Differentiation

We have, $f(x)=\sqrt{x}$

$ (x \geq 0) $

$ \begin{aligned}and\,\, g(x) & =1+x^2 \\ f \circ g(x) & =\sqrt{x^2+1} \\ f^{\prime} g^{\prime}(x) & =\frac{1}{2 \sqrt{x^2+1}} \times 2 x \\ f \circ g^{\prime}(1) & =\frac{1}{2 \sqrt{1^2+1}} \times 2 \times 1=\frac{1}{\sqrt{2}} \end{aligned} $

$ \begin{aligned} & \text { (i) } x=a(\theta-\sin \theta), y=a(1-\cos \theta) \\ & \begin{aligned} & \frac{d x}{d \theta}=a(1-\cos \theta), \frac{d y}{d \theta}=a \sin \theta \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\sin \theta}{1-\cos \theta} \\ &=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}}=\cot \frac{\theta}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}}=\cot \frac{\pi}{6}=\sqrt{3}\\ &\text { }\\ &\begin{aligned}(ii) & x=3 \cos \theta-2 \cos ^3 \theta \\ & \frac{d x}{d \theta}=-3 \sin \theta+6 \cos ^2 \theta \sin \theta \\ & y=3 \sin \theta-2 \sin ^3 \theta \\ & \frac{d y}{d \theta}=3 \cos \theta-6 \sin ^2 \theta \cos \theta \\ & \frac{d y}{d x}=\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta} \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(\cos ^2 \theta-1\right)}=\cot \theta \\ \left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}} & =\cot \frac{\pi}{3}=\frac{1}{\sqrt{3}} \\ \text { (iii) } x & =3 \cos \theta-\cos ^3 \theta \\ \frac{d x}{d \theta} & =-3 \sin \theta+3 \cos ^2 \theta \sin \theta \\ y & =3 \sin \theta-\sin ^3 \theta \\ \frac{d y}{d \theta} & =3 \cos \theta-3 \sin ^2 \theta \cos \theta \\ \frac{d y}{d x} & =\frac{3 \cos \theta\left(1-\sin ^2 \theta\right)}{3 \sin \theta\left(\cos ^2 \theta-1\right)} \\ & =\cot \theta \times \frac{\cos ^2 \theta}{-\left(\sin ^2 \theta\right)}=\cot ^3 \theta \end{aligned} $

$ \begin{aligned} \left.\frac{d y}{d x}\right|_{\theta=\frac{\pi}{3}} & =-\cot ^3 \frac{\pi}{3} \\ & =-\left(\frac{1}{\sqrt{3}}\right)^3=-\frac{1}{3 \sqrt{3}} \end{aligned} $

(iv) $\frac{d y}{d x}=\frac{a \sec ^2 \theta}{\frac{a}{\sin \theta} \cos \theta}=\frac{\sin \theta}{\cos ^3 \theta}$

At $\theta=\frac{\pi}{3}$,

$ \frac{d y}{d x}=\frac{\sqrt{3} / 2}{1 / 8}=4 \sqrt{3} $

Hence, option (c) is correct.

$ \begin{aligned} &\\ &\begin{aligned} & \text { } y=(\sin x) x \\ & \frac{d y}{d x}=x \cos x+\sin x \\ & \text { Now, } \frac{\frac{d y}{d x}-\frac{y}{x}}{x \frac{d y}{d x}-y}=\frac{\frac{d y}{d x}-\frac{y}{x}}{x\left(\frac{d y}{d x}-\frac{y}{x}\right)}=\frac{1}{x}=\frac{1}{\alpha}=1 \\ & \therefore \quad \alpha=1 \end{aligned} \end{aligned} $

Given,

$ 2 x^2-3 x y+y^2+x+2 y-8=0 $

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

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

$ \begin{aligned}Now, \frac{g(2,2)}{f(1,1)} & =\frac{-4(2)+3(2)-1}{-3(1)+2(1)+2} \\ & =\frac{-8+6-1}{-3+2+2} \\ & =\frac{-3}{1}=-3 \end{aligned} $

