Find the derivative with respect to $x$ 

(1) $y=\sin\sqrt{x^{2} +a^{2}}$

let $y=\sin u$

$\frac{dy}{du} =\cos u$

and $u=\sqrt{v} \ \ \frac{du}{dv} =\frac{1}{2\sqrt{v}}$

$v=x^{2} +a^{2} \ \ \frac{dv}{dx} =2x$

$\frac{dy}{dx} =\frac{dy}{du} \ \cdotp \frac{du}{dv} \cdotp \frac{dv}{dx}$

$=\cos u\cdotp \frac{1}{2\sqrt{v}} \cdotp 2x$

$=\frac{x\cos\sqrt{x^{2} +a^{2}}}{\sqrt{x^{2} +a^{2}}}$ Ans   

(2)  $y=\log_{2}\left(\sin x^{3}\right)$

$=\log_{e}\left(\sin x^{3}\right) \times \log_{2} e$

let $y=\log_{e} u\times \log_{2} e$

$\frac{dy}{du} =\log_{2} e\cdotp \frac{1}{u}$

$u=\sin v\ \ \ \ \ \frac{du}{dv} =cosv$

$v=x^{3} \ \ \ \ \frac{dv}{dx} =3x^{2}$

$\frac{dy}{dx} =\frac{dy}{du} \cdotp \ \frac{du}{dv} \cdotp \frac{dv}{dx}$

$=\log_{2} e\cdotp \frac{1}{u} \cdotp cosv\cdotp 3x^{2}$

$=3x^{2}\cot\left( x^{3}\right)\log_{2} e$ Ans

(3)    $y=e^{\sqrt{cosx}}$

let, $y=e^{u} \ \ \ \therefore y=e^{u} \ \ $

$\frac{dy}{du} =e^{u} \ $

$u=\sqrt{v} \ \ \ \ \frac{du}{dv} =\frac{1}{2\sqrt{v}}$

$v=\cos x\ \ \ \frac{dv}{dx} =-\sin x$

$\frac{dy}{dx} =\frac{dy}{du} \cdotp \ \frac{du}{dv} \cdotp \frac{dv}{dx}$

$=e^{u} \ \cdotp \frac{1}{2\sqrt{v}} \cdotp ( -\sin x)$

$=-\frac{e^{\sqrt{cosx}}\sin x}{2\sqrt{\cos x}}$ Ans

(4)     $y=\frac{1}{\sqrt{\log(\sec x)}}$

let, $y=\frac{1}{u} \ \ \ \ \ \frac{dy}{du} =-\frac{1}{u^{2}}$

$u=\sqrt{v} \ \ \ \ \ \ \ \frac{du}{dv} =\frac{1}{2\sqrt{v}}$

$v=\log(\sec x) \ \ \ \ \ \frac{dv}{dx} =\tan x$ 

$\frac{dy}{dx} =\frac{dy}{du} \cdotp \ \frac{du}{dv} \cdotp \frac{dv}{dx}$

$=-\frac{1}{u^{2}} \cdotp \frac{1}{2\sqrt{v}} \cdotp \tan x$

$=-\frac{\tan x}{2\log(\sec x)\sqrt{\log(\sec x)} \ }$

$=-\frac{1}{2}\tan x[\log(\sec x)]^{3/2}$ Ans

(5)    $y=\cos^{-1}\sqrt{2x-3}$

let  $y=\cos^{-1} u\ \ \ \ \ \ \frac{dy}{du} =-\frac{1}{\sqrt{1-u^{2}}}$

$u=\sqrt{v} \ \ \ \ \ \ \frac{du}{dv} =\frac{1}{2\sqrt{v}}$

$v=2x-3\ \ \ \frac{dv}{dx} =2$

$\frac{dy}{dx} =\frac{dy}{du} \cdotp \ \frac{du}{dv} \cdotp \frac{dv}{dx}$

$=-\frac{1}{\sqrt{1-u^{2}}} \cdotp \frac{1}{2\sqrt{v}} \cdotp 2$

$=\frac{1}{\sqrt{1-2x+3}} \cdotp \frac{1}{\sqrt{2x-3}}$

$=-\frac{1}{\sqrt{4-2x}\sqrt{2x-3}}$

$=-\frac{1}{\sqrt{2}} \cdotp \frac{1}{\sqrt{2-x}\sqrt{2x-3}}$ Ans

(6)         $y=x{^{x}}^{2} +a{^{x}}^{2}$

let,  $y=u+v\ \ \ \ \frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=x{^{x}}^{2}$

