Differentiation

$ \begin{aligned} &\text { Let } f^{\prime}(1)=a\\ &\begin{aligned} & f^{\prime}(2)=b \\ & f^{\prime}(3)=c \\ & f(x)=x^3+a x^2+b x+c \\ & f^{\prime}(x)=3 x^2+2 a x+b \\ & f^{\prime}(x)=a=3+2 a+b \Rightarrow a+b=3 \ldots . . (1)\\ & f^{\prime \prime}(x)=6 x+2 a \\ & f^{\prime \prime}(2)=12+2 a=\frac{b}{2} \Rightarrow 4 a-b=-24 \\ & f^{\prime \prime}(x)=6 \\ & f^{\prime \prime}(3)=c=6 \\ & \Rightarrow a=\frac{-27}{5}, b=\frac{12}{5} \\ & f^{\prime}(5)=75+10 a+b \\ & =75-54+\frac{12}{5}=21+\frac{12}{5}=\frac{117}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} & y=\tan ^{-1}\left(\frac{\frac{3}{4}-\tan x}{1+\frac{3}{4} \tan x}\right)+2 \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right) \\ & \text { In the first term, divide numerator and denominator by } \cos x: \\ & \frac{3\cos x-4\sin x}{4\cos x+3\sin x}=\frac{3-4\tan x}{4+3\tan x} =\frac{\frac{3}{4}-\tan x}{1+\frac{3}{4}\tan x} \\ & \text{Now use } \tan^{-1}\left(\frac{a-b}{1+ab}\right)=\tan^{-1}a-\tan^{-1}b: \\ & \tan ^{-1}\left(\frac{\frac{3}{4}-\tan x}{1+\frac{3}{4} \tan x}\right)=\tan^{-1}\left(\frac{3}{4}\right)-\tan^{-1}(\tan x) \\ & \text { For } \quad 2 \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right) \\ & \text { put } \quad x=\sin \theta \quad \Rightarrow \quad \sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta \\ & \therefore 2 \tan ^{-1}\left(\frac{\sin \theta}{1+\cos \theta}\right)=2\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\theta=\sin ^{-1} x \end{aligned} $

Hence

$ \begin{aligned} & y=\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}(\tan x)+\sin ^{-1} x \\ & \text{Differentiate term by term: } \tan^{-1}\left(\frac{3}{4}\right) \text{ is constant, so its derivative is } 0. \\ & \frac{d}{dx}\left[-\tan^{-1}(\tan x)\right]=-1 \quad (\text{taking } \tan^{-1}(\tan x)=x \text{ in the principal range}) \\ & \frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}} \\ & \therefore \frac{d y}{d x}=-1+\frac{1}{\sqrt{1-x^2}} \\ & x=\frac{\sqrt{3}}{2} \Rightarrow \sqrt{1-x^2}=\sqrt{1-\frac{3}{4}}=\frac{1}{2} \\ & \frac{d y}{d x}=-1+\frac{1}{1/2}=-1+2=1 \end{aligned} $

To solve for $ y(x) $, we start with the determinant of the given matrix:

$ \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix} $

Using the formula for a $3 \times 3$ determinant, we expand along the first row:

$ f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} $

Calculating the smaller $2 \times 2$ determinants:

$ \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1 $

$ \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0 $

$ \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1 $

Substituting these into the determinant calculation gives:

$ f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1) $

$ f(x) = \sin x - (\sin x + \cos x + 1) $

$ f(x) = \sin x - \sin x - \cos x - 1 $

$ f(x) = -\cos x - 1 $

Now, calculate the derivatives:

First derivative $ f'(x) $:

$ f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x $

Second derivative $ \frac{d^2 f}{dx^2} $:

$ \frac{d^2 f}{dx^2} = \frac{d}{dx}(\sin x) = \cos x $

Finally, compute $ \frac{d^2 y}{dx^2} + y $:

$ \frac{d^2 f}{dx^2} + f(x) = \cos x - \cos x - 1 = -1 $

Thus, $ \frac{d^2 y}{dx^2} + y $ is equal to $-1$.

$\begin{aligned} & \sin (x-y) f(2 x+2 y)=f(2 x-2 y) \sin (x+y) \\ & \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)}=k(\text { say }) \\ & f(2 x+2 y)=k \sin (x+y) \\ & f(2 x)=5 \sin x \quad(\because y=0) \\ & f(x)=k \sin \frac{x}{2} \\ & f^{\prime}(x)=\frac{k}{2} \cos \frac{x}{2} \\ & f^{\prime}(0)=\frac{1}{2} \Rightarrow k=1 \\ & f(x)=\sin \frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2} \\ & f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2} \\ & 24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=-3 \end{aligned}$

$\begin{aligned} & x^2 f^{\prime}(x)-2 x f(x)=3 \\ & \left(\frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}\right)=\frac{3}{\left(x^2\right)^2} \\ & \Rightarrow \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4} \end{aligned}$

Integrating both sides

$\begin{aligned} & \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^2}=-\frac{1}{\mathrm{x}^3}+\mathrm{C} \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+\mathrm{Cx}^2 \\ & \text { put } \mathrm{x}=1 \\ & 4=-1+\mathrm{C} \Rightarrow \mathrm{C}=5 \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^2 \end{aligned}$

Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^2\right]$

$=39$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{d x}{d t}=1-\cos t \\ & \frac{d y}{d t}=\sin t \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{\sin t}{1-\cos t}=\frac{2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{2 \sin ^2\left(\frac{t}{2}\right)} \\ & \Rightarrow \quad \frac{d y}{d x}=\cot \left(\frac{t}{2}\right) \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=-\operatorname{cosec}^2\left(\frac{t}{2}\right) \times \frac{1}{2} \times \frac{d t}{d x} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{d^2 y}{d x^2}=-\frac{1}{2 \sin ^2\left(\frac{t}{2}\right)} \times \frac{1}{1-\cos t} \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{-1}{(1-\cos t)^2} \\ & \Rightarrow \quad \frac{-1}{(1-\cos k)^2}=-1 \\ & \Rightarrow \quad(1-\cos k)^2=1 \\ & \Rightarrow \quad 1-\cos k= \pm 1 \\ & \Rightarrow \quad \cos k=0 \Rightarrow k=\frac{\pi}{2} \\ & \Rightarrow\quad \lim _{t \rightarrow \frac{\pi}{2}} \frac{1-\cos t}{t-\sin t}=\frac{1-0}{\frac{\pi}{2}-1}=\frac{2}{\pi-2} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=\tan ^2\left(\cos ^{-1} \sqrt{\frac{1+x^2}{2}}\right) \\ & \text { Put } \frac{1+x^2}{2}=\cos ^2(2 \theta) \\ & \qquad x^2=2 \cos ^2(2 \theta)-1=\cos 4 \theta \\ & \text { Also, } 0 \leq \frac{1+x^2}{2} \leq 1 \\ & \quad 0 \leq 1+x^2 \leq 2 \\ & \Rightarrow \quad x^2 \leq 1 \Rightarrow-1 \leq x \leq 1 \\ & \Rightarrow \quad-1 \leq \cos 4 \theta \leq 1 \end{aligned} \end{aligned} $