Given, $f(x)=e^x$ and $h(x)=f \circ f(x)$

$ \begin{aligned} h(x) & =f \circ f(x)=f(f(x))=f\left(e^x\right)=e^{e^x} \\ \Rightarrow \log h(x) & =\log e^{e^x}=e^x \end{aligned} $

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

$ \frac{1}{h(x)} \cdot h^{\prime}(x)=e^x \Rightarrow \frac{h^{\prime}(x)}{h(x)}=\log h(x) $

$\sin y=\sin 3 t \Rightarrow \cos y \frac{d y}{d t}=3 \cos 3 t$

and $x=\sin t \Rightarrow d x / d t=\cos t$

$ \begin{aligned}Then, \frac{d y}{d x} & =\frac{3 \cos 3 t}{\cos y \cdot \cos t} \\ & =\frac{3 \cos 3 t}{\cos 3 t \cos t} \quad[\because \sin y=\sin 3 t] \\ & =\frac{3}{\cos t} \quad\left[\because \cos \theta=\sqrt{1-\sin ^2 \theta}\right] \\ & =\frac{3}{\sqrt{1-\sin ^2 t}} \\ & =\frac{3}{\sqrt{1-x^2}} \quad \quad[\because x=\sin t] \end{aligned} $

$ \begin{aligned} &\text {Given, }\\ &\begin{aligned} & f(x)=\sqrt{\log \left(x^2+x+1\right)+\sqrt{\cosh (2 x-3)}} \\ & f(x)=\left[\log \left(x^2+x+1\right)+(\cosh (2 x-3))^{1 / 2}\right]^{1 / 2} \\ & f^{\prime}(x)=\frac{1}{2}\left[\log \left(x^2+x+1\right)\right. \\ & \left.\quad+(\cosh (2 x-3))^{1 / 2}\right]^{-1 / 2} \\ & {\left[\frac{d}{d x} \log \left(x^2+x+1\right)+\frac{d}{d x}(\cosh (2 x-3))^{1 / 2}\right]} \\ & =\frac{1}{2}\left(\log \left(x^2+x+1\right)+(\cosh (2 x-3))^{1 / 2}\right)^{-1 / 2} \end{aligned} \end{aligned} $

$ \begin{aligned} & \quad\left[\begin{array}{r} \frac{2 x+1}{x^2+x+1}+\frac{1}{2}(\cosh (2 x-3))^{-1 / 2} \\ \times \sinh (2 x-3) \times 2 \end{array}\right] \\ & f^{\prime}(0)=\frac{1}{2}\left[\log (0+0+1)+(\cosh (-3))^{1 / 2}\right]^{-1 / 2} \\ & \quad\left[\frac{1}{1}+\frac{1}{2}(\cosh (-3))^{-1 / 2} \times \sinh (-3) \times 2\right] \\ & =\frac{1}{2}\left[0+(\cosh 3)^{1 / 2}\right]^{-1 / 2} \times\left[1+(\cosh 3)^{-1 / 2}\right.\times(-\sin h 3)] \\ & =\frac{1}{2}(\cosh 3)^{-1 / 4}\left(1-\frac{\sinh 3}{(\cosh 3)^{1 / 2}}\right) \\ & =\frac{\cosh (3)-\sinh (3)}{2(\cosh (3))^{3 / 4}} \end{aligned} $

Given, $x=\cos ^3 \theta-\sin ^3 \theta$ and $y=(\cos \theta)^{1 / 3}-(\sin \theta)^{1 / 3}$

