Differentiation

$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

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

$ \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

Given, x² + y² + sin y = 4

After differentiating the above equation  w.r.t.x  we get

2x + 2y ${{dy} \over {dx}}$ + cos y ${{dy} \over {dx}}$ = 0          . . . . (1)

$ \Rightarrow $  2x + (2y + cos y) ${{dy} \over {dx}}$ = 0

$ \Rightarrow $  ${{dy} \over {dx}}$ = ${{ - 2x} \over {2y + \cos y}}$

At ($-$ 2, 0), ${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$ = ${{ - 2x - 2} \over {2 \times 0 + \cos 0}}$

$ \Rightarrow $  ${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$ = ${4 \over {0 + 1}}$

$ \Rightarrow $   ${\left( {{{dy} \over {dx}}} \right)_{\left( { - 2,0} \right)}}$ = 4          . . . . .(2)

Again differentiating equation (1) w.r.t  to  x, we get

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

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

$ \Rightarrow $  (2y + cos y) ${{{d^2}y} \over {d{x^2}}}$ = $-$ 2 $-$ (2 $-$ sin y)${\left( {{{dy} \over {dx}}} \right)^2}$

$ \Rightarrow $   ${{{d^2}y} \over {d{x^2}}}$ = ${{ - 2 - \left( {2 - \sin y} \right){{\left( {{{dy} \over {dx}}} \right)}^2}} \over {2y + \cos y}}$

So, at ($-$ 2, 0),

${{{d^2}y} \over {d{x^2}}}$ = ${{ - 2 - \left( {2 - 0} \right) \times {4^2}} \over {2 \times 0 + 1}}$

$ \Rightarrow $  ${{{d^2}y} \over {d{x^2}}}$ = ${{ - 2 - 2 \times 16} \over 1}$

$ \Rightarrow $$\,\,\,$ ${{{d^2}y} \over {d{x^2}}}$ = $-$ 34

Let $f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 1}} + \,\,....\,\, + {a_{n - 1}}x + {a_n}$

$f'(x) = {a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,.....\,\, + {a_{n - 1}}$

$f''(x) = {a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}$

Now,

$f(3x) = {3^n}{a_0}{x^n} + {3^{n - 1}}{a_1}{x^{n - 1}} + {3^{n - 2}}{a_2}{x^{n - 2}} + \,\,....\,\, + 3{a_{n - 1}} + {a_n}$

$f'(x)\,.\,f''(x) = [{a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,....\,\, + {a_{n - 1}}]$

$[{a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}]$

Comparing highest powers of x, we get

${3^n}{a_0}{x_n} = a_0^2(n - 1){x^{n - 1 + n - 2}} = a_0^2{n^2}(n - 1){x^{2n - 3}}$

Therefore, $2n - 3 = n$

$\Rightarrow$ n = 3 and ${3^n}{a_0} = a_0^2{n^2}(n - 1)$

$ \Rightarrow {a_0} = 27 = {3 \over 2}$

Therefore, $f(x) = {3 \over 2}{x^3} + {a_1}{x^2} + {a_2}x + {a_3}$

$f'(x) = {9 \over 2}{x^2} + 2{a_1}x + {a_2}$

$f''(x) = 9x + 2{a_1}$

$f(3x) = {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$

Now, $f(3x) = f'(x)\,.\,f''(x)$

$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}$

$ \Rightarrow \left( {{9 \over 2}{x^2} + 2{a_1}x + {a_2}} \right)(9x + 2{a_1})$

$ \Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3} = {{81} \over 2}{x^3} + [9{a_1} + 18{a_1}]{x^2} + [4a_1^2 + 9{a_2}]x + 2{a_1}{a_2}$

Comparing the coefficients, we get

$9{a_1} = 27{a_1}$

$ \Rightarrow {a_1} = 0,\,3{a_2} = 4a_1^2 + 9{a_2} = 9{a_2}$

$ \Rightarrow {a_2} = 0$

Therefore, $f(x) = {3 \over 2}{x^3}$

$f'(x) = {9 \over 2}{x^2}$

$f''(x) = 9x$

Hence, $f''(2) - f'(x) = 18 - 18 = 0$

The given equation is

$y = {({x^2} + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}}$

Differentiating w.r.t. x, we get

${{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) + 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {1 - {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$

Here, we have used the standard differentiatials

${d \over {dx}}{x^n} = n\,{x^{n - 1}}$

That is, ${d \over {dx}}(\sqrt {f(x)} ) = {1 \over {2\sqrt {f(x)} }} \times {d \over {dx}}(f(x))$

Therefore

${{dy} \over {dx}} = {{15{{(x + \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} + {{15{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow \sqrt {{x^2} - 1} {{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{15}} - 15{(x - \sqrt {{x^2} - 1} )^{15}}$

Differentiating w.r.t. x, we get

${{1(2x)} \over {2\sqrt {{x^2} + 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 15 \times 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) - 15 \times 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {{{1 - 1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)$

$ \Rightarrow {x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 225{(x + \sqrt {{x^2} - 1} )^{14}}{{(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} - {{225{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow \sqrt {{x^2} - 1} \left[ {{x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}}} \right] = 225{(x + \sqrt {{x^2} - 1} )^{15}} + 225{(x - \sqrt {{x^2} - 1} )^{15}}$

