Higher order derivatives Questions in English

(B) Let $f(x) = x^5 - 5x^3 + 5x^2 - 1$.
To find equal roots,we check for common roots between $f(x)$ and its derivative $f'(x) = 5x^4 - 15x^2 + 10x$.
Setting $f'(x) = 0$:
$5x(x^3 - 3x + 2) = 0$.
By inspection,$x=1$ is a root of $f'(x)$ since $1-3+2=0$.
Checking $f(1) = 1 - 5 + 5 - 1 = 0$.
Since $f(1) = 0$ and $f'(1) = 0$,$x=1$ is a repeated root.
Dividing $f(x)$ by $(x-1)^2$:
$f(x) = (x-1)^2(x^3 + 2x^2 - 2x - 1)$.
Checking the derivative of $g(x) = x^3 + 2x^2 - 2x - 1$:
$g'(x) = 3x^2 + 4x - 2$.
$g(1) = 1 + 2 - 2 - 1 = 0$.
Since $g(1) = 0$ and $g'(1) = 3+4-2 = 5 \neq 0$,$x=1$ is a root of multiplicity $3$.
Thus,the equation has $3$ equal roots.

(B) Given $y = \operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}\right)$.
For $|x| > 1$,we use the substitution $x = \tan \theta$. Since $x=2 > 1$,we use the identity $\operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}(x)$.
Thus,$y = \pi - 2\tan^{-1}(x)$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{2}{1+x^2}$.
Differentiating again,$\frac{d^2y}{dx^2} = -2 \cdot (-1)(1+x^2)^{-2} \cdot (2x) = \frac{4x}{(1+x^2)^2}$.
At $x=2$,$k = \left(\frac{d^2y}{dx^2}\right)_{x=2} = \frac{4(2)}{(1+2^2)^2} = \frac{8}{25}$.
Therefore,$25k = 25 \cdot \frac{8}{25} = 8$.
Comparing with the options,$8 = (-2)^3$.

(B) Given $f(x) = e^x \sin x$.
First derivative: $f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.
Second derivative: $f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
Third derivative: $f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.
Fourth derivative: $f^{(4)}(x) = 2e^x(\cos x - \sin x) + 2e^x(-\sin x - \cos x) = -4e^x \sin x$.
Fifth derivative: $f^{(5)}(x) = -4e^x \sin x - 4e^x \cos x = -4e^x(\sin x + \cos x)$.
Sixth derivative: $f^{(6)}(x) = -4e^x(\sin x + \cos x) - 4e^x(\cos x - \sin x) = -4e^x \sin x - 4e^x \cos x - 4e^x \cos x + 4e^x \sin x = -8e^x \cos x$.

(A) Given $y = \sin(\sin x)$ . . . $(i)$
Differentiating with respect to $x$ using the chain rule:
$y' = \cos(\sin x) \cdot \cos x$ . . . $(ii)$
Differentiating again with respect to $x$:
$y'' = -\sin(\sin x) \cdot \cos^2 x - \sin x \cdot \cos(\sin x)$
From equation $(ii)$,we have $\cos(\sin x) = \frac{y'}{\cos x}$. Substituting this into the expression for $y''$:
$y'' = -\sin(\sin x) \cdot \cos^2 x - \sin x \cdot \left(\frac{y'}{\cos x}\right)$
$y'' = -y \cdot \cos^2 x - \tan x \cdot y'$
Rearranging the terms gives:
$y'' + \tan x \cdot y' + \cos^2 x \cdot y = 0$
Comparing this with $y'' + f(x) \cdot y' + g(x) \cdot y = 0$,we get $f(x) = \tan x$ and $g(x) = \cos^2 x$.
Therefore,$f(x) \cdot g(x) = \tan x \cdot \cos^2 x = \frac{\sin x}{\cos x} \cdot \cos^2 x = \sin x \cdot \cos x$.
Using the identity $\sin(2x) = 2 \sin x \cos x$,we get:
$f(x) \cdot g(x) = \frac{1}{2} \sin(2x)$.