On differentiating $x$ and $y$ separately w.r.t. θ, we get

$ \begin{aligned} & \frac{d x}{d \theta}=3 \cos ^2 \theta(-\sin \theta)-3 \sin ^2 \theta(\cos \theta) \\ & \Rightarrow \frac{d x}{d \theta}=-3 \cos \theta \sin \theta(\cos \theta+\sin \theta)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { and } \frac{d y}{d \theta}=\frac{1}{3}(\cos \theta)^{-2 / 3}(-\sin \theta) \\ & \left.\qquad-\frac{1}{3}(\sin \theta)^{-2 / 3}\right] \cos \theta \end{aligned} $

$ \begin{aligned} & =-\frac{1}{3}\left[\frac{\sin \theta}{(\cos \theta)^{2 / 3}}+\frac{\cos \theta}{(\sin \theta)^{2 / 3}}\right] \\ & =-\frac{1}{3} \sin \theta \cos \theta\left[(\cos \theta)^{-5 / 3}+(\sin \theta)^{-5 / 3}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Now, $\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}$

$ \begin{aligned} = & \frac{-\frac{1}{3} \sin \theta \cos \theta\left[(\cos \theta)^{-5 / 3}+(\sin \theta)^{-5 / 3}\right]}{-3 \sin \theta \cos \theta[\sin \theta+\cos \theta]} \\ \frac{d y}{d x}= & \frac{1}{9}\left[\frac{(\cos \theta)^{-5 / 3}+(\sin \theta)^{-5 / 3}}{(\sin \theta+\cos \theta)}\right] \end{aligned} $

At $\quad \theta=\frac{\pi}{4}$

$ \begin{aligned} \left.\frac{d y}{d x}\right|_{\theta=\pi / 4} & =\frac{1}{9}\left[\frac{\left(\cos \frac{\pi}{4}\right)^{-5 / 3}+\left(\sin \frac{\pi}{4}\right)^{-5 / 3}}{\sin \frac{\pi}{4}+\cos \frac{\pi}{4}}\right] \\ & =\frac{1}{9}\left[\frac{\left(\frac{1}{\sqrt{2}}\right)^{-5 / 3}+\left(\frac{1}{\sqrt{2}}\right)^{-5 / 3}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}\right] \\ & =\frac{1}{9}\left[\frac{2\left(\frac{1}{\sqrt{2}}\right)^{-5 / 3}}{2\left(\frac{1}{\sqrt{2}}\right)}\right]=\frac{1}{9}\left(\frac{1}{\sqrt{2}}\right)^{-\frac{5}{3}-1} \\ & =\frac{1}{9}\left(\frac{1}{\sqrt{2}}\right)^{-8 / 3} \\ & =\frac{(\sqrt{2})^{8 / 3}}{9}=\frac{(2)^{8 / 3 \times 1 / 2}}{9}=\frac{2^{4 / 3}}{9} \\ & =\frac{2 \sqrt[3]{2}}{9} \end{aligned} $

Given,

$ 2 x^2+3 x y-y^2+4 x-5 y+6=0 $

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

$ \begin{aligned} & 4 x+3 y+3 x \frac{d y}{d x}-2 y \frac{d y}{d x}+4-\frac{5 d y}{d x}=0 \\ & \Rightarrow \quad-3 x \frac{d y}{d x}+2 y \frac{d y}{d x}+5 \frac{d y}{d x}=4 x+3 y+4 \\ & \Rightarrow \quad(-3 x+2 y+5) \frac{d y}{d x}=4 x+3 y+4 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{4 x+3 y+4}{-3 x+2 y+5} \end{aligned} $

$ \begin{aligned} \left(\frac{d y}{d x}\right)_{(1,-2)} & =\frac{4 \times 1+3 \times-2+4}{-3 \times 1+2 \times-2+5} \\ & =\frac{4-6+4}{-3-4+5}=\frac{2}{-2}=-1 \end{aligned} $

$ \begin{aligned} & f(x)=|x-1|+|x-2| \\ & f(x)=\left\{\begin{array}{cc} -2 x+3, & x<1 \\ 1, & 1 \leq x<2 \\ 2 x-3, & 2 \leq x \end{array}\right. \end{aligned} $

