Differentiation

$ \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} $

To find the derivative $\frac{dy}{dx}$ for $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$, we use the formula for the derivative of the inverse hyperbolic sine function:

$ \frac{d}{dx}\left(\sinh^{-1}x\right) = \frac{1}{\sqrt{1+x^2}} $

Applying this to our function:

Identify $u = \frac{1-x}{1+x}$.

Differentiate $u$ with respect to $x$:

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

Now apply the chain rule:

$ \frac{dy}{dx} = \frac{1}{\sqrt{1+\left(\frac{1-x}{1+x}\right)^2}} \times \frac{du}{dx} $

Simplify the expression:

Compute $1 + \left(\frac{1-x}{1+x}\right)^2$:

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

Note that $(1+x)^2 + (1-x)^2 = 2(1+x^2)$

Plug this into the derivative formula:

$ \frac{dy}{dx} = \frac{1}{\sqrt{\frac{2(1+x^2)}{(1+x)^2}}} \times \frac{-2}{(1+x)^2} $

Simplify further to get the final derivative:

$ = \frac{-2}{\sqrt{2(1+x^2)} \cdot (1+x)} = \frac{-\sqrt{2}}{|1+x| \sqrt{1+x^2}} $

Thus, the derivative $\frac{dy}{dx}$ is:

$ \frac{-\sqrt{2}}{|1+x| \sqrt{1+x^2}} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & y=(x-1)(x+2)\left(x^2+5\right)\left(x^4+8\right) \\ & \Rightarrow y=\left(x^4+x^3+3 x^2+5 x-10\right)\left(x^4+8\right) \\ & \Rightarrow \frac{d y}{d x}=\left(x^4+x^3+3 x^2+5 x-10\right)\left(4 x^3\right) \\ & \quad+\left(4 x^3+3 x^2+6 x+5\right)\left(x^4+8\right) \\ & \begin{aligned} \therefore \lim _{x \rightarrow-1}\left(\frac{d y}{d x}\right) & =(1-1+3-5-10)(-4) \\ & \quad+(-4+3-6+5)(1+8) \\ = & (-12)(-4)+(-2)(9) \\ = & 48-18=30 \end{aligned} \end{aligned} \end{aligned} $

To find the expression for $(1+4x^2)^2 y'' - 16$, we start with

$ y = \left(\tan^{-1} 2x\right)^2 + \left(\cot^{-1} 2x\right)^2. $

Using the identity $\cot^{-1} u = \frac{\pi}{2} - \tan^{-1} u$, we can rewrite the equation as

$ y = \left(\tan^{-1} 2x\right)^2 + \left(\frac{\pi}{2} - \tan^{-1} 2x\right)^2. $

Simplifying,

$ y = 2\left(\tan^{-1} 2x\right)^2 + \frac{\pi^2}{4} - \pi \tan^{-1} 2x. $

Next, we differentiate $y$ with respect to $x$:

$ y' = \frac{4 \tan^{-1} 2x}{1 + 4x^2} \cdot 2 - \frac{\pi}{1 + 4x^2} \cdot 2. $

This simplifies to

$ y' = \frac{8 \tan^{-1} 2x - 2\pi}{1 + 4x^2}. $

Multiplying both sides by $(1 + 4x^2)$, we get

$ (1 + 4x^2) y' = 8 \tan^{-1} 2x - 2\pi. $

Differentiating this equation with respect to $x$ gives:

$ (1 + 4x^2) y'' + y'(8x) = \frac{16}{1 + 4x^2}, $

Rearranging this result, we find:

$ (1 + 4x^2)^2 y'' + 8x y'(1 + 4x^2) = 16. $

Thus,

$ (1 + 4x^2)^2 y'' - 16 = -8x y'(1 + 4x^2). $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \begin{array}{l} y=\tan ^{-1} \frac{x}{1+2 x^2}+\tan ^{-1} \frac{x}{1+6 x^2} \\ \\ +\tan ^{-1} \frac{x}{1+12 x^2} \\ \Rightarrow \frac{d y}{d x}=\frac{1}{1+\left(\frac{x}{1+2 x^2}\right)^2} \cdot \frac{\left(1+2 x^2\right) \cdot 1-x \cdot 4 x}{\left(1+2 x^2\right)^2} \end{array} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & +\frac{1}{1+\left(\frac{x}{1+6 x^2}\right)^2} \times \frac{\left(1+6 x^2\right) \cdot 1-\times(12,x)}{\left(1+6 x^2\right)^2} \\ & +\frac{1}{1+\left(\frac{x}{1+12 x^2}\right)^2} \cdot \frac{\left(1+12 x^2\right) \cdot 1-x \cdot 24x}{\left(1+12 x^2\right)^2} \end{aligned}\\ &\text { put, } x=\frac{1}{2} \text { and simplify }\\ &\left.\frac{d y}{d x}\right|_{x=\frac{1}{2}}=0 \end{aligned} $