$ \Rightarrow \quad 0 \leq 4 \theta \leq \pi \Rightarrow 0 \leq \theta \leq \frac{\pi}{4} $

Differentiating with resped to $x$

$ \begin{array}{ll} \Rightarrow & 2 x \frac{d x}{d \theta}=-4 \sin 4 \theta \\ \Rightarrow & \frac{d x}{d \theta}=\frac{-2 \sin 4 \theta}{\sqrt{\cos 4 \theta}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

and $\quad y=\tan ^2 \cos ^{-1}(\cos 2 \theta)=\tan ^2(2\theta)$

$ \begin{aligned} \Rightarrow \quad \frac{d y}{d \theta} & =2 \tan (2 \theta) \times \sec ^2(2 \theta) \times 2 \ldots \text { (i) } \\ \Rightarrow \quad \frac{d y}{d \theta} & =\frac{4 \sin (2 \theta)}{\cos ^3(2 \theta)} \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{4 \sin (2 \theta) \times \sqrt{\cos 4 \theta}}{\cos ^3(2 \theta)(-4) \sin 2 \theta \cdot \cos 2 \theta} \\ & =-\frac{\sqrt{\cos 4 \theta}}{\cos ^4(2 \theta)} \\ & =\frac{-\sqrt{x^2}}{\left(\frac{1+\cos 4 \theta}{2}\right)^2}=\frac{-4 x}{\left(1+x^2\right)^2} \end{aligned} $

$ \begin{aligned} y= & (x)^{\ln x}+(\ln x)^x \\ y= & e^{\ln x \ln x}+e^{x \ln (\ln x)} \\ \frac{d y}{d x}= & (x)^{\ln x}\left(\frac{2 \ln x}{x}\right) +(\ln x)^x\left(\ln (\ln x)+\frac{x}{\ln x} \times \frac{1}{x}\right) \end{aligned} $

$ \begin{aligned} \left(\frac{d y}{d x}\right)_{\text {at } x=0} & =(e)\left(\frac{2}{e}\right)+1^e\left(0+\frac{e}{1} \times \frac{1}{e}\right) \\ & =2+1=3 \end{aligned} $

$ \begin{aligned} &\begin{aligned} y & =\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)+\sqrt{\log \left(x^2+1\right)}}} \ldots \infty \\ & \Rightarrow y=\sqrt{\log \left(x^2+1\right)+y} \\ & \Rightarrow y^2=\log \left(x^2+1\right)+y \\ & \Rightarrow y^2-y=\log \left(x^2+1\right) \end{aligned}\\ &\text { Differentiating wrt } x \text {, we get }\\ &\begin{aligned} & (2 y-1) \frac{d y}{d x}=\frac{2 x}{x^2+1} \\ & \frac{d y}{d x}=\frac{2 x}{(2 y-1)\left(x^2+1\right)} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } x=\sqrt{1-\tan y} \\ & \tan y=1-x^2 \end{aligned} $

Differentiating w.r.t $x$, we get

$ \begin{gathered} \sec ^2 y \frac{d y}{d x}=-2 x \\ \Rightarrow \quad \frac{d y}{d x}=\frac{-2 x}{\sec ^2 y}=\frac{-2 x}{1+\left(1-x^2\right)^2} \\ =\frac{-2 x}{x^4-2 x^2+2} \end{gathered} $

$ \begin{aligned} &\\ &\begin{aligned} x & =\sin 2 \theta \cos 3 \theta, y=\sin 3 \theta \cos 2 \theta \\ \Rightarrow 2 x & =2 \sin 2 \theta \cos 3 \theta, 2 y=2 \sin 3 \theta \cdot \cos 2 \theta \\ \Rightarrow 2 x & =\sin 5 \theta-\sin \theta, 2 y=\sin 5 \theta+\sin \theta \\ \frac{d y}{d x} & =\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{5 \cos 5 \theta+\cos \theta}{5 \cos 5 \theta-\cos \theta} \\ & =\frac{4 \cos 5 \theta+\cos 5 \theta+\cos \theta}{4 \cos 5 \theta+\cos 5 \theta-\cos \theta} \\ & =\frac{2 \cos 5 \theta+\cos 3 \theta \cdot \cos 2 \theta}{2 \cos 5 \theta-\sin 3 \theta \cdot \sin 2 \theta} \end{aligned} \end{aligned} $

$ \text { } 3^x y^x=x^{3 y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

At $x=1 \Rightarrow 3 y=1 \Rightarrow y=\frac{1}{3}$

Taking $\log$ both sides in Eq. (i), we get

$ \begin{aligned} & \Rightarrow \log 3^x+\log y^x=\log x^{3 y} \\ & \Rightarrow x \log 3+x \log y=3 y \log x \end{aligned} $

Differential w.r.t. $x$, we get

$ \log 3+\frac{x}{y} \frac{d y}{d x}+\log y=\frac{3 y}{x}+3 \log x \cdot \frac{d y}{d x} $

Put $x=1, y=\frac{1}{3}$.

