Differentiation

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

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

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

We have,

$ y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) $

$ =\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|} $

$ \left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right. $

and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$

$ \begin{aligned} & =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\ & \end{aligned} $

$ =\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right) $

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

$ \begin{aligned} & \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\ & \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right) \end{aligned} $

$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$

${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $

or $f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$

$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$

$f(x) = {{1 - x} \over {1 + x}}$

Now, $f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$

or $f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$

or ${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$.

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

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

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. $x$

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

${\tan ^{ - 1}}\left( {{{\cos x} \over {1 + \sin x}}} \right) = {\tan ^{ - 1}}\left( {{{\sin \left( {{\pi \over 2} - x} \right)} \over {1 + \cos \left( {{\pi \over 2} - x} \right)}}} \right)$

$ = {\tan ^{ - 1}}\left[ {{{2\sin \left( {{\pi \over 4} - {\pi \over 2}} \right)\cos \left( {{\pi \over 4} - {\pi \over 2}} \right)} \over {2{{\cos }^2}\left( {{\pi \over 4} - {\pi \over 2}} \right)}}} \right]$

$ = {\tan ^{ - 1}}\left[ {\tan \left( {{\pi \over 4} - {\pi \over 2}} \right)} \right]$

${\tan ^{ - 1}}\left( {{{\cos x} \over {1 + \sin x}}} \right) = {\pi \over 4} - {\pi \over 2}$

$\therefore$ ${d \over {dx}}\left( {{{\tan }^{ - 1}}\left( {{{\cos x} \over {1 + \sin x}}} \right)} \right) = {d \over {dx}}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {{ - 1} \over 2}$

$x^2+y^2=1$

Again differentiating w.r.t. $x$

$2x+2yy'=0$

or $x+yy'=0$

Again differentiating w.r.t. x

$\begin{aligned} & 1+y\left(y^{\prime \prime}\right)+\left(y^{\prime}\right)^2=0 \\ & y\left(y^{\prime \prime}\right)+\left(y^{\prime}\right)^2+1=0 \end{aligned}$

$y=x+\frac{1}{x}$ ..... (i)

Differentiating with respect to x, we get

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

or $x^2y'=x^2-1$ .... (ii)

From Eq. (i), we get

$x^2=xy-1$

$\therefore$ From Eq. (ii), we have

$x^2y'=xy-1-1$ or $x^2y'-xy+2=0$

$3 \sin x y+4 \cos x y=5$

Differentiating with respect to $x$, we get

$\begin{aligned} & 3 \cos x y \cdot\left(x \frac{d y}{d x}+y\right)-4 \sin x y\left(x \frac{d y}{d x}+y\right)=0 \\ & \Rightarrow x \frac{d y}{d x}(3 \cos x y-4 \sin x y) \\ & +y(3 \cos x y-4 \sin x y)=0 \\ & \Rightarrow(3 \cos x y-4 \sin x y)\left(x \frac{d y}{d x}+y\right)=0 \\ & \Rightarrow \quad x \frac{d y}{d x}+y=0 \\ & \text {or}\quad\frac{d y}{d x}=\frac{-y}{x} \\ \end{aligned}$

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

So, $\quad h^{\prime}\left(g^{\prime}(x)\right)=2$

$\therefore f^{\prime}\left(h^{\prime}\left(g^{\prime}(x)\right)\right)=f^{\prime}(2)=\frac{2}{\sqrt{2^2+1}}=\frac{2}{\sqrt{5}}$

$f(x)=-x^3+4 a x^2+2 x-5$

$f^{\prime}(x)=-3 x^2+8 a x+2$

Discriminant of $f^{\prime}(x)=(8 a)^2-4(-3)(2)$

$=64 a^2+24>0, \forall a \in R$

Hence, $f(x)$ is increasing for every real value of $a$.