Given that

$ f(x)=5 \cos ^3 x-3 \sin ^2 x $

and $g(x)=4 \sin ^3 x+\cos ^2 x$

Now, $\frac{d f(x)}{d x}=-15 \cos ^2 x \cdot \sin x-6 \sin x \cos x $

$ \begin{aligned} & =-\frac{15 \cos ^2 x \sin x-6 \sin x \cos x}{12 \sin ^2 x \cos x-2 \cos x \sin x} \\ & =\frac{-\cos x \sin x(15 \cos x+6)}{-\cos x \sin x(-12 \sin x+2)} \\ & =-\left(\frac{15 \cos x+6}{12 \sin x-2}\right) \end{aligned} $

Given the series:

$ y = 1 + x + x^2 + x^3 + \ldots $

and noting that $ |x| < 1 $, we recognize this as a geometric series. The sum of this series can be expressed in closed form as:

$ y = \frac{1}{1-x} $

Now, let's find the first derivative of $ y $ with respect to $ x $:

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

Next, differentiate $ y' $ to find the second derivative $ y'' $:

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

Since we have $ y = \frac{1}{1-x} $, it follows that:

$ y'' = 2 \cdot \frac{1}{1-x} \cdot \frac{1}{(1-x)^2} $

This can be rewritten using the previously calculated derivatives:

$ y'' = 2y \cdot y' $

Thus, the second derivative of $ y $, $ y'' $, is equal to:

$ y'' = 2 y \cdot y' $

Given the expression:

$ y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}} $

We will evaluate $\frac{d^2 y}{d x^2}$ at the point $(\pi, 1)$.

Step 1: Determine $ y $

Squaring both sides of the equation:

$ y^2 = \sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}} $

Therefore, from the self-similarity of the expression, we have:

$ y^2 - \sin x = y \quad \Rightarrow \quad y^2 - y = \sin x $

Step 2: First Derivative $\frac{d y}{d x}$

Differentiating both sides with respect to $x$:

$ 2y \frac{d y}{d x} - \cos x = \frac{d y}{d x} $

Rearranging, we get:

$ \frac{d y}{d x} (2y - 1) = \cos x \quad \Rightarrow \quad \frac{d y}{d x} = \frac{\cos x}{2y - 1} $

Step 3: Second Derivative $\frac{d^2 y}{d x^2}$

Differentiating the equation for $\frac{d y}{d x}$ again:

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

Rearranging:

$ 2y \frac{d^2 y}{d x^2} + 2 \left( \frac{\cos x}{2y - 1} \right)^2 + \sin x = \frac{d^2 y}{d x^2} $

Simplifying:

$ (2y - 1)\frac{d^2 y}{d x^2} = -\sin x - \frac{2 \cos^2 x}{(2y - 1)^2} $

Thus:

$ \frac{d^2 y}{d x^2} = \frac{-\sin x - \frac{2 \cos^2 x}{(2y - 1)^2}}{2y - 1} $

Final Calculation at $(\pi, 1)$

At the point $(\pi, 1)$:

$\sin \pi = 0$

$\cos \pi = -1$

$y = 1$

Substituting these values into the formula for the second derivative:

$ \frac{d^2 y}{d x^2} = \frac{0 - \frac{2 \times (-1)^2}{(2 \times 1 - 1)^2}}{1} = \frac{-2}{1} = -2 $

Therefore, $\frac{d^2 y}{d x^2}$ at $(\pi, 1)$ is $-2$.

We are given the function $ y = f(f(f(f(f(x))))) $ and need to find its derivative at $ x = 0 $, where we know $ f(0) = 0 $ and $ f^{\prime}(0) = 3 $.

To find $ y^{\prime}(0) $, apply the chain rule:

$ y^{\prime} = f^{\prime}(f(f(f(f(x))))) \cdot f^{\prime}(f(f(f(x)))) \cdot f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) $

Substituting $ x = 0 $, we find:

$ y^{\prime}(0) = f^{\prime}(f(f(f(f(0))))) \cdot f^{\prime}(f(f(f(0)))) \cdot f^{\prime}(f(f(0))) \cdot f^{\prime}(f(0)) \cdot f^{\prime}(0) $

Since $ f(0) = 0 $, this simplifies to:

$ y^{\prime}(0) = f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) $

Given $ f^{\prime}(0) = 3 $, this becomes:

$ y^{\prime}(0) = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5 = 243 $

Thus, the derivative of $ y $ at $ x = 0 $ is $ 243 $.