$ \begin{aligned} & \log 3+3 \frac{d y}{d x}+\log \frac{1}{3}=3 \times \frac{1}{3}+3 \log 1 \frac{d y}{d x} \\ & \Rightarrow \quad 3 \frac{d y}{d x}=1 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{1}{3} \end{aligned} $

$ \begin{aligned} &\\ &\text { } \begin{aligned} y & =\left(1-x^2\right) \tanh ^{-1} x \\ \frac{d y}{d x} & =\left(1-x^2\right) \cdot \frac{1}{1-x^2}+\tanh ^{-1} x(-2 x) \\ & =1-2 x \tanh ^{-1}(x) \\ \frac{d^2 y}{d x^2} & =-2 \tanh ^{-1} x-\frac{2 x}{1-x^2} \\ & =-2 \frac{\left(1-x^2\right) \tanh ^{-1} x+x}{1-x^2} \\ & =-2 \frac{(y+x)}{1-x^2} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} f(x) & =\log _{\left(x^2-2 x+1\right)} x^2-3 x+2 \\ & =\log _{(x-1)^2}(x-2)(x-1) \\ & =\frac{1}{2} \log _{(x-1)}(x-2)(x-1) \\ \Rightarrow f(x) & =\frac{1}{2}\left(1+\frac{\log (x-2)}{\log (x-1)}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow f^{\prime}(x)=\frac{1}{2}\binom{\log (x-1) \cdot \frac{1}{x-2}}{\frac{-\log (x-2) \frac{1}{(x-1)}}{(\log (x-1))^2}} \\ & \therefore f^{\prime}(3)=\frac{\frac{1}{2}\left(\log 2-\frac{\log 1}{2}\right)}{(\log 2)^2} \\ & =\frac{1}{2 \log 2}=\frac{1}{\log 4}=\log _4 e \end{aligned} $

Given, $\frac{d}{d x}\left\{\left(\frac{x-1}{x-\sqrt{x}}\right) e^{2 x+1}\right\}$

$ =\frac{x-1}{x-\sqrt{x}} e^{2 x+1} f(x) $

Here, $\frac{x-1}{x-\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}$

$ \begin{gathered} =\frac{\sqrt{x}+1}{\sqrt{x}}=1+\frac{1}{\sqrt{x}} \\ \therefore \frac{d}{d x}\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1}=\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1} f(x) \end{gathered} $

Now, $\frac{d}{d x}\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1}$

$ \begin{aligned} & =\frac{-1}{2 x \sqrt{x}} e^{2 x+1}+\left(1+\frac{1}{\sqrt{x}}\right) e^{2 x+1}(2) \\ & =e^{2 x+1}\left[\frac{-1}{2 x \sqrt{x}}+2\left(1+\frac{1}{\sqrt{x}}\right)\right] \end{aligned} $

$ \begin{aligned} \therefore f(x) & =\frac{\frac{-1}{2 x \sqrt{x}}+2\left(1+\frac{1}{\sqrt{x}}\right)}{1+\frac{1}{\sqrt{x}}} \\ & =\frac{-1+4 x \sqrt{x}+4 x}{2 x \sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}+1} \\ \Rightarrow f(x) & =\frac{-1+4 x \sqrt{x}+4 x}{2 x(\sqrt{x}+1)} \\ \therefore f(4) & =\frac{-1+4(4)(2+4(4)}{2(4)(2+1)} \\ & =\frac{-1+32+16}{24}=\frac{47}{24} \end{aligned} $

Given, $y=f(\cos h x)$ and

$ f^{\prime}(x)=\log \left(x+\sqrt{x^2-1}\right) $

Let $u=\cos h x$, so $y=f(u)$

$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =f^{\prime}(u) \cdot \frac{d}{d x}(\cos h x) \\ & =f^{\prime}(\cos h x) \cdot \sin h x \end{aligned} $

Since, $f^{\prime}(x)=\log \left(x+\sqrt{x^2-1}\right)$

$ \begin{aligned} \therefore f^{\prime}(\cosh x) & =\log \left(\cosh x+\sqrt{\cosh h^2 x-1}\right) \\ & =\log (\cosh x+|\sinh x|) \end{aligned} $

Since, $x>0, \sin h>0$

So, $f^{\prime}(\cos h x)=\log (\cos h x+\sin h x)$

$ \Rightarrow \log e^x=x \quad\left[\because \cosh x+\sinh x=e^x\right] $

So, $\frac{d y}{d x}=x \cdot \sin h x$

$ \begin{aligned} & \frac{d^2 y}{d x^2}=x \cdot \frac{d}{d x}(\sin h x)+\sin h x \cdot \frac{d}{d x}(x) \\ \Rightarrow & x \cdot \cos h x+\sin h x \cdot 1 \\ \Rightarrow & x \cos h x+\sin h x \end{aligned} $

Given equation

$ \left(x^2-3 x+2\right) e^{\frac{y}{x-1}}=x+2 $

Differentiate both sides w.r.t., we get

$ \begin{array}{r} \frac{d}{d x}\left(\left(x^2-3 x+2\right) \cdot e^{\frac{y}{x-1}}\right)=\frac{d}{d x}(x+2) \\ \Rightarrow(2 x-3) e^{\frac{y}{x-1}}+\left(x^2-3 x+2\right) \cdot e^{\frac{y}{x-1}} \\ \cdot \frac{d}{d x}\left(\frac{y}{x-1}\right)=1 \\ \Rightarrow(2 x-3) e^{\frac{y}{x-1}}+\left(x^2-3 x+2\right) \cdot e^{\frac{y}{x-1}} \\ {\left[\frac{\frac{d y}{d x}(x-1)-y}{(x-1)^2}\right]=1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Now, substitute $x=0$ into the given equation

$ \begin{array}{ll} & \left(0^2-3(0)+2\right) e^{\frac{y}{(0-1)}}=0+2 \\ \Rightarrow & 2 e^{-y}=2 \Rightarrow e^{-y}=1 \\ \Rightarrow & -y=\ln (1) \Rightarrow-y=0 \\ \Rightarrow & y=0 \end{array} $

Substitute $x=0$ and $y=0$ in Eq. (i), we get

$ \begin{aligned} & (2(0)-3) e^{\frac{0}{0-1}}+\left(0^2-3(0)+2\right) e^{\frac{0}{0-1}} \\ & \cdot\left[\frac{\frac{d y}{d x}(0-1)-0}{(0-1)^2}\right]=1 \\ & \Rightarrow-3 e^0+2 e^0\left(\frac{-d y}{d x}\right)=1 \\ & \Rightarrow \quad-3+2\left(\frac{-d y}{d x}\right)=1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad-2 \frac{d y}{d x}=1+3=4 \Rightarrow \frac{d y}{d x}=\frac{4}{-2}=-2 \\ & \therefore \quad\left(\frac{d y}{d x}\right)_{x=0}=-2 \end{aligned} $

Given $x=\frac{t^2}{1+t^3}, y=\frac{2 t^3}{1+t^5}$ and

$t \neq-1$

So, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Now, $\frac{d x}{d t}=\frac{(2 t)\left(1+t^5\right)-t^2 \cdot 5 t^4}{\left(1+t^5\right)^2}$