Let f = ${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$

Put x = tan $\theta $ $ \Rightarrow $ $\theta $ = tan^–1 x

f = ${\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)$

$ \Rightarrow $ f = ${\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)$ = ${\theta \over 2}$

$ \Rightarrow $ f = ${{{{\tan }^{ - 1}}x} \over 2}$

$ \therefore $ ${{df} \over {dx}}$ = ${1 \over {2\left( {1 + {x^2}} \right)}}$ ....(1)

Let g = ${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$

Put x = sin $\theta $ $ \Rightarrow $ $\theta $ = sin^–1 x

$ \Rightarrow $ g = ${\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)$

$ \Rightarrow $ g = tan^–1 (tan 2$\theta $) = 2$\theta $

$ \Rightarrow $ g = 2sin^-1 x

$ \Rightarrow $ ${{dg} \over {dx}}$ = ${2 \over {\sqrt {1 - {x^2}} }}$ ...(2)

Using (i) and (ii),

$ \therefore $ ${{df} \over {dg}}$ = ${1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}$

At x = ${1 \over 2}$, ${\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}$ = ${{\sqrt 3 } \over {10}}$

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

$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$

Differentiating both sides :

$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$

$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$

At $\left( {{\pi \over 4},{\pi \over 4}} \right)$ :

$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$

$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$; a, b > 0

Given y² + log_e (cos²x) = y .....(1)

Put x = 0, we get

y² + log_e (1) = y

$ \Rightarrow $ y² = y

$ \Rightarrow $ y = 0, 1

Differentiating (1) we get

2yy' + ${1 \over {\cos x}}\left( { - \sin x} \right)$ = y'

$ \Rightarrow $ 2yy' - 2tanx = y' ....(2)

From (2) when x = 0, y = 0 then y'(0) = 0

From (2) when x = 0, y = 1 then

2y' = y'

$ \Rightarrow $ y'(0) = 0

Again differentiating (2) we get

2(y')² + 2yy'' – 2sec²x = y''

from (2) when x = 0, y = 0, y’(0) = 0 then

y”(0) = -2

Also from (2) when x = 0, y = 1, y’(0) = 0 then

y”(0) = 2

$ \therefore $ |y''(0)| = 2

Given the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x.

$ \Rightarrow $ ƒ'(g(x)) g'(x) = 1

put x = a

$ \Rightarrow $ ƒ'(b) g'(a) = 1

$ \Rightarrow $ ƒ'(b) = ${1 \over 5}$

$x = 2\sin \theta - \sin 2\theta $

$ \Rightarrow $ ${{dx} \over {d\theta }}$ = $2\cos \theta - 2\cos 2\theta $

$y = 2\cos \theta - \cos 2\theta $

$ \Rightarrow $ ${{dy} \over {d\theta }}$ = –2sin$\theta $ + 2sin2$\theta $

${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$ = ${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$

This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression:

$ \frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$

This is the rate of change of y with respect to x, as a function of θ.

Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$, yielding :

$ \frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$

Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes :

$ \frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$

Finally, we evaluate this expression at $\theta = \pi$, yielding :

$ \frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$

Therefore, the correct answer is (A) $\frac{3}{8}$.

Alternate Method :

First, let's find the derivatives of x and y with respect to θ :

1) $ \frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta $

2) $ \frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta $

We know that $ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $

So, we substitute 1) and 2) into this equation :

We have

$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$

For simplification, let's denote the numerator as $N = \sin 2\theta - \sin \theta$ and the denominator as $D = \cos \theta - \cos 2\theta$.

We have to compute $\frac{d}{d\theta}(\frac{dy}{dx})$ which is $\frac{d}{d\theta}(\frac{N}{D})$.

We can use the quotient rule for differentiation, which states that if we have a function of the form $\frac{u}{v}$, then its derivative is given by $\frac{vu' - uv'}{v^2}$.

So here, $N' = 2\cos 2\theta - \cos \theta$ and $D' = -\sin \theta + 2\sin 2\theta$.

Applying the quotient rule :

$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$

Substituting the expressions for $N'$, $D'$, $N$, and $D$ we get :

$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$

Now $ \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x} $

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

$ \begin{aligned} \left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\ & =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8} \end{aligned} $

Given ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1

= (sin(tan^–1x) + sin(${\pi \over 2}$ - tan^–1x))² – 1

= (sin(tan^–1x) + cos(tan^–1x))² – 1

= sin²(tan^–1x) + cos²(tan^–1x) + 2sin(tan^–1x)cos(tan^–1x) + 1

= 1 + sin(2tan^–1x) - 1

= sin(2tan^–1x)