$\log u=x^{2}\log x$

$\frac{1}{u}\frac{du}{dx} =2x\log x+x$

$\frac{du}{dx} =x{^{x}}^{2}[ 2x\log x+x]$

$v=a{^{x}}^{2}$

$\frac{dv}{dx} =2xa{^{x}}^{2}\log_{e} a$

$\therefore \frac{dy}{dx} =x{^{x}}^{2}[ 2x\log x+x] +2xa{^{x}}^{2}\log_{e} a$ Ans

(7)     $y=(\sin x)^{\cos x} +e^{3x}$

let,   $\ \ y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=(\sin x)^{\cos x}$

$\log u=\cos x\log\sin x$

$\frac{1}{u}\frac{du}{dx} =-\sin x\log(\sin x) +\cos x\cdotp \cot x$

$\frac{du}{dx} =(\sin x)^{\cos x}[\cos x\cdotp \cot x-\sin x\log(\sin x)]$

$\frac{dv}{dx} =3e^{3x}$

$\therefore \frac{dy}{dx} =(\sin x)^{\cos x}[\cos x\cdotp \cot x-\sin x\log(\sin x)] +3e^{3x}$ Ans

(8)      $y=x^{x} +(\sin x)^{x}$

let,   $y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=x^{x}$

$\log u=x\log x$

$\frac{1}{u}\frac{du}{dx} =\log x+1$

$\frac{du}{dx} =x^{x}[\log x+1]$

$v=(\sin x)^{x}$

$\log v=x\log(\sin x)$

$\frac{1}{v}\frac{dv}{dx} =\log(\sin x) +x\cot x$

$\frac{dv}{dx} =(\sin x)^{x}[\log(\sin x) +x\cot x]$

$\therefore \frac{dy}{dx} =x^{x}[\log x+1] +(\sin x)^{x}[\log(\sin x) +x\cot x]$ Ans

(9)   $y=(\tan x)^{\cot x} +(\cot x)^{\tan x}$

let         $y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=(\tan x)^{\cot x}$

$\log u=\cot x\log(\tan x)$

$\Longrightarrow \frac{1}{u} \cdotp \frac{du}{dx} =-\mathrm{cosec}^{2} x\log(\tan x) +\frac{\sec^{2} x}{\tan^{2} x}$

$=(\tan x)^{\cot x}\left[ -\mathrm{cosec}^{2} x\log(\tan x) +\frac{\sec^{2} x}{\tan^{2} x}\right]$

$v=(\cot x)^{\tan x}$

$\log v=\tan x\log(\cot x)$

$\Longrightarrow \frac{1}{v} \cdotp \frac{dv}{dx} =\sec^{2} x\log(\cot x) -\frac{\mathrm{cosec} ^{2} x}{\cot^{2} x}$

$\frac{dv}{dx} =(\cot x)^{\tan x}\left[\sec^{2} x\log(\cot x) -\frac{\mathrm{cosec}^{2} x}{\cot^{2} x}\right]$

$\therefore \frac{dy}{dx} =(\tan x)^{\cot x}\left[ -\mathrm{cosec}^{2} x\log(\tan x) +\frac{\sec^{2} x}{\tan^{2} x}\right]$

        $+(\cot x)^{\tan x}\left[\sec^{2} x\log(\cot x) -\frac{\mathrm{cosec}^{2} x}{\cot^{2} x}\right]$ Ans

(10)     $y=(\sin x)^{\cos x} +(\cos x)^{\sin x}$

let       $y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=(\sin x)^{\cos x}$

$\log u=\cos x\log(\sin x)$

$\Longrightarrow \frac{1}{u} \cdotp \frac{du}{dx} =-\sin x\log(\sin x) +\cos x\cot x$