To find $ (a, b) $ such that

$ \frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right) = ax + b $

we start by simplifying the expression in the numerator.

Simplification

The numerator $ 1 + x^2 + x^4 $ can be factored:

$ \begin{aligned} 1 + x^2 + x^4 &= x^4 + 2x^2 + 1 - x^2 \\ &= (x^2 + 1)^2 - x^2 \\ &= (x^2 + 1 - x)(x^2 + 1 + x). \end{aligned} $

Substituting Back

Using the factorized form of the numerator, rewrite the original expression:

$ \frac{d}{d x}\left(\frac{(x^2-x+1)(x^2+x+1)}{1+x+x^2}\right) = ax + b. $

We notice that $ (x^2+x+1) $ cancels out with the denominator:

$ \frac{d}{d x}(x^2 - x + 1) = ax + b. $

Differentiation

Calculating the derivative of the simplified expression $ x^2 - x + 1 $:

$ \frac{d}{d x}(x^2 - x + 1) = 2x - 1. $

Identifying Constants

Set the expression for derivative equal to $ ax + b $:

$ 2x - 1 = ax + b. $

By comparing coefficients, we get:

$ a = 2, \quad b = -1. $

Thus, the values are:

$ (a, b) = (2, -1). $

To find the rate of change of $ x^{\sin x} $ with respect to $ (\sin x)^x $, we proceed as follows:

Let $ y = x^{\sin x} $.

Taking the natural logarithm on both sides, we have:

$ \log y = \sin x \log x $

Differentiating with respect to $ x $, we get:

$ \frac{1}{y} \frac{dy}{dx} = \cos x \log x + \frac{\sin x}{x} $

Let $ z = (\sin x)^x $.

Taking the logarithm similarly, we have:

$ \log z = x \log \sin x $

Differentiating with respect to $ x $, we have:

$ \frac{1}{z} \frac{dz}{dx} = \log \sin x + x \frac{\cos x}{\sin x} $

Simplifying this gives:

$ \frac{1}{z} \frac{dz}{dx} = \log \sin x + x \cot x $

Next, we find the derivative of $ y $ with respect to $ z $ by dividing the derivatives:

$ \frac{dz}{dy} = \frac{yddx}{zddx} = \frac{\cos x \log x + \frac{\sin x}{x}}{\log \sin x + x \cot x} $

Thus, the rate of change is:

$ \frac{dy}{dz} = \frac{x^{\sin x}}{(\sin x)^x} \left( \frac{\log x \cos x + \frac{1}{x} \sin x}{\log \sin x + x \cot x} \right) $

$y=\frac{\alpha x+\beta}{\gamma x+\delta}$ differentiation w.r.t. $x$, we get

$ y_1=\frac{\alpha \delta-\beta \gamma}{(\gamma x+\delta)^2} $

Again, differentiation with respect to $x$, we get

$ y_2=(\alpha \delta-\beta \gamma)\left(\frac{-2 \gamma}{(\gamma x+\delta)^3}\right) $

Again, differentiation with respect to $x$, we get

$ \begin{aligned} & y_3=(\alpha \delta-\beta \gamma)\left(\frac{6 \gamma^2}{(\gamma x+\delta)^4}\right) \\ & y_3=\frac{(\alpha \delta-\beta \gamma)}{(\gamma x+\delta)^2} \cdot \frac{6 \gamma}{(\gamma x+\delta)^2} \end{aligned} $

On squaring both sides of Eq. (ii), we get

$ \begin{aligned} & y_2^2=(\alpha \delta-\beta \gamma)^2 \cdot \frac{4 \gamma^2}{(\gamma x+\delta)^6} \\ & y_2^2=\frac{(\alpha \delta-\beta \gamma)}{(\gamma x+\delta)^2} \cdot \frac{4 \gamma^2}{(\gamma x+\delta)^4} \\ & y_2^2=\frac{(\alpha \delta-\beta \gamma)}{(\gamma \alpha+\delta)^2} \cdot \frac{(\alpha \delta-\beta \gamma) \cdot 6 \gamma^2}{(\gamma x+\delta)^4} \cdot \frac{4}{6} \end{aligned} $