$ =\frac{2 t-3 t^6}{\left(1+t^5\right)^2} $

And, $\frac{d y}{d t}=\frac{\left(6 t^2\right)\left(1+t^5\right)-\left(2 t^3\right)\left(5 t^4\right)}{\left(1+t^5\right)^2}$

$ =\frac{6 t^2-4 t^7}{\left(1+t^5\right)^2} $

$ \begin{aligned}So,\,\, \frac{d y}{d x} & =\frac{\left[\frac{6 t^2-4 t^7}{\left(1+t^5\right)^2}\right]}{\left[\frac{\left(2 t-3 t^6\right)}{\left(1+t^5\right)^2}\right]}=\frac{6 t^2-4 t^7}{2 t-3 t^6} \\ & =\frac{2 t^2\left(3-2 t^5\right)}{t\left(2-3 t^5\right)}=\frac{2 t\left(3-2 t^5\right)}{\left(2-3 t^5\right)} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \because \sin x \sqrt{\cos y}-\cos y \sqrt{\sin x}=0\,\,\,\, \ldots \text { (i) } \\ & \Rightarrow \frac{d}{d x}[\sin x \sqrt{\cos y}-\cos y \cdot \sqrt{\sin x}]=0 \\ & \Rightarrow \cos x \sqrt{\cos y}-\frac{\sin x \sin y}{2 \sqrt{\cos y}} \cdot \frac{d y}{d x} +\sin y \cdot \sqrt{\sin x} \cdot \frac{d y}{d x}-\cos y \cdot \frac{\cos x}{2 \sqrt{\sin x}}=0 \\ & \Rightarrow \frac{d y}{d x}\left(\sin y \cdot \sqrt{\sin x}-\frac{\sin x \cdot \sin y}{2 \sqrt{\cos y}}\right) \\ & =\frac{\cos x \cdot \cos y}{2 \sqrt{\sin x}}-\cos x \cdot \sqrt{\cos y} \end{aligned} \end{aligned} $

$ (\cos x \cdot \cos y-2 \cos x \sqrt{\sin x \cos y}) $

$ \begin{aligned} &\begin{array}{r} \Rightarrow \frac{d y}{d x}=\frac{(2 \sqrt{\cos y})}{(2 \sqrt{\sin x})(2 \sin y \sqrt{\cos y \cdot \sin x}} \\ -\sin x \cdot \sin y) \end{array}\\ &\text { By Eq. (i), we get }\\ &\begin{aligned} & \sin x \sqrt{\cos y}=\cos y \sqrt{\sin x} \\ \Rightarrow & \cos y=\sin x \\ \therefore & \frac{d y}{d x}=\frac{\sin x \cos x-2 \cos x \sqrt{\sin ^2 x}}{2 \sin y \sqrt{\sin ^2 x}-\sin x \cos x} \\ \Rightarrow & \frac{d y}{d x}=\frac{-(2 \cos x \sin x-\sin x \cos x)}{(2 \sin x \cos x-\sin x \cos x)}=-1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{array}{r} y=\left(\log _x \sin x\right)^x=\left(\frac{\log (\sin x)}{\log x}\right)^x \\ \Rightarrow \quad \ln y=x \cdot \ln \left(\frac{\log (\sin x)}{\log x}\right) \\ \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\ln \left(\frac{\log (\sin x)}{\log x}\right)+ x \cdot \frac{d}{d x}\left[\ln \left(\frac{\log (\sin x)}{\log x}\right)\right] \\ \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\ln \left(\frac{\ln (\sin x)}{\ln x}\right)+x \frac{d}{d x}\left[\ln \left(\frac{\ln (\sin x)}{\ln x}\right)\right] \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\ln \left(\frac{\ln (\sin x)}{\ln x}\right) +x\left[\frac{\ln x}{\ln (\sin x)}\left\{\frac{\ln x \cdot \cot x-\frac{\ln (\sin x)}{x}}{(\ln x)^2}\right\}\right] \\ & \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\ln \left(\frac{\ln (\sin x)}{\ln x}\right) +\frac{x}{\ln (\sin x)}\left\{\frac{\ln x \cdot \cot x-\frac{\ln (\sin x)}{x}}{(\ln x)}\right\} \\ & \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}=\ln \left(\frac{\ln (\sin x)}{\ln x}\right) +\frac{x \cdot \cot x}{\ln (\sin x)}-\frac{1}{\ln x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \frac{d y}{d x}=y\left[\begin{array}{l} \frac{x \cdot \cot x}{\ln (\sin x)}+\ln (\ln (\sin x)) \\ -\ln (\ln x)-\frac{1}{\ln x} \end{array}\right] \\ & \Rightarrow \frac{d y}{d x}=y\left[\begin{array}{l} \frac{x \cdot \cot x}{\log (\sin x)}+\log (\log (\sin x)) \\ -\frac{1}{\log x}-\log (\log x) \end{array}\right] \end{aligned} $

$ \begin{aligned} x & =\sqrt{2^{\operatorname{cosec}^{-1} t}} \quad y=\sqrt{2^{\sec ^{-1} t}} \\ x^2 & =2^{\operatorname{cosec}^{-1} t} \quad y^2=2^{\sec ^{-1} t} \\ \because x^2 y^2 & =2^{\operatorname{cosec}^{-1} t} \cdot 2^{\sec ^{-1} t} t \\ \Rightarrow x^2 y^2 & =2^{\operatorname{cosec}^{-1} t+\sec ^{-1} t} \\ \Rightarrow x^2 y^2 & =2^{\frac{\pi}{2}} \end{aligned} $

Differentiating w.r.t. $x$, we get

$ \begin{aligned} & 2 x y^2+x^2 2 y\left(\frac{d y}{d x}\right)=0 \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{-2 x y^2}{2 x^2 y}=-\frac{y}{x} \end{aligned} $

We have,

$ (a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2 $

Differentiating w.r.t $x$, we get $(-\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)$

