Differentiation

$ \begin{aligned} &\text { Given, }\\ &f(x)=\frac{e^{-x} \sin x}{\log _e x} \text { and } f^{\prime}(x)=f(x) \cdot g(x) \end{aligned} $

$ f^{\prime}(x)=\frac{\left(\log _e x\right)\left(e^{-x} \cos x-e^{-x} \sin x\right)-e^{-x} \frac{\sin x}{x}}{\left(\log _e x\right)^2} $

$ \begin{aligned} &= \frac{e^{-x}(\cos x-\sin x)}{\left(\log _e x\right)}-e^{-x} \frac{\sin x / x}{\left(\log _e x\right)^2} \\ & \text { Now, } g(x)=\frac{f^{\prime}(x)}{f(x)} \\ &=\frac{\cos x-\sin x}{\sin x}-\frac{1}{x \log _e x} \\ &=\cot x-1-\frac{1}{x \log _e x} \\ & \Rightarrow \quad g^{\prime}(x)=-\operatorname{cosec}^2 x+\frac{1}{\left(x \log _e x\right)^2} \\ & \,\,\,\,\,\,\,\, \quad\left(x \cdot \frac{1}{x}+\log x\right) \\ & \therefore \quad g^{\prime}(e)=-\operatorname{cosec}^2(e)+\frac{1}{e^2}(1+1) \\ &=2 e^{-2}-\operatorname{cosec}^2(e) \end{aligned} $

$ \text {} \begin{aligned} Given, \,\, y & =\frac{e^{\sin x}+\sinh ^3 x}{\cosh x-\tan x} \\ & =\frac{e^{\sin x}+\left[\frac{1}{2}\left(e^x-e^{-x}\right)\right]^3}{\frac{1}{2}\left(e^x+e^{-x}\right)-\tan x} \end{aligned} $

$ \begin{aligned} & \frac{d y}{d x}= \\ & {\left[\frac{1}{2}\left(e^x+e^{-x}\right)-\tan x\right]} \\ & {\left[\cos x e^{\sin x}+\frac{3}{8}\left(e^x-e^{-x}\right)^2\left(e^x+e^{-x}\right)\right]-} \\ & \frac{\left[e^{\sin x}+\frac{\left(e^x-e^{-x}\right)^3}{8}\right]\left[\frac{e^x-e^{-x}}{2}-\sec ^2 x\right]}{\left(\frac{e^x+e^{-x}}{2}-\tan x\right)^2} \end{aligned} $

$ \begin{aligned} &\text { On putting } x=0 \text {, }\\ &\frac{d y}{d x}=\frac{(1-0)(1+0)-(1+0)(0-1)}{\left(\frac{1+1}{2}-0\right)^2}=2 \end{aligned} $

$ \begin{aligned} & \text {Given, } \frac{d}{d x}\left(\frac{2 x+1}{(x+1)^2(x-2)}\right) \\ & =\frac{A}{(x-2)^2}+\frac{B}{(x+1)^3}+\frac{C}{(x+1)^2} \\ & \text { Let } \frac{2 x+1}{(x+1)^2(x-2)} \\ & =\frac{P}{x+1}+\frac{Q}{(x+1)^2}+\frac{R}{x-2} \\ & \Rightarrow 2 x+1=P(x+1)(x-2)+Q(x-2) +R(x+1)^2 \\ & \Rightarrow 2 x+1=P\left(x^2-x-2\right)+Q x-2 Q +R x^2+R+2 R x \\ & \Rightarrow 2 x+1=P x^2-P x-2 P+Q x-2 Q +R x^2+R+2 R x \end{aligned} $