(D) Given $x=3 \cos t$ and $y=4 \sin t$.
Squaring and adding,we get $\frac{x^2}{9} + \frac{y^2}{16} = \cos^2 t + \sin^2 t = 1$.
Differentiating with respect to $x$,we get $\frac{2x}{9} + \frac{2y}{16} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{16x}{9y}$.
Now,differentiating again with respect to $x$,we get $\frac{d^2y}{dx^2} = -\frac{16}{9} \left[ \frac{y - x \frac{dy}{dx}}{y^2} \right]$.
Substituting $\frac{dy}{dx} = -\frac{16x}{9y}$,we have $\frac{d^2y}{dx^2} = -\frac{16}{9} \left[ \frac{y - x(-\frac{16x}{9y})}{y^2} \right] = -\frac{16}{9} \left[ \frac{9y^2 + 16x^2}{9y^3} \right]$.
Since $\frac{x^2}{9} + \frac{y^2}{16} = 1$,we have $16x^2 + 9y^2 = 144$.
Thus,$\frac{d^2y}{dx^2} = -\frac{16}{9} \left[ \frac{144}{9y^3} \right] = -\frac{256}{9y^3}$.
At $(x_0, y_0) = (\frac{3\sqrt{2}}{2}, 2\sqrt{2})$,we have $y = 2\sqrt{2}$.
So,$\frac{d^2y}{dx^2} = -\frac{256}{9(2\sqrt{2})^3} = -\frac{256}{9(16\sqrt{2})} = -\frac{16}{9\sqrt{2}} = -\frac{8\sqrt{2}}{9}$.

(B) Given,$y = \sin(\log(x^2 + 2x + 1))$.
Since $x^2 + 2x + 1 = (x+1)^2$,we have $y = \sin(2 \log(x+1))$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \cos(2 \log(x+1)) \times \frac{2}{x+1}$.
Multiplying by $(x+1)$:
$(x+1) \frac{dy}{dx} = 2 \cos(2 \log(x+1))$.
Differentiating again with respect to $x$ using the product rule:
$(x+1) \frac{d^2y}{dx^2} + \frac{dy}{dx} = -2 \sin(2 \log(x+1)) \times \frac{2}{x+1}$.
Multiplying the entire equation by $(x+1)$:
$(x+1)^2 \frac{d^2y}{dx^2} + (x+1) \frac{dy}{dx} = -4 \sin(2 \log(x+1))$.
Since $y = \sin(2 \log(x+1))$,we get:
$(x+1)^2 \frac{d^2y}{dx^2} + (x+1) \frac{dy}{dx} = -4y$.

(C) Given,$y=x \log \left(\frac{x}{2-3 x}\right)$.
Applying the product rule and chain rule to differentiate with respect to $x$:
$\frac{dy}{dx} = x \cdot \frac{d}{dx} \left[ \log \left( \frac{x}{2-3x} \right) \right] + \log \left( \frac{x}{2-3x} \right) \cdot \frac{d}{dx}(x)$
$\frac{dy}{dx} = x \left( \frac{2-3x}{x} \right) \cdot \frac{(2-3x)(1) - x(-3)}{(2-3x)^2} + \log \left( \frac{x}{2-3x} \right)$
$\frac{dy}{dx} = x \left( \frac{2-3x}{x} \right) \cdot \frac{2}{(2-3x)^2} + \log \left( \frac{x}{2-3x} \right) = \frac{2}{2-3x} + \log \left( \frac{x}{2-3x} \right)$.
Now,differentiate again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( 2(2-3x)^{-1} \right) + \frac{d}{dx} \left[ \log \left( \frac{x}{2-3x} \right) \right]$
$\frac{d^2y}{dx^2} = -2(2-3x)^{-2}(-3) + \frac{2}{x(2-3x)} = \frac{6}{(2-3x)^2} + \frac{2}{x(2-3x)}$.
At $x = \frac{1}{2}$,$2-3x = 2 - 3(\frac{1}{2}) = 2 - 1.5 = 0.5 = \frac{1}{2}$.
Substituting $x = \frac{1}{2}$ and $2-3x = \frac{1}{2}$:
$\frac{d^2y}{dx^2} = \frac{6}{(1/2)^2} + \frac{2}{(1/2)(1/2)} = \frac{6}{1/4} + \frac{2}{1/4} = 24 + 8 = 32$.