$ \begin{aligned} & \Rightarrow f^{\prime}(x)=\left\{\begin{array}{cc} -2, & x<1 \\ 0, & 1 \leq x<2 \\ 2, & 2 \leq x \end{array}\right. \\ & \Rightarrow f^{\prime}(-2023)=-2 \Rightarrow f^{\prime}\left(\frac{2024}{2023}\right)=0 \\ & \Rightarrow f^{\prime}(2023)=2 \\ & \therefore f^{\prime}(-2023)+f^{\prime}\left(\frac{2024}{2023}\right)+f^{\prime}(2023) \\ & =-2+0+2=0 \end{aligned} $

$ \begin{aligned} & \text { } f(x)=\frac{e^{2 x}-e^{-2 x}}{e^{3 x}+e^{-3 x}} \\ & f^{\prime}(x)=\frac{\left(e^{3 x}+e^{-3 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)}{-\left(e^{2 x}-e^{-2 x}\right)\left(3 e^{3 x}-3 e^{-3 x}\right)} \\ & \left(e^{3 x}+e^{-3 x}\right)^2 \\ & f^{\prime}(0)=\frac{2(4)-(0)}{2^2}=2 \end{aligned} $

$f(x)=x^{\tan x}+(\tan x)^x$

Let $\quad u=x^{\tan x}$ and $v=(\tan x)^x$

$ \begin{aligned} \log u & =\tan x \log x \\ \frac{1}{u} \frac{d u}{d x} & =\frac{\tan x}{x}+\sec ^2 x \log x \\ \frac{d u}{d x} & =x^{\tan x}\left(\frac{\tan x}{x}+\sec ^2 x \log x\right) \end{aligned} $

and $\log v=x \log \tan x$

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

$ \begin{aligned} f^{\prime}\left(\frac{\pi}{4}\right) & =\frac{\pi}{4}\left(\frac{4}{\pi}+2 \log \frac{\pi}{4}\right)+\left(2 \cdot \frac{\pi}{4}+\log 1\right) \\ & =1+\frac{\pi}{2} \log \frac{\pi}{4}+\frac{\pi}{2} \\ & =1+\frac{\pi}{2}\left(1+\log \frac{\pi}{4}\right) \\ & =1+\frac{\pi}{2}\left(\log e+\log \frac{\pi}{4}\right) \\ & =1+\frac{\pi}{2} \log \frac{e \pi}{4} \end{aligned} $

Given, $\sec \left(\log _2 y^2\right)=\operatorname{cosec}\left(\log _2 x^2\right)$

$ \begin{aligned} & \Rightarrow \sec \left(\log _2 y^2\right)=\sec \left(90^{\circ}-\log _2 x^2\right) \\ & \Rightarrow \quad \log _2 y^2=90^{\circ}-\log _2 x^2 \\ & \Rightarrow \quad 2 \log _2 y=90^{\circ}-2 \log _2 x \end{aligned} $

Differentiating on both sides w.r.t. $x$, we get

$ 2 \cdot \frac{1}{y} \frac{1}{\ln 2} \frac{d y}{d x}=-\frac{2}{x} \frac{1}{\ln 2} \Rightarrow \frac{d y}{d x}=-\frac{y}{x} $

Given, $e^x=y+\sqrt{y^2-1}$

Differentiating on both side w.r.t. $x$, we get

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

Now, $\frac{e^x-e^{-x}}{2}=\sqrt{y^2-1}$

So, $\frac{d y}{d x}=\frac{e^x-e^{-x}}{2}=\sin \mathrm{h} x$

Given:

$ x = \log p $

Differentiating $ x $ with respect to $ p $ gives:

$ \frac{dx}{dp} = \frac{1}{p} $

We also have:

$ y = \frac{1}{p} $

Differentiating $ y $ with respect to $ p $ yields:

$ \frac{dy}{dp} = -\frac{1}{p^2} $

Next, we need to find $ \frac{dy}{dx} $:

$ \frac{dy}{dx} = \frac{\frac{dy}{dp}}{\frac{dx}{dp}} = \frac{-\frac{1}{p^2}}{\frac{1}{p}} = -\frac{1}{p} $

Since $ p = e^x $, we can substitute:

$ \frac{dy}{dx} = -\frac{1}{e^x} = -e^{-x} $

Given the equation:

$ \tan y = \cot \left(\frac{\pi}{4} - x\right) $

We need to differentiate both sides with respect to $ x $.