$ \begin{aligned} & \Rightarrow P+R=0,-P+Q+2 R=2 \\ & \text { and } \quad-2 P-2 Q+R=1 \\ & \Rightarrow \quad P=-R, R+Q+2 R=2 \\ & \Rightarrow \quad Q+3 R=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } 2 R-2 Q+R=1 \\ & \Rightarrow \quad-2 Q+3 R=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} &\text { Subtracting Eq. (ii) from Eq. (i), }\\ &\begin{aligned} & \quad 3 Q=1 \Rightarrow Q=\frac{1}{3} \\ & \therefore \quad \frac{1}{3}+3 R=2 \Rightarrow 3 R=2-\frac{1}{3} \\ & \Rightarrow \quad 3 R=\frac{5}{3} \Rightarrow R=\frac{5}{9} \\ & \therefore \quad P=-\frac{5}{9} \\ & \frac{2 x+1}{(x+1)^2(x-2)}=-\frac{5}{9} \times \frac{1}{x+1} \\ & \quad+\frac{1}{3} \times \frac{1}{(x+1)^2}+\frac{5}{9} \times \frac{1}{(x-2)} \\ & \frac{d}{d x}\left(\frac{2 x+1}{(x+1)^2(x-2)}\right)=\frac{5}{9} \frac{1}{(x+1)^2} \\ & \quad+\left(-\frac{2}{3}\right) \times \frac{1}{(x+1)^3}+\left(-\frac{5}{9}\right) \frac{1}{(x-2)^2} \\ & \text { Here, } A=-\frac{5}{9}, B=-\frac{2}{3} \text { and } C=\frac{5}{9} \\ & \therefore \quad A+B+C=-\frac{5}{9}-\frac{2}{3}+\frac{5}{9}=-\frac{2}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} \text { } \begin{array}{r} \frac{d}{d x}\left[\left(x^{\frac{5}{2}}-x^{\frac{3}{2}}+1\right)\left(x^2-3 x+5\right)\right] \\ =\frac{d}{d x}\left[x^{\frac{9}{2}}-x^{\frac{7}{2}}+x^2-3 x^{\frac{7}{2}}+3 x^{\frac{5}{2}}-3 x\right. \left.+5 x^{\frac{5}{2}}-5 x^{\frac{3}{2}}+5\right] \\ =\frac{d}{d x}\left[x^{\frac{9}{2}}-4 x^{\frac{7}{2}}+8 x^{\frac{5}{2}}-5 x^{\frac{3}{2}}+x^2\right. -3 x+5] \\ =\frac{9}{2} x^{\frac{7}{2}}-4 \times \frac{7}{2} x^{\frac{5}{2}}+8 \times \frac{5}{2} x^{\frac{3}{2}} -5 \times \frac{3}{2} x^{\frac{1}{2}}+2 x-3 \\ =\frac{9}{2} x^{\frac{7}{2}}-14 x^{\frac{5}{2}}+20 x^{\frac{3}{2}}-\frac{15}{2} x^{\frac{1}{2}} +2 x-3 \end{array} \end{aligned} $

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

$ \begin{aligned} & =\frac{1}{2} \cot \sqrt{\frac{x^2+1}{x^2+2}} \times \sqrt{\frac{x^2+2}{x^2+1}} \times \frac{2 x}{\left(x^2+4^2\right.} \\ & \therefore \frac{d}{d x}\left[\log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)\right] \\ & \text { at } x=\sqrt{2} \\ & \quad=\frac{1}{2} \cot \sqrt{\frac{2+1}{2+2}} \times \sqrt{\frac{2+2}{2+1}} \times \frac{2 \sqrt{2}}{(2+2)^2} \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \cot \left(\frac{\sqrt{3}}{2}\right) \times \frac{2}{\sqrt{3}} \times \frac{2 \sqrt{2}}{16} \\ & =\frac{\sqrt{2}}{8 \sqrt{3}} \cot \left(\frac{\sqrt{3}}{2}\right) \end{aligned} $

$ f(x)=\frac{1+\sec x}{2(\sec x-1)} $

$ \Rightarrow \quad f^{\prime}(x)=\frac{1}{2}\left[\begin{array}{l} (\sec x-1) \sec x \tan x \\ \frac{-(1+\sec x) \sec x \tan x}{(\sec x-1)^2} \end{array}\right] $

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

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

$ \begin{aligned} & =-\frac{\sec x \tan x}{(\sec x-1)^2} \div \frac{1+\sec x}{2(\sec x-1)} \\ & =-\frac{\sec x \tan x}{(\sec x-1)^2} \times \frac{2(\sec x-1)}{(\sec x+1)} \\ & =-\frac{2 \sec x \tan x}{\sec ^2 x-1} \\ & =-\frac{2 \sec x \tan x}{\tan ^2 x}=-\frac{2 \sec x}{\tan x} \\ & =-2 \operatorname{cosec} x \end{aligned} $