$ \begin{gathered} \quad+(a+\sqrt{2} b \cos x)(\sqrt{2} b \sin y) \frac{d y}{d x}=0 \\ \Rightarrow \quad \frac{d y}{d x}=\frac{\sin x(a-\sqrt{2} b \cos y)}{\sin y(a+\sqrt{2} b \cos x)} \\ \left.\Rightarrow \frac{d y}{d x}\right|_{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)}=\frac{\frac{1}{\sqrt{2}}\left(a-\sqrt{2} b\left(\frac{1}{\sqrt{2}}\right)\right)}{\frac{1}{\sqrt{2}}\left(a+\sqrt{2} b \frac{1}{\sqrt{2}}\right)} \\ =\frac{a-b}{a+b} \end{gathered} $

We have,

$ f(x)=x^{\sec ^{-1} x} \Rightarrow \ln f=\sec ^{-1} x \cdot \ln x $

Differentiate w.r.t. $x$, we get

$ \begin{aligned} \frac{1}{f} \cdot f^{\prime}(x) & =\frac{\sec ^{-1} x}{x}+\ln x\left(\frac{1}{x \sqrt{x^2-1}}\right) \\ \Rightarrow \frac{f^{\prime}(2)}{f(2)} & =\frac{\sec ^{-1} 2}{2}+\ln 2\left(\frac{1}{2 \sqrt{4-1}}\right) \\ \Rightarrow f^{\prime}(2) & =f(2)\left(\frac{\pi}{6}+\frac{\ln 2}{2 \sqrt{3}}\right) \\ & =2^{\pi / 6}\left(\frac{\pi}{6}+\frac{\ln 2}{2 \sqrt{3}}\right) \\ & =\frac{2^{\pi / 6}}{6}(\pi+\sqrt{3} \log 2 \end{aligned} $

Given, $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{7 x}{1-12 x^2}\right)$

Let $x=\tan \theta$

$ \text { } \begin{aligned}So, & =\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)+\tan ^{-1}\left(\frac{7 \tan \theta}{1-12 \tan ^2 \theta}\right) \\ & =3 \tan ^{-1} x+\tan ^{-1}\left(\frac{7 x}{1-12 x^2}\right) \end{aligned} $

Now, differentiate $y$ w.r.t $x$, we get

$ \begin{aligned} \frac{d y}{d x} & =3 \cdot \frac{1}{1+x^2}+\frac{1}{1+\left(\frac{7 x}{1-12 x^2}\right)^2} \cdot \frac{d}{d x}\left(\frac{7 x}{1-12 x^2}\right) \\ & =\frac{3}{1+x^2}+\frac{1}{1+\frac{49 x^2}{\left(1-12 x^2\right)^2}} \cdot \frac{\left(1-12 x^2\right) \cdot 7-7 x(-24 x)}{\left(1-12 x^2\right)^2} \\ & =\frac{3}{1+x^2}+\frac{\left(1-12 x^2\right)^2}{\left(1-12 x^2\right)^2+49 x^2} \cdot \frac{7-84 x^2+168 x^2}{\left(1-12 x^2\right)^2} \\ & =\frac{3}{1+x^2}+\frac{7+84 x^2}{\left(1-12 x^2\right)^2+49 x^2} \end{aligned} $

Now, At $x=0$

$ \left.\frac{d y}{d x}\right|_{x=0}=\frac{3}{1+0}+\frac{7+84(0)}{\left(1-12(0)^2\right)^2+49.0^2}=3+\frac{7}{1+0}=3+7=10 $

$y=\sqrt{\frac{x^4 \sqrt{3 x-5}}{\left(x^2-3\right)(2 x-3)}}$

Take the natural $\log$ to both sides, we get

$ \begin{aligned} & \ln (y)=\frac{1}{2} \ln \left(\frac{x^4 \sqrt{3 x-5}}{\left(x^2-3\right)(2 x-3)}\right) \\ & \ln (y)=\frac{1}{2}\left[\ln \left(x^4\right)+\ln (\sqrt{3 x-5})-\ln \left(x^2-3\right)-\ln (2 x-3)\right] \end{aligned} $

$ =\frac{1}{2}\left[4 \ln (x)+\frac{1}{2} \ln (3 x-5)-\ln \left(x^2-3\right)-\ln (2 x-3)\right] $

Now, differentiating w.r.t $x$, we get

$ \begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2}\left[\frac{4}{x}+\frac{1}{2} \cdot \frac{3}{3 x-5}-\frac{2 x}{x^2-3}-\frac{2}{2 x-3}\right] \\ & \frac{d y}{d x}=\frac{y}{2}\left[\frac{4}{x}+\frac{3}{2(3 x-5)}-\frac{2 x}{x^2-3}-\frac{2}{2 x-3}\right] \end{aligned} $

Now, at $x=2$

$ y=\sqrt{\frac{2^4 \sqrt{3 \cdot 2-5}}{\left(2^2-3\right)(4-3)}}=\sqrt{\frac{16 \cdot 1}{1 \cdot 1}}=4 $

So, $\left.\quad \frac{d y}{d x}\right|_{x=2}=\frac{4}{2}\left[\frac{4}{2}+\frac{3}{2(3 \cdot 2-5)}-\frac{2.2}{2^2-3}-\frac{2}{2 \cdot 2-3}\right]$

$ =2\left[2+\frac{3}{2}-\frac{4}{1}-2\right]=2\left[\frac{3-8}{2}\right]=-5 $

$\left.\therefore \quad \frac{d y}{d x}\right|_{x=2}=-5$

Given, $x^2+y^2+\sin y=4$

Differentiate both sides w.r.t $x$, we get

$ \begin{aligned} & 2 x+2 y \cdot \frac{d y}{d x}+\cos y \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}(2 y+\cos y)=-2 x \\ & \frac{d y}{d x}=\frac{-2 x}{2 y+\cos y} \end{aligned} $

Again, differentiate both sides w.r.t $x$, we get

$ \begin{aligned} \frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(\frac{-2 x}{2 y+\cos y}\right) \\ & =\frac{(2 y+\cos y) \cdot(-2)-(-2 x) \cdot(2-\sin y) \cdot \frac{d y}{d x}}{(2 y+\cos y)^2} \\ & =\frac{-4 y-2 \cos y+2 x \cdot(2-\sin y) \cdot\left(\frac{-2 x}{2 y+\cos y}\right)}{(2 y+\cos y)^2} \\ & =\frac{-4 y-2 \cos y-\frac{4 x^2(2-\sin y)}{2 y+\cos y}}{(2 y+\cos y)^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Put $x=-2$ in original equation, we get

$ \begin{array}{ll} & (-2)^2+y^2+\sin y=4 \\ \Rightarrow \quad & y^2+\sin y=0 \Rightarrow y=0 \end{array} $