Here, from Eqs. (i) and (iii)

$ y_2^2=y_1 y_3 \cdot\left(\frac{2}{3}\right) $

$ \begin{aligned} &\begin{aligned} & y_2^2=\frac{2}{3} y_1 y_3 \\ & y_2^2=\frac{2}{3} y_1 y_3 \end{aligned}\\ &\text { Now, } 2 y_1 y_3=3 y_2^2 \end{aligned} $

The following explanations clarify the derivatives being calculated:

(A) $\frac{d}{d x}\left[\sec ^{-1}(\cosh x)\right]=\operatorname{sech}(x)$

For the inverse secant function, the derivative is given by:

$ \frac{d}{d x}\left[\sec ^{-1}(\theta)\right] = \frac{1}{|\theta| \sqrt{\theta^2 - 1}} $

Substituting $\theta = \cosh x$:

$ \frac{d}{d x}\left[\sec ^{-1}(\cosh x)\right] = \frac{1}{|\cosh x| \sqrt{(\cosh x)^2 - 1}} \cdot \sinh x $

Since $|\cosh x| = \cosh x$ for all $x \in \mathbb{R}$ and $(\cosh x)^2 - 1 = \sinh^2 x$, we find:

$ \frac{d}{d x}\left[\sec ^{-1}(\cosh x)\right] = \frac{\sinh x}{\cosh x \sinh x} = \frac{1}{\cosh x} = \operatorname{sech}(x) $

This statement is true.

(B) $\frac{d}{d x}\left[\cos ^{-1}(\operatorname{sech} x)\right]=\operatorname{sech} x$

The derivative of the inverse cosine function is:

$ \frac{d}{d x}\left[\cos ^{-1}(\theta)\right] = -\frac{1}{\sqrt{1 - \theta^2}} $

Substituting $\theta = \operatorname{sech} x$:

$ \frac{d}{d x}\left[\cos ^{-1}(\operatorname{sech} x)\right] = -\frac{\operatorname{sech} x \tanh x}{\sqrt{1 - (\operatorname{sech} x)^2}} $

Since $1 - \operatorname{sech}^2 x = \tanh^2 x$, we have:

$ \frac{d}{d x}\left[\cos ^{-1}(\operatorname{sech} x)\right] = \operatorname{sech} x $

This statement is true.

(C) $\frac{d}{d x}\left[\tan ^{-1}(\sinh x)\right]=\operatorname{sech} x$

The derivative of the inverse tangent function is:

$ \frac{d}{d x} \tan ^{-1}(\theta) = \frac{1}{1 + \theta^2} $

Substituting $\theta = \sinh x$:

$ \frac{d}{d x}\left[\tan ^{-1}(\sinh x)\right] = \frac{1}{1 + (\sinh x)^2} $

Knowing that $\cosh^2 x - \sinh^2 x = 1$, it simplifies to:

$ \frac{\cosh x}{1 + \sinh^2 x} = \operatorname{sech} x $

This statement is true.

(D) $\frac{d}{d x}\left[\tan ^{-1}\left(\tan \frac{x}{2}\right)\right]=\sec x$

The expression simplifies as:

$ \tan^{-1}\left(\tan \frac{x}{2}\right) = \frac{x}{2} \quad \text{for } x \in (-\pi, \pi) $

Thus, the derivative is:

$ \frac{d}{d x}\left[\tan ^{-1}\left(\tan \frac{x}{2}\right)\right] = \frac{d}{d x}\left(\frac{x}{2}\right) = \frac{1}{2} $

And $\frac{1}{2} \neq \sec x$, rendering this statement false.

Given the functions $y = t^2 + t^3$ and $x = t - t^4$, we need to find $\frac{d^2 y}{d x^2}$ at $t = 1$.

First, find the derivatives with respect to $t$:

$ \frac{d y}{d t} = 2t + 3t^2 $

$ \frac{d x}{d t} = 1 - 4t^3 $

Next, use these derivatives to find $\frac{d y}{d x}$:

$ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2t + 3t^2}{1 - 4t^3} $

To find $\frac{d^2 y}{d x^2}$, differentiate $\frac{d y}{d x}$ with respect to $t$ and multiply by $\frac{d t}{d x}$:

$ \frac{d^2 y}{d x^2} = \frac{\left(1 - 4t^3\right)(2 + 6t) - (2t + 3t^2)(-12t^2)}{\left(1 - 4t^3\right)^2} \cdot \frac{dt}{dx} $