$ \begin{aligned} & \frac{3 x+5}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3} \\ & 3 x+5=A\left(2 x^2+3\right)+(B x+C)(x+1) \\ & 3 x+5=2 A x^2+3 A+B x^2+B x+C x+C \end{aligned} $

Therefore,  $2 A+B=0, B+C=3$

$ \begin{gathered} 3 A+C=5 \\ B=-2 A,-2 A+C=3,3 A+C=5 \\ \Rightarrow \quad A=2 / 5, C=19 / 5 \Rightarrow B=-4 / 5 \\ f(x)=\frac{2}{5} x^3-\frac{4}{5} x^2+7 x+\frac{19}{5} \\ f^{\prime}(x)=\frac{6}{5} x^2-\frac{8}{5} x+7 \\ f^{\prime}(-2)=\frac{6}{5}(-2)^2-\frac{8}{5}(-2)+7 \\ =\frac{24}{5}+\frac{16}{5}+7 \end{gathered} $

Given, $f(x)=\sin x, g(x)=\cos x$,

$ h(x)=x^2 $

Let $P(x)=f(g(h(x)))=\sin \left(\cos x^2\right)$

$ \begin{aligned} & \lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=p^{\prime}(1) \\ & p^{\prime}(x)=\cos \left(\cos x^2\right) \times\left(-\sin x^2\right)(2 x) \\ & p^{\prime}(1)=-2 \cos (\cos 1) \sin (1) \end{aligned} $

Given, $x \cos (k+y)=\cos y$

$ \Rightarrow \frac{\cos y}{\cos (k+y)}=x $

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

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

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{\cos ^2(k+y)}{\sin (k+y-y)} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{\cos ^2(k+y)}{\sin k} \\ & \left.\therefore \frac{d y}{d x}\right|_{y=\pi / 2}=\frac{\cos ^2\left(\frac{\pi}{2}+k\right)}{\sin k}=\frac{\sin ^2 k}{\sin k} \\ & =\sin k \end{aligned} $

Given, $x=a(\cos \theta+\theta \sin \theta)$

On differentiating w.r.t. $\theta$,

$ \frac{d x}{d \theta}=a(-\sin \theta+\theta \cos \theta+\sin \theta) $

$ \Rightarrow \quad d x / d \theta=a \theta \cos \theta $

$ \begin{array}{ll} \therefore & y=f(\theta) \Rightarrow d y / d \theta=f^{\prime}(\theta) \\ \therefore & \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{f^{\prime}(\theta)}{a \theta \cos \theta} \end{array} $

$ \begin{array}{ll} \Rightarrow & \frac{\tan \theta}{\theta}=\frac{f^{\prime}(\theta)}{a \theta \cos \theta} \Rightarrow f^{\prime}(\theta)=a \sin \theta \\ \Rightarrow & f(\theta)=-a \cos \theta+c \Rightarrow f(2 \pi)=0 \\ & -a \cos (2 \pi)+c=0 \\ \Rightarrow& \quad c=a \Rightarrow f(\theta)=a-a \cos \theta \\ & f\left(\frac{\pi}{3}\right)=a-a \cos \frac{\pi}{3}=a-\frac{a}{2}=\frac{a}{2} \end{array} $

$a f(x)+b f\left(\frac{1}{x}\right)=x+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Substitute $x$ by $1 / x$

$ \begin{aligned} &a f\left(\frac{1}{x}\right)+b f(x)=\frac{1}{x}+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { From Eq. (i) } \times a-\text { Eq. (ii) } \times b \text {, } \end{aligned} $

$ \begin{aligned} \Rightarrow x^2 f(x)= & \frac{1}{\left(a^2-b^2\right)}\left[a x^3+(a-b) x^2-b x\right] \\ \frac{d}{d x}\left(x^2 f(x)\right)= & \frac{1}{a^2-b^2}[ \left. 3 a x^2+2(a-b) x-b\right] \end{aligned} $