Starting with the left side:

$ \frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx} $

For the right side, we differentiate:

$ \frac{d}{dx}\left(\cot \left(\frac{\pi}{4} - x\right)\right) = -\csc^2\left(\frac{\pi}{4} - x\right) \cdot \left(-1\right) = \csc^2\left(\frac{\pi}{4} - x\right) $

Setting the derivatives equal gives:

$ \sec^2 y \cdot \frac{dy}{dx} = \csc^2\left(\frac{\pi}{4} - x\right) $

Thus, solving for $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{\csc^2\left(\frac{\pi}{4} - x\right)}{\sec^2 y} $

Next recognize that:

$ \sec^2 y = 1 + \tan^2 y $

Given that $\tan y = \cot \left(\frac{\pi}{4} - x\right)$, we recognize that:

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

Therefore:

$ \frac{dy}{dx} = \frac{\csc^2\left(\frac{\pi}{4} - x\right)}{1 + \tan^2\left(\frac{\pi}{4} + x\right)} $

Given the expressions:

$ x = 3 \sqrt{2} \cos^3 \theta $

and

$ y = 4 \tan^2 \theta, $

we want to find the derivative $\left(\frac{d y}{d x}\right)_{\theta=\pi / 4}$.

To do this, we will use the chain rule to compute $\frac{d y}{d \theta}$ and $\frac{d x}{d \theta}$, and then find $\frac{d y}{d x}$ by dividing the two derivatives.

First, compute $\frac{d y}{d \theta}$:

$ y = 4 \tan^2 \theta $

Differentiating with respect to $\theta$:

$ \frac{d y}{d \theta} = 4 \cdot 2 \tan \theta \sec^2 \theta = 8 \tan \theta \sec^2 \theta. $

Next, compute $\frac{d x}{d \theta}$:

$ x = 3 \sqrt{2} \cos^3 \theta $

Differentiating with respect to $\theta$ and using the chain rule:

$ \frac{d x}{d \theta} = 3 \sqrt{2} \cdot 3 \cos^2 \theta (-\sin \theta) = -9 \sqrt{2} \cos^2 \theta \sin \theta. $

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

$ \frac{d y}{d x} = \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} = \frac{8 \tan \theta \sec^2 \theta}{-9 \sqrt{2} \cos^2 \theta \sin \theta}. $

Evaluate this derivative at $\theta = \frac{\pi}{4}$:

At $\theta = \frac{\pi}{4}$:

$\tan \frac{\pi}{4} = 1$

$\sec \frac{\pi}{4} = \sqrt{2}$

$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

Substitute these values in:

$ \frac{d y}{d x}\bigg|_{\theta=\frac{\pi}{4}} = \frac{8 \cdot 1 \cdot (\sqrt{2})^2}{-9 \sqrt{2} \left(\frac{1}{\sqrt{2}}\right)^2 \cdot \frac{1}{\sqrt{2}}} = \frac{8 \cdot 2}{-9 \sqrt{2} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{2}}}. $

Simplifying gives:

$ = \frac{16}{-9 \sqrt{2} \cdot \frac{1}{2\sqrt{2}}} = \frac{-32}{9}. $

Thus, $\left(\frac{d y}{d x}\right)_{\theta=\pi / 4} = -\frac{32}{9}$.