Substitute $\frac{dt}{dx} = \frac{1}{\frac{d x}{d t}}$:

$ = \frac{(2 + 6t)(1 - 4t^3) + 12t^2(2t + 3t^2)}{\left(1 - 4t^3\right)^2} \cdot \frac{1}{1 - 4t^3} $

Evaluate at $t = 1$:

$ \frac{d^2 y}{d x^2} = \frac{(2 + 6)(1 - 4) + 12(2 + 3)}{(1 - 4)^2} \cdot \frac{1}{1 - 4} $

Simplify the expression:

$ = \frac{(-24 + 60)}{-27} = \frac{36}{-27} = -\frac{4}{3} $

Therefore, $\frac{d^2 y}{d x^2}$ at $t = 1$ is $-\frac{4}{3}$.

To find the second derivative of $ y = \tan (\log x) $, follow these steps:

First Derivative:

Start by differentiating $ y = \tan (\log x) $ with respect to $ x $. Apply the chain rule:

$ \frac{dy}{dx} = \frac{d}{dx}[\tan(\log x)] = \sec^2(\log x) \cdot \frac{d}{dx}[\log x] = \frac{\sec^2(\log x)}{x} $

Second Derivative:

To find the second derivative, apply the quotient rule to $\frac{dy}{dx} = \frac{\sec^2(\log x)}{x}$. The quotient rule is given by:

$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} $

Here, let $ u = \sec^2(\log x) $ and $ v = x $.

Then,

$ \frac{du}{dx} = 2 \sec^2(\log x) \tan(\log x) \cdot \frac{1}{x} = \frac{2 \sec^2(\log x) \tan(\log x)}{x} $

$ \frac{dv}{dx} = 1 $

Substitute into the quotient rule:

$ \frac{d^2y}{dx^2} = \frac{x \cdot \left(\frac{2 \sec^2(\log x) \tan(\log x)}{x}\right) - \sec^2(\log x) \cdot 1}{x^2} $

Simplify the expression:

$ \frac{d^2y}{dx^2} = \frac{2 \sec^2(\log x) \tan(\log x) - \sec^2(\log x)}{x^2} $

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

Thus, the second derivative is:

$ \frac{d^2y}{dx^2} = \frac{\sec^2(\log x)[2 \tan(\log x) - 1]}{x^2} $

For $ x < 0 $, we need to find the derivative $\frac{d}{dx}\left[|x|^x\right]$.

Start by letting $ y = (-x)^x $. Since $ x < 0 $, we have $ |x| = -x $.

We proceed by taking the natural logarithm of both sides:

$ \log y = x \log (-x) $

Now, differentiate both sides with respect to $ x $:

$ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log (-x)) $

Use the product rule to differentiate the right side:

$ \frac{d}{dx}(x \log (-x)) = \log (-x) + x \cdot \frac{d}{dx}(\log (-x)) $

Since $\frac{d}{dx}(\log (-x)) = \frac{-1}{-x} = \frac{1}{x}$, we have:

$ \frac{d}{dx}(x \log (-x)) = \log (-x) - 1 $

Therefore:

$ \frac{1}{y} \frac{dy}{dx} = \log (-x) - 1 $

Multiplying both sides by $ y = (-x)^x $:

$ \frac{dy}{dx} = (-x)^x (\log (-x) - 1) $

Simplify to:

$ \frac{dy}{dx} = (-x)^x [1 + \log (-x)] $

To find the rate of change of $ y^2 $ with respect to $ x^2 $, we begin with the given function:

$ y = x - x^2 $

We introduce new variables $ u = y^2 $ and $ v = x^2 $. Thus:

$ u = (x - x^2)^2 $

First, we find the derivatives $ \frac{du}{dx} $ and $ \frac{dv}{dx} $:

$ \begin{aligned} \frac{du}{dx} & = 2(x - x^2)(1 - 2x) \\ \frac{dv}{dx} & = 2x \end{aligned} $

Now, the rate of change of $ y^2 $ with respect to $ x^2 $ is:

$ \frac{dy^2}{dx^2} = \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{2(x - x^2)(1 - 2x)}{2x} = (1 - x)(1 - 2x) $

We evaluate this expression at $ x = 2 $:

$ (1 - 2)(1 - 4) = (-1)(-3) = 3 $

Thus, the derivative of $ y^2 $ with respect to $ x^2 $ at $ x = 2 $ is 3.

If $ y = f(x) $ is a thrice differentiable function and a bijection, we want to evaluate the expression $ \frac{d^2 x}{dy^2}\left(\frac{dy}{dx}\right)^3 + \frac{d^2 y}{dx^2} $.