Also given ${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$

Integrating both sides we get

y = ${1 \over 2}$ sin^-1 (f(x)) + C

= ${1 \over 2}$ sin^-1 (sin(2tan^–1x)) + C

Given $y\left( {\sqrt 3 } \right) = {\pi \over 6}$ mean x = $\sqrt 3 $ and y = ${\pi \over 6}$

$ \therefore $ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (sin(2tan^–1$\sqrt 3 $)) + C

$ \Rightarrow $ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (sin(2$ \times $${\pi \over 3}$)) + C

$ \Rightarrow $ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (${{\sqrt 3 } \over 2}$) + C

$ \Rightarrow $ ${\pi \over 6}$ = ${1 \over 2}$ $ \times $ ${\pi \over 3}$ + C

$ \Rightarrow $ C = 0

Now y(${ - \sqrt 3 }$) means when x = ${ - \sqrt 3 }$ then find y.

y = ${1 \over 2}$ sin^-1 (sin(2tan^–1x))

= ${1 \over 2}$ sin^-1 (sin(2tan^–1(${ - \sqrt 3 }$)))

= ${1 \over 2}$ sin^-1 (sin(-2tan^–1(${ \sqrt 3 }$)))

= ${1 \over 2}$ sin^-1 (sin(-2$ \times $${\pi \over 3}$))

= ${1 \over 2}$ sin^-1 (-sin(2$ \times $${\pi \over 3}$))

= ${1 \over 2}$ sin^-1 (-${{\sqrt 3 } \over 2}$)

= ${1 \over 2}$ $ \times $ -sin^-1 (${{\sqrt 3 } \over 2}$)

= ${1 \over 2}$ $ \times $ -${\pi \over 3}$

= -${\pi \over 6}$

$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $ ....(1)

On differentiating both side of eq. (1) w.r.t. x we get,

${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$

= 0 - $\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$

Put x = ${1 \over 2}$ and y = $ - {1 \over 4}$, we get

${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$

= $ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$

$ \therefore $ ${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$

x^k + y^k = a^k

$ \Rightarrow $ kx^{k - 1} + ky^{k - 1}${{{dy} \over {dx}}}$ = 0

$ \Rightarrow $ ${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$ = 0 ...(1)

Given ${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$ ...(2)

Comparing (1) and (2), we get

k - 1 = $ - {1 \over 3}$

$ \Rightarrow $ k = ${2 \over 3}$

$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$

= $\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}} $

= $\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} $

= $\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}} $

= ${{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}$

At $\alpha $ = ${{5\pi } \over 6}$

${\left| {\sin \alpha + \cos \alpha } \right|}$ = -(sin$\alpha $ + cos$\alpha $)

and |sin$\alpha $| = sin$\alpha $

$ \therefore $ y($\alpha $) = ${{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}$

= -1 - cot$\alpha $

$ \therefore $ ${{dy} \over {d\alpha }}$ = cosec²$\alpha $

So ${{{dy} \over {d\alpha }}}$ at $\alpha $ = ${{5\pi } \over 6}$,

= cosec²${{5\pi } \over 6}$ = 4

(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} $

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

$ \Rightarrow $ $ - {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$

$ \Rightarrow $ $ - {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)$

$ \Rightarrow $ $ - \left( {{\pi \over 4} - x} \right)$

$ \Rightarrow $ ${{dy} \over {dx}} = 1$

Now if differentiation of ${x \over 2}$ w.r.t is ${1 \over 2}$

$ \Rightarrow $ So differentiation of y w.r.t ${x \over 2}$ is $1 \over {1 \over 2}$ = 2

y = 1 $ \Rightarrow $ x = 0

${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$

$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$

$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0$

x = 0, y = 1

$ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0$

$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}$

f(x) = ln(sin x), g(x) = sin^–1 (e^–x)

f(g(x)) = ln(sin(sin^–1 e^–x)) = -x

f(g($\alpha $)) = – $\alpha $ = b

As f(g(x)) = – x

$ \therefore $ (f(g(x)))' = – 1

$ \Rightarrow $ (f(g($\alpha $)))' = – 1 = a

$ \therefore $ b = – $\alpha $, a = – 1

$ \therefore $ a$\alpha $² - b$\alpha $ - a = - $\alpha $² + $\alpha $² + 1 = 1

Given ƒ(1) = 1, ƒ'(1) = 3

Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))²

On differentiating both sides with respect to x we get,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x)