We need to find the derivative of

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

with respect to

$ u(x) = \frac{2x}{1+x^2} $

at $ x = 2 $. In other words, we want to compute

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

evaluated at $ x = 2 $.

Here’s how to do it step by step:

Differentiate $ y(x) $ with respect to $ x $:

We have

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

Using the quotient rule, where if $ f(x) = \frac{v(x)}{w(x)} $ then

$ f'(x) = \frac{v'(x)w(x) - v(x)w'(x)}{[w(x)]^2}, $

let:

$ v(x) = 1-x^2 $ so that $ v'(x) = -2x $,

$ w(x) = 1+x^2 $ so that $ w'(x) = 2x $.

Therefore,

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

Simplify the numerator:

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

Thus,

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

Differentiate $ u(x) $ with respect to $ x $:

We have

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

Again, using the quotient rule:

$ v(x) = 2x $ so that $ v'(x) = 2 $,

$ w(x) = 1+x^2 $ so that $ w'(x) = 2x $.

Therefore,

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

Simplify the numerator:

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

Thus,

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

Compute $ \frac{dy}{du} $:

By the chain rule:

$ \frac{dy}{du} = \frac{y'(x)}{u'(x)} = \frac{\frac{-4x}{(1+x^2)^2}}{\frac{2(1-x^2)}{(1+x^2)^2}}. $

Notice that the factor $ (1+x^2)^2 $ cancels out, so:

$ \frac{dy}{du} = \frac{-4x}{2(1-x^2)} = \frac{-2x}{1-x^2}. $

Evaluate at $ x = 2 $:

Substitute $ x = 2 $:

$ \frac{dy}{du}\Bigg|_{x=2} = \frac{-2(2)}{1-2^2} = \frac{-4}{1-4} = \frac{-4}{-3} = \frac{4}{3}. $

Thus, the derivative of $ \frac{1-x^2}{1+x^2} $ with respect to $ \frac{2x}{1+x^2} $ at $ x=2 $ is $ \frac{4}{3} $.

This matches Option B.

To find the value of $2a - 3b$, consider the function $y = e^{a + bx^2}$.

Given Point on the Curve:

The point $P(1, 1)$ lies on the curve. At $x = 1$, $y = 1$:

$ 1 = e^{a + b} \implies a + b = 0 $

Slope of the Tangent:

The slope of the tangent at $(1, 1)$ is given as $-2$. We find the derivative:

$ \frac{dy}{dx} = \frac{d}{dx}(e^{a + bx^2}) = e^{a + bx^2} \cdot \frac{d}{dx}(a + bx^2) $

$ = e^{a + bx^2} \cdot (2bx) $

Evaluate the Derivative at $x = 1$:

At $x = 1$:

$ \left. \frac{dy}{dx} \right|_{(1,1)} = e^{a + b} \cdot 2b \cdot 1 = e^{a + b} \cdot 2b = -2 $

Since $e^{a+b} = 1$, we have:

$ 1 \cdot 2b = -2 \implies 2b = -2 \implies b = -1 $

Find $a$:

Substituting $b = -1$ into $a + b = 0$:

$ a - 1 = 0 \implies a = 1 $

Calculate $2a - 3b$:

$ 2a - 3b = 2(1) - 3(-1) = 2 + 3 = 5 $

Thus, the value of $2a - 3b$ is 5.

$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $

$\therefore$ ${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$

$\therefore$ ${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)$

and ${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$

$\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)$

$\therefore$ ${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$

Let $f(x) = {\log _{\cos x}}\cos ec\,x$

$ = {{\log \cos ec\,x} \over {\log \cos x}}$

$ \Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}$

at $x = {\pi \over 4}$

$f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}$

$\therefore$ ${\log _e}2f'(x)$ at $x = {\pi \over 4} = 4$

${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$

Differentiating on both side

$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$

${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} $

Square on both side

${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$

Diff on both side

$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$

$xy' + y''{x^2} + 25y = 0$

$f(x) = 3{x^2} + 1$

f'(x) is bijective function

and $f(g(x)) = x \Rightarrow g(x)$ is inverse of f(x)