According to the question,

$ \begin{aligned} & & \frac{3 a}{a^2-b^2} & =2, \frac{a-b}{a^2-b^2}=1,-\frac{b}{a^2-b^2}=\frac{1}{3} \\ \Rightarrow & & \frac{3 a}{2} & =a-b=-3 b=a^2-b^2 \\ & \Rightarrow & 3 a & =2 a-2 b \Rightarrow a=-2 b \end{aligned} $

Also $\frac{a-b}{a^2-b^2}=1$ and $\frac{1}{a+b}=1$

$ \begin{array}{ll} \Rightarrow & a+b=1 \Rightarrow \\ \Rightarrow & b=-1 \Rightarrow \quad a+b=1 \\ \therefore & a-b=2-(-1)=3 \end{array} $

$ \begin{aligned} & \text { } f(x)=\sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right) \\ & f^{\prime}(x)=\cos \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right) \\ & \times \sinh \left(\frac{x^2+1}{x^2+2}\right) \times \frac{2 x}{\left(x^2+2\right)^2} \\ & \therefore f^{\prime}(1)=\cos \left(\cosh \left(\frac{2}{3}\right)\right) \times \sinh \left(\frac{2}{3}\right) \times \frac{2}{9} \\ & =\frac{2}{9} \sinh \left(\frac{2}{3}\right) \cos \left(\cosh \left(\frac{2}{3}\right)\right) \end{aligned} $

$ \begin{aligned} & \text { Given, } \\ & f(x)=\log _e\left[e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right], x \neq \frac{-5}{3}, \frac{5}{3} \\ & f(x)=\log _e e^{2 x}+\log _e\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}} \\ & f(x)=2 x+\frac{2}{3}\left[\log _e(3 x+5)-\log (5-3 x)\right] \\ & \frac{d f}{d x}=2+\frac{2}{3}\left[\frac{3}{3 x+5}-\frac{(-3)}{5-3 x}\right] \end{aligned} $

At $x=1$,

$ \begin{aligned} \frac{d f}{d x}=2+\frac{2}{3}\left[\frac{3}{8}+\frac{3}{2}\right] & =2+\frac{2}{3} \times \frac{15}{8} \\ & =2+\frac{5}{4}=\frac{13}{4} \end{aligned} $

$ \begin{aligned} &\text { Here, } x=\operatorname{cosec} \theta-\sin \theta\\ &\begin{gathered} x^2+4=(\operatorname{cosec} \theta+\sin \theta)^2 \\ y=\operatorname{cosec}^{2022} \theta-\sin ^{2022} \theta \\ y^2+4=\left(\operatorname{cosec}^{2022} \theta+\sin ^{2022} \theta\right)^2 \\ \frac{d y}{d \theta}=2022 \operatorname{cosec}^{2021} \theta(-\operatorname{cosec} \theta \cot \theta) \\ -2022 \sin ^{2021} \theta \cos \theta \\ =-2022 \cos \theta\left(\frac{\operatorname{cosec}^{2022} \theta}{\sin \theta}+\sin ^{2021} \theta\right) \\ \frac{d y}{d \theta}=-2022 \cot \theta\left(\operatorname{cosec}^{2022} \theta+\sin ^{2022} \theta\right) \\ \frac{d x}{d \theta}=-\operatorname{cosec} \theta \cot \theta-\cos \theta \\ =-\cot \theta(\operatorname{cosec} \theta+\sin \theta) \\ \frac{d y}{d x}=\frac{2022\left(\operatorname{cosec} \operatorname{cosi}^{2022} \theta+\sin \cos ^{2022} \theta\right)}{\operatorname{cosec} \theta+\sin \theta} \end{gathered} \end{aligned} $

$ \begin{aligned} & \quad\left(\frac{d y}{d x}\right)^2=\frac{(2022)^2\left(y^2+4\right)}{x^2+4} \\ & \Rightarrow \frac{k\left(y^2+4\right)}{g(x)}=\frac{(2022)^2\left(y^2+4\right)}{x^2+4} \\ & \quad k=(2022)^2, g(x)=x^2+4 \\ & \therefore \quad 10+k-g(2022) \\ & =10+(2022)^2-\left((2022)^2+4\right)=6 \end{aligned} $