Now at x = 1,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1)

= ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= 3$ \times $3$ \times $3 + 2$ \times $3

= 33

$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$

As x $ \in $ $\left( {0,{\pi \over 2}} \right)$ then

${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$ = ${\left( {{\pi \over 3} + x} \right)}$

$ \Rightarrow $ 2y = ${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{\pi \over 6} - x} \right)^2}$

$ \therefore $ $2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$

$ \Rightarrow $ ${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$

(2x)^2y = 4e^2x-2y

2y$\ell $n2x = $\ell $n4 + 2x $-$ 2y

y = ${{x + \ell n2} \over {1 + \ell n2x}}$

y ' = ${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$

y '${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$

Differentiating with respect to x,

$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$

at   $x = e$  we get

$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$

$ \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$

as   $y(e) = \sqrt {4 + {e^2}} $

f(x) = x³ + x²f '(1) + xf ''(2) + f '''(3)

$ \Rightarrow $  f '(x) = 3x² + 2xf '(1) + f ''(x)     . . . . . (1)

$ \Rightarrow $  f ''(x) = 6x + 2f '(1)     . . . . . . (2)

$ \Rightarrow $  f '''(x) = 6      . . . . . .(3)

put x = 1 in equation (1) :

f '(1) = 3 + 2f '(1) + f ''(2)     . . . . .(4)

put x = 2 in equation (2) :

f ''(2) = 12 + 2f '(1)     . . . . .(5)

from equation (4) & (5) :

$-$3 $-$ f '(1) = 12 + 2f'(1)

$ \Rightarrow $  3f '(1) = $-$ 15

$ \Rightarrow $  f '(1) = $-$ 5 $ \Rightarrow $  f ''(2) = 2      . . . . .(2)

put x = 3 in equation (3) :

f ''' (3) = 6

$ \therefore $  f(x) = x³ $-$ 5x² + 2x + 6

f(2) = 8 $-$ 20 + 4 + 6 = $-$ 2

x = 3 tan t and y = 3 sec t

So that ${{dx} \over {dt}}$ = 3sec²t and ${{dy} \over {dt}}$ = 3 sec t tan t

${{dy} \over {dx}}$ = ${{dy/dt} \over {dx/dt}}$ = sin t

${{{d^2}y} \over {d{x^2}}}$ = (cos t)$.{{dt} \over {dx}}$

${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$

${{{d^2}y} \over {d{x^2}}}$ = ${1 \over 3}$(cos³ t)

${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$ = ${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$

x = $\sqrt {{2^{\cos e{c^{ - 1}}t}}} $

$\therefore\,\,\,\,$ ${{dx} \over {dt}}$ = ${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$ $ \times $ (${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$) $ \times $ ${{ - 1} \over {t\sqrt {{t^2} - 1} }}$

${{dy} \over {dt}}$ = ${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$ $ \times $ $\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$ $ \times $ ${1 \over {t\sqrt {{t^2} - 1} }}$

$\therefore\,\,\,\,$ ${{dy} \over {dx}}$

= ${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$

= ${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$ $ \times $ ${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$

= $ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $

= $-$ ${y \over x}$

Since f(x) = sin$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)$

Suppose 3^x = tan t

$ \Rightarrow $   f(x) = sin^$-$1 $\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)$ = sin^$-$1 (sin2t) = 2t

                                         = 2tan^$-$1 (3x)

So, f'(x) = ${2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3$

$ \therefore $   f '$\left( { - {1 \over 2}} \right)$ = ${2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}$ . log_e 3

=  ${1 \over 2} \times \sqrt 3 \times {\log _e}3$   =  $\sqrt 3 \times {\log _e}\sqrt 3 $

Given,

$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|$

= cosx(x² - 2x²) - x(2 sinx - 2x tanx) + (2x sinx - x² tanx)
= x² (tanx - cosx)
$ \therefore $ ${f^{'}}(x)$ = 2x (tanx - cosx) + x²(sec²x + sinx)

$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$

= $\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$

= $\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$

= 2 (0-1) + 0

= -2