
Since, $y = x^{\sin x}$+$\sqrt{x}$
Let $y_{1} = x^{\sin x}\ \ \ and\ \ \ \ y_{2} = \sqrt{x}$
Now, $y_{1} = {x\ }^{\sin x}\ \ \ \ \ \Rightarrow \ \ \ \ \ \log{y_{1} = \sin{x\log x}}$
On differentiating w.r.t. $x,$ we get
$\frac{1}{y_{1}}.\frac{dy_{1}}{dx} = \cos{x\log{\ x + \frac{1}{x}\sin x}}$
$\Rightarrow \ \ \ \frac{dy_{1}}{dx} = x^{\sin x}\lbrack\cos{x\log{x + \frac{1}{x}\sin{x\rbrack}}}$
$\Rightarrow \ \left( \frac{dy_{1}\ }{dx} \right)_{x = \frac{\pi}{2}} = \left( \frac{\pi}{2} \right)^{\sin\frac{\pi}{2}}\ \left\lbrack \cos\frac{\pi}{2}\log{\frac{\pi}{2} + \frac{2}{\pi}\sin\frac{\pi}{2}} \right\rbrack$
$= \frac{\pi}{2} \times \frac{2}{\pi} = 1$
Now, $y_{2} = \sqrt{x}\ \ \ \Rightarrow \ \ \ \ \frac{dy_{2}}{dx} = \frac{1}{2\sqrt{x}}$
$\Rightarrow \left( \frac{dy_{2}}{dx} \right)_{x = \frac{\pi}{2}} = \frac{1}{2\sqrt{\frac{\pi}{2}}} = \sqrt{\frac{1}{2\pi}}\ \ \ \ \$
Since, $y = y_{1} + y_{2}$
$\therefore\ \ At\ \ \ \ x = \frac{\pi}{2},\ \ \ \ \ \frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx}\ \ \ \ \Rightarrow \ \ \ \frac{dy}{dx} = 1 + \frac{1}{\sqrt{2\pi}}$