(a) $\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}},|x|>1$

(b) $\frac{d}{d x}\left(\tanh ^{-1} x\right)=\frac{1}{1-x^2}, x \in(-1,1)$

(c) $\frac{d}{d x}\left(\operatorname{coth}^{-1} x\right)=\frac{1}{1-x^2}, x \in R-[-1,1]$

(d) $\frac{d}{d x}\left(\operatorname{cosech}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$

Given that,

$ \begin{equation*} f(x)=\frac{x-1}{e^x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Differentiating Eq. (i) w.r.t. $x$ on both sides, we get

$ \begin{align*} & f^{\prime}(x)=\frac{e^x \frac{d}{d x}(x-1)-(x-1) \frac{d}{d x}\left(e^x\right)}{\left(e^x\right)^2} \\ & {\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d}{d x} u-u \frac{d v}{d x}}{v^2}\right] } \\ &=\frac{e^x \cdot 1-(x-1) e^x}{e^{2 x}} \\ & f^{\prime}(x)=\frac{2-x}{e^x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{align*} $

Putting $x=0$ in Eq. (ii), we get

$ \begin{equation*} f^{\prime}(0)=\frac{2-0}{e^0}=\frac{2}{1} \Rightarrow f^{\prime}(0)=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{equation*} $

Now, differentiating Eq. (ii) w.r.t. $x$ on both sides, we get

$ \begin{align*} f^{\prime \prime}(x) & =\frac{e^x \frac{d}{d x}(2-x)-(2-x) \frac{d}{d x}\left(e^x\right)}{\left(e^x\right)^2} \\ & =\frac{e^x(-1)-(2-x) e^x}{e^{2 x}} \\ f^{\prime \prime}(x) & =\frac{(x-3)}{e^x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $

Putting $x=0$, in Eq. (iv), we get

$ \begin{equation*} f^{\prime \prime}(0)=\frac{0-3}{e^0}=\frac{-3}{1} \Rightarrow f^{\prime \prime}(0)=-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{equation*} $

Adding Eq. (iii) and Eq. (v), we get

$ \begin{aligned} & & f^{\prime}(0)+f^{\prime \prime}(0) & =2+(-3) \\ \Rightarrow & & f^{\prime}(0)+f^{\prime \prime}(0) & =-1 \end{aligned} $

Given that,

$ \begin{gather*} \left(\frac{d y}{d x}\right)=\frac{1}{\left(\frac{d x}{d y}\right)} \text { or } \\ \frac{d y}{d x}=\left(\frac{d x}{d y}\right)^{-1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{gather*} $

Differentiating Eq. (i) on both sides w.r.t. $x$, we get

$ \begin{array}{ll} & \frac{d^2 y}{d x^2}=-\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d}{d x}\left(\frac{d x}{d y}\right)\right\} \\ \Rightarrow & \frac{d^2 y}{d x^2}=-\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d}{d x}\left(\frac{d x}{d y}\right) \frac{d y}{d x}\right\} \text { [from chain rule] }\\ \Rightarrow & \frac{d^2 y}{d x^2}=\left(\frac{d x}{d y}\right)^{-2}\left\{\frac{d^2 x}{d y^2} \cdot \frac{d x}{d y}\right\} \\ \Rightarrow & \frac{d^2 y}{d x^2}=-\left(\frac{d y}{d x}\right)^2 \frac{d^2 x}{d y^2} \\ \Rightarrow & \frac{d^2 x}{d y^2}+\left(\frac{d y}{d x}\right)^3 \frac{d^2 x}{d y^2}=0 \end{array} $

Comparing with given expression, we get

$ k=0 $

Now, $e^{k f(x)}-k f(x)=e^{0 \times f(x)}-0 \times f(x)$

$ e^{k f(x)}-k f(x)=e^0-0=1\left[\because e^0=1\right] $

$ \begin{aligned} & \text { We have, } \frac{d}{d x}\left(\operatorname{cosech} \mathrm{~h}^{-1}(\tan 2 x)\right. \\ & =\frac{-1}{|\tan 2 x|} \times \frac{1}{\sqrt{1+\tan ^2 2 x}} \times 2 \sec ^2 2 x \\ & \quad\left[\frac{d}{d x} \operatorname{cosech} \mathrm{~h}^{-1} x=\frac{-1}{|x| \sqrt{1+x^2}}\right] \\ & =-2|\operatorname{cosec} 2 x| \end{aligned} $