First, we know:

$ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} $

To find the second derivative of $ x $ with respect to $ y $, we differentiate:

$ \frac{d^2 x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{d}{dy}\left(\frac{dy}{dx}\right)^{-1} $

Applying the chain rule:

$ = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dy}{dx}} $

Rearranging terms, we have:

$ = -\left(\frac{dx}{dy}\right)^2 \cdot \frac{d^2 y}{dx^2} \cdot \left(\frac{dx}{dy}\right) $

Thus, simplifying further:

$ = -\left(\frac{dx}{dy}\right)^3 \cdot \frac{d^2 y}{dx^2} $

This simplifies our original expression:

$ \frac{d^2 x}{dy^2} \left(\frac{dy}{dx}\right)^3 = -\frac{d^2 y}{dx^2} $

Therefore:

$ \frac{d^2 x}{dy^2} \left(\frac{dy}{dx}\right)^3 + \frac{d^2 y}{dx^2} = 0 $

$\begin{aligned} & y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right) \\ & \begin{array}{l}\frac{d y}{d x}=\frac{1}{1+\left(\frac{2-3 \sin x}{3-2 \sin x}\right)^2} \\ \left\{\frac{(3-2 \sin x)(-3 \cos x)-(2-3 \sin x)(-2 \cos x)}{(3-2 \sin x)^2}\right\} \\ =\frac{-9 \cos x+6 \sin x \cos x+4 \cos x-6 \sin x \cos x}{(3-2 \sin x)^2+(2-3 \sin x)^2} \\ =\frac{-5 \cos x}{9+4 \sin ^2 x-12 \sin x+4+9 \sin ^2 x-12 \sin x} \\ =\frac{-5 \cos x}{13+13 \sin ^2 x-24 \sin x}\end{array}\end{aligned}$

To determine $\frac{dy}{dx}$, we begin with the expressions given for $x$ and $y$.

$ x = 3\left[\sin t - \log\left(\cot \frac{t}{2}\right)\right] $

Differentiating $x$ with respect to $t$:

$ \begin{align*} \frac{dx}{dt} &= 3\left[\cos t - \frac{1}{\cot \frac{t}{2}} \cdot (-\csc^2 \frac{t}{2}) \cdot \frac{1}{2}\right] \\ &= 3\left[\cos t + \frac{1}{\sin t}\right] \\ &= 3\left[\frac{\cos t \sin t + 1}{\sin t}\right] \end{align*} $

Next, differentiating $y$ with respect to $t$:

$ y = 6\left[\cos t + \log\left(\tan \frac{t}{2}\right)\right] $

$ \begin{align*} \frac{dy}{dt} &= 6\left[-\sin t + \frac{1}{\tan \frac{t}{2}} \cdot \sec^2 \frac{t}{2} \cdot \frac{1}{2}\right] \\ &= 6\left[-\sin t + \frac{1}{\sin t}\right] \\ &= 6\left[\frac{-\sin^2 t + 1}{\sin t}\right] \end{align*} $

Now, to find $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6\left[\frac{-\sin^2 t + 1}{\sin t}\right]}{3\left[\frac{\cos t \sin t + 1}{\sin t}\right]} $

The $\sin t$ terms cancel out:

$ = \frac{6(-\sin^2 t + 1)}{3(\cos t \sin t + 1)} $

Simplifying further:

$ = \frac{2(-\sin^2 t + 1)}{\cos t \sin t + 1} $

Rewriting the numerator, $1 - \sin^2 t = \cos^2 t$:

$ = \frac{2\cos^2 t}{\cos t \sin t + 1} $

Therefore, the expression for $\frac{dy}{dx}$ is:

$ \frac{dy}{dx} = \frac{2\cos^2 t}{1 + \sin t \cos t} $

To find the length of the tangent at the point $ P\left(\frac{\pi}{4}\right) $ on the curve $ x^{2/3} + y^{2/3} = 2^{2/3} $, we can proceed as follows:

The given equation of the curve can be parameterized as:

$ x = 2 \cos^3 \theta, \quad y = 2 \sin^3 \theta $

For the point $ P $, we substitute $ \theta = \frac{\pi}{4} $:

$ x = 2 \cos^3 \frac{\pi}{4}, \quad y = 2 \sin^3 \frac{\pi}{4} $

Calculating the coordinates:

$ x = 2 \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}, \quad y = 2 \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} $

So the point is $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

Next, differentiate the curve equation $ x^{2/3} + y^{2/3} = 2^{2/3} $ with respect to $ x $:

$ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 $

Solving for $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} $

At the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$,

$ \frac{dy}{dx} = -1 $

The formula to find the length of the tangent at a point is:

$ L = y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} $

Given $\frac{dx}{dy} = -1$, the length of the tangent at the point is calculated as:

$ L = \frac{1}{\sqrt{2}} \sqrt{1 + (-1)^2} = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1 $

Thus, the length of the tangent is $1$.