$ \Rightarrow x{{dy} \over {dx}} + ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 225\left[ {{{(x + \sqrt {{x^2} - 1} )}^{15}} + {{(x - \sqrt {{x^2} - 1} )}^{15}}} \right]$

Substituting ${(x + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}} = y$, we get

$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y$

Let y = ${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$

= ${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$

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

$ \therefore $ ${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$

= ${9 \over {1 + 9{x^3}}}.\sqrt x $

$ \therefore $ g(x) = ${9 \over {1 + 9{x^3}}}$

Since $f(x)$ and $g(x)$ are inverse of each other

$\therefore$ $g'\left( {f\left( x \right)} \right) = {1 \over {f'\left( x \right)}}$

$ \Rightarrow g'\left( {f\left( x \right)} \right) = 1 + {x^5}$

$\left( \, \right.$ As $\,f'\left( x \right) = {1 \over {1 + {x^5}}}$ $\left. \, \right)$

Here $x=g(y)$

$\therefore$ $g'\left( y \right) = 1 + \left\{ {g\left( y \right)} \right\}$

$ \Rightarrow g'\left( x \right) = 1 + \left\{ {g\left( x \right)} \right\}$

Let $y = \sec \left( {{{\tan }^{ - 1}}x} \right)$

and ${\tan ^{ - 1}}\,\,x = \theta .$

$ \Rightarrow x = \tan \theta $



Thus, we have $y = \sec \,\theta $

$ \Rightarrow y = \sqrt {1 + {x^2}} $

$\left( {\,\,} \right.$ As $\,\,\,\,\,\,{\sec ^2}\theta = 1 + {\tan ^2}\theta $ $\left. {\,\,} \right)$

$ \Rightarrow {{dy} \over {dx}} = {1 \over {2\sqrt {1 + {x^2}} }}.2x$

At $x = 1,\,\,{{dy} \over {dx}} = {1 \over {\sqrt 2 }}.$

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

$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$

$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$

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

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

$g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)$

$ = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)$

$ \Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)$

$ = 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}$

$ = 4\left( { - 1} \right){\left( 1 \right)^2} = - 4$

${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0$

$ \Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$

$ \Rightarrow 2\,\cot \,y\, = u - {1 \over u}$

where $u = {x^x}$

Differentiating both sides with respect to $x,$

we get $ \Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$

$ = \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}$

where $u = {x^x} \Rightarrow \log \,u = x\,\log \,x$

$ \Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x$

$ \Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)$

$\therefore$ We get $ - 2\cos e{c^2}y{{dy} \over {dx}}$

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

$ \Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)$

Now when

$x=1,$ ${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0,$

gives $1 - 2\,\cot y - 1 = 0$

$ \Rightarrow \,\,\cot y\, = 0$

$\therefore$ From equation $(i),$ at $x=1$

and $\cot \,y = 0,$ we get

$y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1$

${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$

$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$

Differentiating both sides.

$\therefore$ ${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$

$ \Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$

$ \Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$

$ \Rightarrow {{dy} \over {dx}} = {y \over x}$

$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$

Taking log.

$\log \,\,x = y + x$

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

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

$f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1$

$f'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n$

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

$ \Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)$

$\therefore$ ${f^n}\left( x \right) = n!$

$ \Rightarrow {f^n}\left( 1 \right) = n!$

$ = 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}$

$ = {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0$

$f\left( x \right) = a{x^2} + bx + c$

$f\left( 1 \right) = f\left( { - 1} \right)$

$ \Rightarrow a + b + c = a - b + c$

or $b = 0$

$\therefore$ $f\left( x \right) = a{x^2} + c$

or $f'\left( x \right) = 2ax$

Now $f'\left( a \right);f'\left( b \right);$

and $f'\left( c \right)$ are $2a\left( a \right);2a\left( b \right);2a\left( c \right)$

i.e.$\,2{a^2},\,2ab,\,2ac.$

$ \Rightarrow $ If $a,b,c$ are in $A.P.$ then

$f'\left( a \right);f'\left( b \right)$ and

$f'\left( c \right)$ are also in $A.P.$

$y = {\left( {x + \sqrt {1 + {x^2}} } \right)^n}$

${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + {1 \over 2}{{\left( {1 + {x^2}} \right)}^{ - 1/2}}.2x} \right);$

${{dy} \over {dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {\sqrt {1 + {x^2}} + x} \right)} \over {\sqrt {1 + {x^2}} }}$

$ = {{n{{\left( {\sqrt {1 + {x^2}} + x} \right)}^n}} \over {\sqrt {1 + {x^2}} }}$

or $\sqrt {1 + {x^2}} {{dy} \over {dx}} = ny$

or $\sqrt {1 + {x^2}} {y_1} = ny$

$\left( {{y_1} = {{dy} \over {dx}}} \right)$

Squaring, $\left( {1 + {x^2}} \right){y_1}^2 = {n^2}{y^2}$

Differentiating, $\left( {1 + {x^2}} \right)2{y_1}{y_2} + {y_1}^2.2x$

$ = {n^2}.2y{y_1}$

or $\left( {1 + {x^2}} \right){y_2} + x{y_1} = {n^2}y$