Now, from Eq. (i), we get

$ \begin{aligned} \frac{d^2 y}{d x^2} & =\frac{-4(0)-2 \cos (0)-\frac{4(-2)^2(2-\sin (0))}{2 \cdot(0)+\cos (0)}}{(2(0)+\cos (0))^2} \\ & =\frac{-2-\frac{16(2-0)}{1}}{1^2}=\frac{-2-32}{1}=-34 \\ \left.\therefore \quad \frac{d^2 y}{d x^2}\right|_{x=-2} & =-34 \end{aligned} $

$y=\sqrt{\cos h x+\sqrt{\cos h x}}$

Let $\sqrt{\cos h x}=u \Rightarrow y=\sqrt{u^2+u}$

Now, $\frac{d y}{d x}=\frac{1}{2 \sqrt{u^2+u}} \cdot(2 u+1)=\frac{2 u+1}{2 \sqrt{u^2+u}}$

$ \frac{d y}{d u}=\frac{2 u+1}{2 y} $

Now, $\frac{d u}{d x}=\frac{d}{d x}(\sqrt{\cos h x})=\frac{1}{2 \sqrt{\cosh x}}(\sin hx)$

$ =\frac{\sin h x}{2 u} $

Thus, $\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=\frac{2 u+1}{2 y} \cdot \frac{\sin h x}{2 u}$

$ =\frac{(2 u+1) \sin h x}{4 y u} $

put $u=\sqrt{\cos h x}$

Therefore, $\frac{d y}{d x}=\frac{\sin h x(1+2 \sqrt{\cos h x})}{4 y \sqrt{\cos h x}}$

$ \begin{aligned} &\text { } y=\tan ^{-1} \sqrt{x^2-1}+\sin h^{-1} \sqrt{x^2-1}\\ &\begin{aligned} & \therefore \frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1} \sqrt{x^2-1}\right)+\frac{d}{d x}\left(\sin h^{-1} \sqrt{x^2-1}\right) \\ & \begin{aligned} =\frac{1}{1+\left(\sqrt{x^2-1}\right)^2} & \left(\frac{(2 x)}{2 \sqrt{x^2-1}}\right) +\frac{1}{\sqrt{1+\left(\sqrt{x^2-1}\right)^2}}\left(\frac{2 x}{2 \sqrt{x^2-1}}\right) \\ =\frac{2 x}{x^2\left(2 \sqrt{x^2-1}\right)} & +\frac{1}{x}\left(\frac{2 x}{2 \sqrt{x^2-1}}\right) \\ = & \frac{1+x}{x \sqrt{x^2-1}} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } y=\left(\log _{\downarrow} x\right)^{\frac{1}{x}}+{\underset{v}{x}}_v^{\log x} \\ & \text { So, } \ln (u)=\frac{1}{x} \log (\log x) \\ & \frac{1}{u} \cdot \frac{d u}{d x}=\frac{1}{x} \cdot \frac{1}{x \log x}+\log (\log x) \cdot\left(-\frac{1}{x^2}\right) \\ & =\frac{1}{x^2 \log x}-\frac{\log (\log x)}{x^2} \\ & \left.\Rightarrow \frac{d u}{d x}\right|_{x=e}=(\log e)^{\frac{1}{e}}\left[\frac{1}{e^2 \log e}-\frac{\log (\log e)}{e^2}\right] \\ & =1\left[\frac{1}{e^2}-0\right]=\frac{1}{e^2} \end{aligned} $

and $\ln (y)=\log x \cdot \log (x)=(\log x)^2$

$ \begin{aligned} & \Rightarrow \frac{1}{v}\left(\frac{d v}{d x}\right)=2 \log (x) \cdot \frac{1}{x} \\ & \Rightarrow \quad \frac{d v}{d x}=\eta\left[\frac{2 \log x}{x}\right] \end{aligned} $

$ \begin{aligned}at \,\,x & =e \\ & =e^{\log x}\left[\frac{2 \log x}{e}\right] \\ & =e \cdot \frac{2}{e}=2 \end{aligned} $

Hence, $\left.\frac{d y}{d x}\right|_{x=e}=\frac{1}{e^2}+2$

Given, $x=\sqrt{2} e^t(\sin t-\cos t)$

and  $ y=\sqrt{2} e^t(\sin t+\cos t) $

On differentiating both side, we get

$ \begin{array}{ll} \therefore & \frac{d x}{d t}=\sqrt{2} e^t[2 \sin t] \text { and } \frac{d y}{d t}=\sqrt{2} e^t(2 \cos t) \\ \therefore & \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\cot t \end{array} $

Again, differentiating both side,

$ \begin{aligned} \frac{d^2 y}{d x^2} & =\frac{d \cot t}{d x}=\frac{d \cot t}{d t} \times \frac{d t}{d x} \\ & =-\operatorname{cosec}^2 t \times \frac{1}{\sqrt{2} e^t(2 \sin t)}=\frac{-1}{2 \sqrt{2} e^t \cdot \sin ^3 t} \\ \left.\therefore \frac{d^2 y}{d x^2}\right|_{t=\frac{\pi}{4}} & =\frac{-1}{2 \sqrt{2} \times e^{\frac{\pi}{4}} \cdot\left(\frac{1}{\sqrt{2}}\right)^3}=-e^{\frac{-\pi}{4}} \end{aligned} $

$ \begin{aligned} & g(x)=x+\tan x \\ & g^{\prime}(x)=1+\sec ^2 x \end{aligned} $

and $g$ is the inverse of $f(x)$, then

$ \begin{aligned} \Rightarrow f^{\prime}(x) & =\frac{1}{g^{\prime}(f(x))} \\ g(x) & =x+\tan x, g^{\prime}(x)=1+\sec ^2 x \\ f^{\prime}(x) & =\frac{1}{1+\sec ^2(f(x))} \end{aligned} $

Given, $\sqrt{x(1-y)}+\sqrt{y(1-x)}=1$

$ \begin{aligned} & \text { Let } x=\sin ^2 A \text { and } y=\sin ^2 B \\ & \Rightarrow \sqrt{\sin ^2 A\left(1-\sin ^2 B\right)}+\sqrt{\sin ^2 B\left(1-\sin ^2 A\right)} \\ & \Rightarrow \quad \sin A \cos B+\cos A \sin B=1 \\ & \Rightarrow \quad \sin (A+B)=1=\sin \frac{\pi}{2} \\ & \Rightarrow \quad A+B=\frac{\pi}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Now, $A=\sin ^{-1} \sqrt{x}, B=\sin ^{-1} \sqrt{y}$