$\begin{aligned} & \text { Assertion : } \frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right) \\ & =\frac{(\log x)\left[x^2 \cos x+2 x \sin x\right]-x \sin x}{(\log x)^2} \\ & =\frac{x^2 \cos x \log x+2 x \sin x \log x-x \sin x}{(\log x)^2} \\ & =\frac{x^2 \sin x}{\log x}\left[\cot x+\frac{2}{x}-\frac{1}{x \log x}\right] \end{aligned}$

$\text { Reason : } \frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u w}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}+\frac{w^{\prime}}{w}\right]$

Given,

$\begin{aligned} & x=f(\theta) \\ & \text { and } y=g(\theta) \\ & \Rightarrow \frac{d x}{d \theta}=f^{\prime}(\theta), \frac{d y}{d \theta}=g^{\prime}(\theta) \\ & \Rightarrow \frac{d y}{d x}=\frac{g^{\prime}(\theta)}{f^{\prime}(\theta)} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{g^{\prime \prime}(\theta) f^{\prime}(\theta)-f^{\prime \prime}(\theta) g^{\prime}(\theta)}{\left(f^{\prime}(\theta)^2\right)}\right) \frac{d \theta}{d x} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\left(\frac{g^{\prime \prime}(\theta) f^{\prime}(\theta)-f^{\prime \prime}(\theta) g^{\prime}(\theta)}{\left(f^{\prime}(\theta)\right)^3}\right) \end{aligned}$

Given, equation $y=x^3-a x^2+48 x+7$

Differentiating the above equation with respect to the variable $x$, we have

$\frac{d y}{d x}=3 x^2-2 a x+48$

Thus, above function is a quadratic function.

So, $D<0$.

In $y^{\prime}=3 x^2-2 a x+48, A=3, B=-2 a$ and $c=48$

$\therefore \quad D<0$

$\Rightarrow(-2 a)^2-4 \cdot 3 \cdot 48<0 \Rightarrow 4 a^2-4 \cdot 3 \cdot 48<0$

$\Rightarrow 4\left(a^2-144\right)<0 \Rightarrow a^2-(12)^2<0$

$\Rightarrow(a-12)(a+12)<0$

$\Rightarrow a \in(-12,12)$

Given the equation:

$ 8 f(x) + 6 f\left(\frac{1}{x}\right) = x + 5 $

First, let's determine the value of $ f(1) $ by setting $ x = 1 $:

$ 8 f(1) + 6 f(1) = 1 + 5 $

$ 14 f(1) = 6 $

$ f(1) = \frac{6}{14} = \frac{3}{7} $

Next, we'll differentiate the given equation with respect to $ x $:

$ 8 f'(x) + 6 \left(-\frac{1}{x^2}\right) f'\left(\frac{1}{x}\right) = 1 $

$ 8 f'(x) - \frac{6}{x^2} f'\left(\frac{1}{x}\right) = 1 $

By setting $ x = 1 $:

$ 8 f'(1) - 6 f'(1) = 1 $

$ 2 f'(1) = 1 $

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

Now, let us find the derivative of $ x^2 f(x) $ with respect to $ x $:

$ y = x^2 f(x) $

$ \frac{dy}{dx} = 2x f(x) + x^2 f'(x) $

Evaluating at $ x = 1 $:

$ \left.\frac{dy}{dx}\right|_{x=1} = 2(1) f(1) + (1)^2 f'(1) $

$ = 2 \left(\frac{3}{7}\right) + \frac{1}{2} $

$ = \frac{6}{7} + \frac{1}{2} = \frac{12}{14} + \frac{7}{14} $

$ = \frac{19}{14} $

Therefore, the value of $\frac{d}{dx} (x^2 f(x))$ at $ x = 1 $ is $\frac{19}{14}$.