$g(f(x)) = x$

$g'(f(x))\,.\,f'(x) = 1$

$g'(f(x)) = {1 \over {3{x^2} + 1}}$

Put x = 4 we get

$g'(63) = {1 \over {49}}$

Let ${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$

$\therefore$ $y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$

$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$

$\therefore$ $y = {\pi \over 4} - {\theta \over 2}$

$y = {\pi \over 4} - {{{x^3}} \over 2}$

$\therefore$ $y' = {{ - 3{x^2}} \over 2}$

$y'' = - 3x$

$\therefore$ ${x^2}y'' - 6y + {{3\pi } \over 2} = 0$

$ f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\left\{\frac{4}{5} \sin (p x)-\frac{3}{5} \cos (p x)\right\} $

Let $\cos \alpha=\frac{4}{5}$, then $\sin \alpha=\frac{3}{5}$

Here, $\frac{\sin \alpha}{\cos \alpha}=\frac{3}{4} \Rightarrow \alpha=\tan ^{-1}\left(\frac{3}{4}\right)$

$ \begin{aligned} f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\{\cos \alpha \sin (p x) & \\ & -\sin \alpha \cos (p x)\} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\sum_{p=1}^7 p^2 \sin ^{-1}\{\sin (p x-\alpha)\} \\ & =\sum_{p=1}^7 p^2(p x-\alpha) \\ & =\sum_{p=1}^7 p^3 x-p^2 \alpha \\ & =x \cdot \sum_{p=1}^7 p^3-\alpha \sum_{p=1}^7 p^2 \\ & =x\left[\frac{7(7+1)}{2}\right]^2-\alpha\left[\frac{7(7+1)(2 \times 7+1)}{6}\right] \\ & =x(28)^2-\alpha[7 \cdot 4 \cdot 5] \\ & =784 x-140 \alpha \\ f(x) & =784 x-140 \tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}\\ &\text { On differentiating w.r.t. } x \text {, we get }\\ &f^{\prime}(x)=784 \times 1-0=784 \end{aligned} $

Given, $y=\frac{a x+b}{c x+d}$

$ \begin{aligned} \Rightarrow & & y c x+y d & =a x+b \\ \Rightarrow & & y x c-a x & =b-y d \\ \Rightarrow & & x & =\frac{b-y d}{c y-a} \end{aligned} $

On differentiating w.r.t. $y$,

$ \begin{aligned} \frac{d x}{d y} & =\frac{(c y-a)(-d)-(b-y d)(c)}{(c y-a)^2} \\ & =\frac{-c d y+a d-b c+c d y}{(a-c y)^2} \end{aligned} $

Hence, $\frac{d x}{d y}=\frac{a d-b c}{(a-c y)^2}$

$ \begin{aligned} & \text { } \\ & \begin{aligned} \text { } & x^2+y^2=t-\frac{1}{t} \\ & x^4+y^4=t^2+\frac{1}{t^2} \\ & \left(x^2+y^2\right)^2=\left(t-\frac{1}{t}\right)^2 \\ \Rightarrow & x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}-2 \\ \Rightarrow & t^2+\frac{1}{t^2}+2 x^2 y^2=t^2+\frac{1}{t^2}-2 \\ \Rightarrow & 2 x^2 y^2=-2 \\ \Rightarrow & x^2 y^2=-1 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { On differentiating w.r.t. } x \text {, we get }\\ &\begin{gathered} 2 x y^2+x^2\left(2 y y^{\prime}\right)=0 \\ \Rightarrow y^{\prime}=\frac{-2 x y^2}{2 x^2 y} \Rightarrow \frac{d y}{d x}=\frac{-y}{x} \end{gathered} \end{aligned} $

$ \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} $

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

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

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 $c=48$

$\therefore \quad D<0$

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

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

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

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