From Eqs. (i), we get

$ \sin ^{-1} \sqrt{x}+\sin ^{-1} \sqrt{y}=\frac{\pi}{2} $

On differentiating,

$ \begin{aligned} & \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}+\frac{1}{\sqrt{1-y}} \cdot \frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \\ \Rightarrow & \frac{1}{2 \sqrt{y} \sqrt{1-y}} \frac{d y}{d x}=-\frac{1}{2 \sqrt{x} \sqrt{1-x}} \\ \Rightarrow & \frac{d y}{d x}=-\frac{\sqrt{y} \sqrt{1-y}}{\sqrt{x}} \sqrt{1-x}=-\sqrt{\frac{y-y^2}{x-x^2}} \end{aligned} $

$ \begin{aligned} & \text { } x=2 \cos ^3 \theta, \dot{y}=3 \sin ^2 \theta \\ & \Rightarrow \frac{d x}{d y}=\frac{\frac{d x}{d \theta}}{\frac{d y}{d \theta}} \Rightarrow \frac{d x}{d y}=\frac{6 \cos ^2 \theta(-\sin \theta)}{6 \sin \theta(\cos \theta)} \\ & \Rightarrow \frac{d y}{d x}=-\sec \theta \end{aligned} $

$y=f(x)=\left(|x|-|x-1|^2\right)$

Here, $f(x)$ is not differentiable at $x=0$,

$\therefore \mathrm{A}$ is false.

R is definition of derivative $f(x)$ at $x=a$

$\therefore \mathrm{R}$ is true.

Given $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$

$ \begin{aligned} & \because \quad\left(x^2+y^2\right)^2=\left(t-\frac{1}{t}\right)^2 \\ & \Rightarrow \quad x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}-2 \\ & \Rightarrow \quad x^4+y^4+2 x^2 y^2=x^4+y^4-2 \\ & \Rightarrow \quad x^2 y^2=-1 \\ & \Rightarrow \quad x^2 2 y y^{\prime}+y^2 \cdot 2 x=0 \\ & \Rightarrow \quad x y^{\prime}+y=0 \\ & \Rightarrow \quad y^{\prime}=-\frac{y}{x} \\ & \therefore \quad \frac{d y}{d x}=-\frac{y}{x} \end{aligned} $

We have, $y=(a x+b) \cos x$

$ y \sec x=a x+b $

$y \sec x \tan x+\sec x y_1=a$

Again, differentiating w.r.t. $x$, we get $\Rightarrow y \sec x \sec ^2 x+\sec x \tan x \cdot y_1+y \tan x \sec x \tan x+\sec x y_2+y_1 \sec x \tan x=0 $

$ \Rightarrow $$ $y \sec ^3 x+y_1 \sec x \tan x+y \sec x \tan ^2 x$ +y_1 \sec x \tan x+\sec x \cdot y_2=0 $

$\Rightarrow y_2 \sec x+2 y_1 \sec x \tan x$ +y\left(\sec ^2 x+\tan ^2 x\right) \sec x=0 $

On dividing both sides by $\sec ^3 x$, we get $y_2 \cos ^2 x+2 y_1 \sin x \cdot \cos x$ +y\left(1+\sin ^2 x\right)=0 $

$y_2-y_1 \sin ^2 x+y_1 \sin 2 x$ +y\left(1+\sin ^2 x\right)=0 $

$ \therefore y_2+y_1 \sin 2 x+y\left(1+\sin ^2 x\right)=y_2 \sin ^2 x $

Given, $5 f(x)+3 f\left(\frac{1}{x}\right)=x+2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ 5 f\left(\frac{1}{x}\right)+3 f(x)=\frac{1}{x}+2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Multiplying Eq. (i) by 5 and Eq. (ii) by 3, we get

$ \begin{aligned} \because \quad f(x) & =\frac{1}{16}\left(5 x-\frac{3}{x}+4\right) \\ y & =x f(x)=\frac{x}{16}\left(5 x-\frac{3}{x}+4\right) \\ y & =\frac{1}{16}\left(5 x^2-3+4 x\right) \\ \frac{d y}{d x} & =\frac{1}{16}(10 x+4) \\ \left(\frac{d y}{d x}\right)_{\text {at } x=1} & =\frac{10+4}{16} \\ & =\frac{14}{16}=\frac{7}{8} \end{aligned} $

$\begin{aligned} &\log _e y=3 \sin ^{-1} x\\ &\begin{aligned} & y=e^{3 \sin ^{-1} x} \\ & \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}} \end{aligned} \end{aligned}$

$\sqrt{1-x^2} \frac{d y}{d x}=3 y$

Again differentiate

$\begin{aligned} & \sqrt{1-x^2} \cdot y^{\prime \prime}-\frac{2 x}{2 \sqrt{1-x^2}} y^{\prime}=3 y^{\prime} \\ & (1-x)^2 y^{\prime \prime}-x y^{\prime}=3 y^{\prime}\left(\sqrt{1-x^2}\right) \end{aligned}$

So value of $3 y^{\prime}\left(\sqrt{1-x^2}\right)$ at $x=\frac{1}{2}$

$\begin{aligned} & 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\ & =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}} \end{aligned}$

Given the polynomial function:

$f(x) = ax^3 + bx^2 + cx + 41$

We are provided the following conditions from the problem:

1. $f(1) = 40$

2. $f^{\prime}(1) = 2$

3. $f^{\prime \prime}(1) = 4$

First, calculate $f(1)$:

$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$

Simplifying, we get:

$a + b + c + 41 = 40$

Therefore:

$a + b + c = -1$

Next, calculate the first derivative $f^{\prime}(x)$:

$f^{\prime}(x) = 3ax^2 + 2bx + c$

Given $f^{\prime}(1) = 2$:

$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$

Simplifying, we get:

$3a + 2b + c = 2$

Next, calculate the second derivative $f^{\prime \prime}(x)$:

$f^{\prime \prime}(x) = 6ax + 2b$

Given $f^{\prime \prime}(1) = 4$:

$f^{\prime \prime}(1) = 6a(1) + 2b = 4$

Simplifying, we get:

$6a + 2b = 4$

Dividing the entire equation by 2:

$3a + b = 2$

We now have three equations:

1. $a + b + c = -1$

2. $3a + 2b + c = 2$

3. $3a + b = 2$

To solve for $a$, $b$, and $c$, follow these steps:

First, subtract the third equation from the second equation:

$(3a + 2b + c) - (3a + b) = 2 - 2$

Which simplifies to:

$b + c = 0$

So,

$c = -b$

Substitute $c = -b$ into the first equation:

$(a + b - b = -1)$

Simplifying, we get:

$a = -1$

Now substitute $a = -1$ into the third equation:

$3(-1) + b = 2$

Which simplifies to:

$-3 + b = 2$

Therefore:

$b = 5$

Next, since $c = -b$:

$c = -5$

Finally, we need to find $a^2 + b^2 + c^2$:

$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$

Simplifying, we get:

$1 + 25 + 25 = 51$

Therefore, the answer is:

Option B: 51

To determine $g^{\prime}(0)$, we start by applying the chain rule and product rule to find the derivative of the given function $g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$.