$f(x)=\cot ^{-1}\left(\frac{x^x+x^{-x}}{2}\right)$

Let $u=x^x$

$\Rightarrow \quad \log u=x \log x$

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

$\begin{aligned} & \frac{1}{u} \frac{d u}{d x}=x \cdot \frac{1}{x}+\log x \\ & \Rightarrow \quad \frac{d u}{d x}=x^x(1+\log x) \\ \end{aligned}$

Similarly, $v=x^{-x}$ gives

$\begin{aligned} \quad \frac{d v}{d x} & =-x^{-x}(1+\log x) \\ \therefore \quad f^{\prime}(x) & =-\frac{1}{1+\left(\frac{x^x-x^{-x}}{2}\right)^2} \cdot \frac{d}{d x}\left(\frac{x^x-x^{-x}}{2}\right) \end{aligned}$

$\begin{aligned} & =-\frac{4}{\left(x^x\right)^2+\left(x^{-x}\right)^2+2}\left[\frac{1}{2}(1+\log x)\left(x^x-\left(-x^{-x}\right)\right]\right. \\ & =-\frac{2}{x^{2 x}+x^{-2 x}+2}\left[(1+\log x)\left(x^x+x^{-x}\right)\right] \\ & \therefore f^{\prime}(1)=-\frac{2}{1+1+2}[(1+0)(1+1)] \\ & \quad=\left(-\frac{2}{4}\right)(2)=-1 \end{aligned}$

$\begin{aligned} & x= \sec \theta-\cos \theta \text {, then } \\ & \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta=\tan \theta(\sec \theta+\cos \theta) \\ & y=\sec ^n \theta-\cos ^n \theta \text {, then } \\ & \frac{d y}{d \theta}=n \sec ^{n-1} \theta \sec \theta \tan \theta+n \cos ^{n-1} \theta \sin \theta \\ &= n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right) \\ & \therefore \frac{d y}{d x}= \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{n \tan \theta\left(\sec ^n \theta+\cos ^n \theta\right)}{\tan \theta(\sec \theta+\cos \theta)} \\ &= \frac{n\left(\sec ^n \theta+\cos \theta\right)}{\sec \theta+\cos \theta} \\ &\left(\frac{d y}{d x}\right)^2= \frac{n^2(\sec \theta+\cos \theta)^2}{(\sec \theta+\cos \theta)^2} \\ &= \frac{n^2\left\{(\sec \theta-\cos \theta)^2+4\right\}}{(\sec \theta-\cos \theta)^2+4}=\frac{n^2\left(y^2+4\right)}{x^2+4} \\ & \Rightarrow \quad\left(x^2+4\right)\left(\frac{d y}{d x}\right)^2=n^2\left(y^2+4\right)\end{aligned}$

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

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

$\begin{aligned} \frac{d y}{d x} & =\frac{\left(4-x^2\right)^2}{\left(4-x^2\right)^2+16 x^2}\left[\frac{\left(4-x^2\right) 4+(4 x)(2 x)}{\left(4-x^2\right)^2}\right] \\ & =\frac{\left(4-x^2\right)^2}{16+x^4+8 x^2}\left[\frac{16-4 x^2+8 x^2}{\left(4-x^2\right)^2}\right] \\ & =\frac{16+4 x^2}{\left(x^2+4\right)^2}=\frac{4\left(x^2+4\right)}{\left(x^2+4\right)^2}=\frac{4}{x^2+4} \\ \frac{d y}{d x} & =\frac{4}{4+x^2} \end{aligned}$

$f'(x)=4x+3$

$f'(0)=3,f'(-1)=-1$

$\Rightarrow f'(0)+3f'(1)=3-3=0$

$\begin{aligned} & y=\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)\left(1+\frac{3}{x}\right)+\ldots+\left(1+\frac{n}{x}\right) \\ & \frac{d y}{d x}=\left(-\frac{1}{x^2}\right)\left(1+\frac{2}{x}\right) \ldots\left(1+\frac{n}{x}\right) \\ &+\left(1+\frac{1}{x}\right)\left(-\frac{2}{x^2}\right) \ldots\left(1+\frac{n}{x}\right)+\ldots \end{aligned}$

At $x=-1$

$d y / d x=-1(-1)(-2)(-n+1)+0=(-1)^n(n-1) !$

Given, $y \sqrt{1+x^2}=\log \left[\sqrt{1+x^2}-x\right]$

On differentiating both sides with respect to $x$, we get,

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

$\begin{aligned} & y=e^{x^2+e^{x^2+e^{x^2}+\ldots . \infty}} \\ & y=e^{x^2+y} \end{aligned}$

Taking $\log$ on both sides

$\log y=x^2+y$

Differentiating w.r.t. $x$

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