We have, $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$

On differentiating w.r.t. ' $x$ ', $y$ as constant

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

Put, $x=0, y=2 x$ we get

$ \begin{aligned} & \quad f^{\prime}(x)=f^{\prime}(0)=-1 \\ & {\left[\because f^{\prime}(0)=-1\right]} \\ & \therefore f(x)=-x+c \\ & \Rightarrow f(0)=0+c \Rightarrow 1=c \\ & \therefore f(x)=-x+1, f(2)=-2+1=-1 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} f(x) & =\tan ^{-1}\left(\frac{1}{\sin ^2 x+\sin x+1}\right) \\ & +\tan ^{-1}\left(\frac{1}{\sin ^2 x+3 \sin x+3}\right) \\ & +\tan ^{-1}\left(\frac{1}{\sin ^2 x+5 \sin x+7}\right)+\ldots \\ & + \text { upto } 10 \text { terms } \end{aligned} \end{aligned} $

$ \begin{aligned} & =\tan ^{-1}\left(\frac{(\sin x+1)-(\sin x+0)}{1+(\sin x+0)(\sin x+1)}\right) \\ & +\tan ^{-1}\left(\frac{(\sin x+2)-(\sin x+1)}{1+(\sin x+1)(\sin x+2)}\right) \\ & +\tan ^{-1}\left(\frac{(\sin x+3)-(\sin x+2)}{1+(\sin x+2)(\sin x+3)}\right)+\ldots \\ & + \text { upto } 10 \text { terms } \\ & =\tan ^{-1}(\sin x+1)-\tan ^{-1}(\sin x) \\ & +\tan ^{-1}(\sin x+2)-\tan ^{-1}(\sin x+1) \\ & +\tan ^{-1}(\sin x+3)-\tan ^{-1}(\sin x+2)+\ldots \\ & +\operatorname{upto}^{-1} 0 \text { terms } \\ & =\tan ^{-1}(\sin x+10)-\tan ^{-1}(\sin x) \end{aligned} $

$ \begin{aligned} \therefore f^{\prime}(x) & =\frac{\cos x}{1+(\sin x+10)^2}-\frac{\cos x}{1+(\sin x)^2} \\ \therefore f^{\prime}(0) & =\frac{1}{1+(0+10)^2}-\frac{1}{1+0} \\ & =\frac{1}{101}-1=\frac{1-101}{101}=\frac{-100}{101} \end{aligned} $

$ \begin{aligned} &\text {We have, }\\ &\begin{aligned} & f(x)=x^2+3 x-3 \\ \Rightarrow & y=x^2+3 x-3 \\ \Rightarrow & y=x^2+3 x-3+\frac{9}{4}-\frac{9}{4} \\ \Rightarrow & y=\left(x+\frac{3}{2}\right)^2-\frac{21}{4} \\ \Rightarrow & y+\frac{21}{4}=\left(x+\frac{3}{2}\right)^2 \\ \Rightarrow & x+\frac{3}{2}=\sqrt{y+\frac{21}{4}} \\ \Rightarrow & x=\sqrt{y+\frac{21}{4}}-\frac{3}{2} \\ \Rightarrow & f^{-1}(y)=\sqrt{y+\frac{21}{4}}-\frac{3}{2} \end{aligned} \end{aligned} $

$ \Rightarrow f^{-1}(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}=g(x) \text { (given) } $

Minimum value of $f^{-1}(x)$ is $\frac{-3}{2}$

$ \begin{array}{ll} \therefore & \alpha=-\frac{3}{2} \\ \therefore & x=\alpha+\frac{5}{2}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1 \end{array} $

Now, $g(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}$

$ \begin{aligned} & \Rightarrow \quad g^{\prime}(x)=\frac{1}{2 \sqrt{x+\frac{21}{4}}} \\ & \text { at } \quad x=1, g^{\prime}(x)=\frac{1}{2 \sqrt{1+\frac{21}{4}}}=\frac{1}{5} \end{aligned} $

We have, $F(x)=f(x) g(x)$

$ \begin{aligned} & \Rightarrow \quad F^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x) \text { and } \\ & G(x)=f^{\prime}(x) g^{\prime}(x) \end{aligned} $