The product rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their product is given by:

$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $

Let's denote $u(x) = h(\mathrm{e}^x)$ and $v(x) = \mathrm{e}^{h(x)}$.

First, we need to find $u'(x)$ and $v'(x)$. Using the chain rule, we find:

$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $

Now, the derivative of $v(x)$ is:

$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $

Using the product rule, we get the derivative of $g(x)$:

$ g'(x) = u'(x) v(x) + u(x) v'(x) $

Substituting $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the above expression, we get:

$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $

Next, we need to evaluate this at $x = 0$:

First, we know that:

$ h(0) = 0 $

$ h(1) = 1 $

$ h'(0) = 2 $

$ h'(1) = 2 $

Substituting $x = 0$ into the expressions, we get:

$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $

$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $

$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $

$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $

Therefore, evaluating $g'(0)$:

$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $

$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $

$ g'(0) = 2 + 2 = 4 $

Thus, the value of $g^{\prime}(0)$ is 4, which corresponds to Option A.

The correct answer is Option A: 4.

Given the function:

$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $

we need to find its second derivative at specific points.

First, let’s compute the first derivative $ f^{\prime}(x) $:

$ f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right) $

Next, the second derivative $ f^{\prime \prime}(x) $ is:

$ f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right) $

Therefore, evaluating the second derivative at $ x = \frac{2}{\pi} $:

$ f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right) $

Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$, this simplifies to:

$ f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi} $

Finally, note that $ f^{\prime}(0) $ is not defined, as it involves terms like $\frac{1}{x}$ when $ x = 0 $.

Let $f^{\prime}(1)=k$

$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$

$\begin{aligned} & \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\ \Rightarrow & f^{\prime \prime}(0)=2 k \end{aligned}$

Given information is not complete.

$\begin{aligned} & y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\ & y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\ & =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \end{aligned}$

$y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}$

$\begin{aligned} & \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\ & \therefore \quad y+y^{\prime}+y^{\prime \prime}=2 \end{aligned}$

Given that $(g \circ f)(x) = x$ for all $x \in \mathbf{R}$. This means $g(f(x)) = x$ for all $x \in \mathbf{R}$. Differentiating both sides with respect to $x$, we get:

$g'(f(x)) \cdot f'(x) = 1$

Now, we want to find the value of $8g'(2)$. To do this, we need to find a value of $x$ such that $f(x) = 2$. Let's solve for $x$:

$x^5 + 2e^{x/4} = 2$

By inspection, we see that $x = 0$ is a solution. Therefore, $f(0) = 2$. Now, we can substitute this into our differentiated equation:

$g'(f(0)) \cdot f'(0) = 1$

$g'(2) \cdot f'(0) = 1$

Let's find $f'(0)$:

$f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$

$f'(0) = \frac{1}{2}$

Substituting this back into our equation:

$g'(2) \cdot \frac{1}{2} = 1$

$g'(2) = 2$

Finally, we can calculate $8g'(2)$:

$8g'(2) = 8 \cdot 2 = \boxed{16}$

$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$

$\begin{aligned} & \mathrm{f}^{\prime}(1)=2024 \\ & \mathrm{f}(1)=1 \end{aligned}$

Partially differentiating w. r. t. x

$\begin{aligned} & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \mathrm{y} \rightarrow \mathrm{x} \\ & \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0 \end{aligned}$

$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$

Also $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$

$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$

$\begin{aligned} & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x \end{array}\right| \\ & \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\ & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{array}\right| \\ & \mathrm{f}(\mathrm{x})=45 \\ & \mathrm{f}^{\prime}(\mathrm{x})=0 \\ & \end{aligned}$

$\begin{aligned} & y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\ & \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4} \end{aligned}$

Again,

$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$

Again

$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$

at $\mathrm{x}=\frac{1}{2}$,

$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$

Thus $225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$

$\begin{aligned} & f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\ & =\sqrt{\pi} \end{aligned}$

To solve for the derivative of the function $ f $, let's analyze the given functional equation:

$ y f(x+y) + \cos mx y = 1 + y f(x) $

Rearranging the terms, we have:

$ y[f(x+y) - f(x)] = 1 - \cos mx y $

Using the trigonometric identity for cosine, we can express this as:

$ y[f(x+y) - f(x)] = 2 \sin^2 \left(\frac{mxy}{2}\right) $

Now, by dividing both sides by $ y^2 $, we get:

$ \frac{f(x+y) - f(x)}{y} = \frac{2 \sin^2 \left(\frac{mxy}{2}\right)}{y^2} $

Taking the limit as $ y \to 0 $, this expression represents the derivative $ f'(x) $:

$ \lim\limits_{y \to 0} \frac{f(x+y) - f(x)}{y} = \lim\limits_{y \to 0} \frac{2 \sin^2 \left(\frac{mxy}{2}\right)}{y^2} $

Substitute $ m = 2 $, and notice that $\sin^2 \left(\frac{2xy}{2}\right) = \sin^2(xy)$. Therefore, the limit becomes:

$ f'(x) = \lim\limits_{y \to 0} \frac{2 \sin^2(xy)}{y^2} $

Recognizing from trigonometric limits that as $ y \to 0 $, $\sin(xy) \approx xy$, the limit simplifies through:

$ f'(x) = \lim\limits_{y \to 0} \frac{2 (xy)^2}{y^2} = \lim\limits_{y \to 0} 2x^2 = 2x^2 $

Thus, the derivative of the function is:

$ f'(x) = 2x^2 $