$\frac{du}{dx} =(\sin x)^{\cos x}[ -\sin x\log(\sin x) +\cos x\cot x]$

$v=(\cos x)^{\sin x}$

$\log v=\sin x\log(\cos x)$

$\Longrightarrow \frac{1}{v} \cdotp \frac{dv}{dx} =\cos x\log(\cos x) -\sec x\tan x$

$\frac{dv}{dx} =(\cos x)^{\sin x}[\cos x\log(\cos x) -\sec x\tan x]$

$\therefore \frac{dy}{dx} =(\sin x)^{\cos x}[ -\sin x\log(\sin x) +\cos x\cot x]$

                      $+(\cos x)^{\sin x}[\cos x\log(\cos x) -\sec x\tan x]$ Ans

(11)    $y=e^{\cos^{-1} x} +x^{\sqrt{x}}$

let                 $y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=e^{\cos^{-1} x}$

$\Longrightarrow \frac{du}{dx} =-\frac{e^{\cos^{-1} x}}{\sqrt{1-x^{2}}}$

$v=x^{\sqrt{x}}$

$\log v=\sqrt{x} \cdotp \log x$

$\Longrightarrow \frac{1}{v} \cdotp \frac{dv}{dx} =\frac{1}{2\sqrt{x}} \cdotp \log x+\frac{1}{\sqrt{x}}$

$\frac{dv}{dx} =x^{\sqrt{x}}\left[\frac{1}{2\sqrt{x}} \cdotp \log x+\frac{1}{\sqrt{x}}\right]$

$\therefore \frac{dy}{dx} =-\frac{e^{\cos^{-1} x}}{\sqrt{1-x^{2}}}$

             $+x^{\sqrt{x}}\left[\frac{1}{2\sqrt{x}} \cdotp \log x+\frac{1}{\sqrt{x}}\right]$ Ans

(12)   $y=(\sin x)^{\tan x} +(\cos x)^{\sec x}$

let     $y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=(\sin x)^{\tan x}$

$\log u=\tan x\log(\sin x)$

$\Rightarrow \frac{1}{u} \cdotp \frac{du}{dx} =\sec^{2} x\log(\sin x) +1$

$\frac{du}{dx} =(\sin x)^{\tan x}\left[\sec^{2} x\log(\sin x) +1\right]$

$v=(\cos x)^{\sec x}$

$\log v=\sec x\log(\cos x)$

$\Rightarrow \frac{1}{v} \cdotp \frac{dv}{dx} =\sec x\tan x\log(\cos x) -\sec x\tan x$

$\frac{dv}{dx} =(\cos x)^{\sec x}[\sec x\tan x\log(\cos x) -\sec x\tan x]$

$\therefore \frac{dy}{dx} =(\sin x)^{\tan x}\left[\sec^{2} x\log(\sin x) +1\right]$

            $+(\cos x)^{\sec x}[\sec x\tan x\log(\cos x) -\sec x\tan x]$ Ans

(13)     $y=x^{\sin x} +(\sin x)^{\cos x}$

let          $y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$

$u=x^{\sin x}$

$\log u=\sin x\log x$

$\Rightarrow \frac{1}{u} \cdotp \frac{du}{dx} =\cos x\log x+\frac{\sin x}{x}$

$\frac{du}{dx} =x^{\sin x}\left[\cos x\log x+\frac{\sin x}{x}\right]$

$v=(\sin x)^{\cos x}$

$\log v=\cos x\log(\sin x)$

$\Longrightarrow \frac{1}{v} \cdotp \frac{dv}{dx} =-\sin x\log(\sin x) +\cos x\cot x$

$\frac{dv}{dx} =(\sin x)^{\cos x}[ -\sin x\log(\sin x) +\cos x\cot x]$

$\therefore \frac{dy}{dx} =x^{\sin x}\left[\cos x\log x+\frac{\sin x}{x}\right]$

           $+(\sin x)^{\cos x}[ -\sin x\log(\sin x) +\cos x\cot x]$ Ans