Now, $F^{\prime}(x)=G(x) H(x)$

$ \begin{array}{ll} \therefore & H(x)=\frac{F^{\prime}(x)}{G(x)} \\ & =\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f^{\prime}(x) g^{\prime}(x)}=\frac{g}{g^{\prime}}+\frac{f}{f^{\prime}} \end{array} $

Again, $F^{\prime}(x)=F(x) K(x)$

$ \begin{aligned} & \Rightarrow \quad K(x)=\frac{F^{\prime}(x)}{F(x)} \\ & \quad=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f(x) g(x)}=\frac{f^{\prime}}{f}+\frac{g^{\prime}}{g} \\ & \therefore H(x)+K(x)=\frac{f^{\prime}}{f}+\frac{g}{g^{\prime}}+\frac{f}{f^{\prime}}+\frac{g^{\prime}}{g} \end{aligned} $

$ \text { We have, } y=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\log \sqrt{1-x^2} $

$ \begin{aligned} & \therefore \\ & \quad\left[1 \cdot \sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right] \end{aligned} $

$ \begin{array}{r} \frac{d y}{d x}=\frac{\left.\sqrt{1-x^2}-x \sin ^{-1} x\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)\right]}{\left(1-x^2\right)} +\frac{1}{2} \frac{1}{\left(1-x^2\right)}(-2 x) \end{array} $

$ \begin{array}{r} =\frac{1}{1-x^2}\left[\sqrt{1-x^2} \sin ^{-1} x+x+\frac{x^2 \sin ^{-1} x}{\sqrt{1-x^2}}\right. -x] \end{array} $

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

We have,

$ \begin{aligned} & & f(x) & =x^2+x g^{\prime}(1)+g^{\prime \prime}(2) \\ \Rightarrow & & f^{\prime}(x) & =2 x+g^{\prime}(1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \Rightarrow & & f^{\prime \prime}(x) & =2 \Rightarrow f^{\prime \prime}(x)=0 \end{aligned} $

Again, $g(x)=f(1) x^2+x f^{\prime}(x)+f^{\prime \prime}(x)$

$ \Rightarrow g^{\prime}(x)=2 x f(1)+f^{\prime}(x)+x f^{\prime \prime}(x)+f^{\prime \prime \prime}(x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Put $x=1$

$ \Rightarrow \quad g^{\prime}(\mathrm{l})=2 f(\mathrm{l})+f^{\prime}(\mathrm{l})+f^{\prime \prime}(\mathrm{l})+0 $

Put $x=1$ in Eq. (i), we get

$ f^{\prime}(1)=2+g^{\prime}(1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

From Eq. (ii) and Eq. (iii), we get

$ \begin{aligned} g^{\prime}(\mathrm{l}) & =2 f(\mathrm{l})+2+g^{\prime}(\mathrm{l})+2 \\ 2 f(\mathrm{l})+4 & =0 \Rightarrow f(\mathrm{l})=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Now, again,

$ \begin{aligned} & g^{\prime \prime}(x)=2 f(1)+f^{\prime \prime}(x)+f^{\prime \prime}(x)+x f^{\prime \prime}(x) \\ & =2 f(1)+4=-4+4 \\ & \\ & \,\,\,\,\ \quad g^{\prime \prime}(x)=0 \\ & \therefore\quad f(x)=x^2+x g^{\prime}(1)+g^{\prime \prime}(2) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad=x^2+x g^{\prime}(1) \end{aligned} $

$ \begin{aligned} &\text { Put, } x=1\\ &\begin{array}{ll} & f(1)=1+g^{\prime}(1) \Rightarrow-2=1+g^{\prime}(1) \\ \Rightarrow & g^{\prime}(1)=-3 \\ \therefore & f(x)= \\ \therefore & f(x)=f(1) x^2+x f^{\prime}(x)+f^{\prime \prime}(x) \\ & =-2 x^2+x(2 x-3)+2 \\ & =-2 x^2+2 x^2-3 x+2 \\ & =2-3 x \\ \therefore & f(x)-g(x)=x^2-3 x-2+3 x \\ & = x^2-2 \end{array} \end{aligned} $