(D) Given $y = a \cos x + (b + 2x) \sin x$.
First,find the first derivative $y^{\prime}$ using the product rule:
$y^{\prime} = -a \sin x + (b + 2x) \cos x + 2 \sin x$.
Now,find the second derivative $y^{\prime \prime}$:
$y^{\prime \prime} = -a \cos x - (b + 2x) \sin x + 2 \cos x + 2 \cos x$.
$y^{\prime \prime} = -(a \cos x + (b + 2x) \sin x) + 4 \cos x$.
Since $y = a \cos x + (b + 2x) \sin x$,we substitute $y$ into the expression:
$y^{\prime \prime} = -y + 4 \cos x$.
Therefore,$y^{\prime \prime} + y = 4 \cos x$.

(D) Given $y = \tan(\log x)$.
First,differentiate with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \sec^2(\log x) \cdot \frac{1}{x} = \frac{\sec^2(\log x)}{x}$.
Now,differentiate again with respect to $x$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$\frac{d^2y}{dx^2} = \frac{x \cdot \frac{d}{dx}(\sec^2(\log x)) - \sec^2(\log x) \cdot 1}{x^2}$.
Using the chain rule for $\frac{d}{dx}(\sec^2(\log x)) = 2 \sec(\log x) \cdot \sec(\log x) \tan(\log x) \cdot \frac{1}{x} = \frac{2 \sec^2(\log x) \tan(\log x)}{x}$.
Substituting this back:
$\frac{d^2y}{dx^2} = \frac{x \cdot \frac{2 \sec^2(\log x) \tan(\log x)}{x} - \sec^2(\log x)}{x^2}$.
$\frac{d^2y}{dx^2} = \frac{2 \sec^2(\log x) \tan(\log x) - \sec^2(\log x)}{x^2}$.
Factoring out $\sec^2(\log x)$:
$\frac{d^2y}{dx^2} = \frac{\sec^2(\log x)[2 \tan(\log x) - 1]}{x^2}$.

(A) Given $y = \frac{\log x}{x}$.
First,find the first derivative $\frac{dy}{dx}$ using the quotient rule:
$\frac{dy}{dx} = \frac{x(\frac{1}{x}) - \log x(1)}{x^2} = \frac{1 - \log x}{x^2}$.
Next,find the second derivative $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4}$.
Now,substitute these into the expression $x^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + y$:
$= x^2 \left( \frac{-3x + 2x \log x}{x^4} \right) + 3x \left( \frac{1 - \log x}{x^2} \right) + \frac{\log x}{x}$
$= \frac{-3x + 2x \log x}{x^2} + \frac{3 - 3 \log x}{x} + \frac{\log x}{x}$
$= \frac{-3 + 2 \log x}{x} + \frac{3 - 3 \log x}{x} + \frac{\log x}{x}$
$= \frac{-3 + 2 \log x + 3 - 3 \log x + \log x}{x} = \frac{0}{x} = 0$.
Thus,the value is $0$ for all $x > 0$.

(A) Given $y = \log(\cosh x)$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\cosh x} \cdot \frac{d}{dx}(\cosh x)$
Since $\frac{d}{dx}(\cosh x) = \sinh x$,we get:
$\frac{dy}{dx} = \frac{\sinh x}{\cosh x} = \tanh x$
Now,differentiate again with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(\tanh x)$
Since $\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$,we have:
$\frac{d^2 y}{dx^2} = \operatorname{sech}^2 x$
Thus,the correct option is $A$.

(B) $y=\tan \left(3 \tan ^{-1} x\right)$
Let $\tan ^{-1} x=\theta$,so $\tan \theta=x$.
Using the formula $\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}$,we get $y = \frac{3x-x^3}{1-3x^2}$.
Thus,$(1-3x^2)y = 3x-x^3$.
Differentiating both sides with respect to $x$:
$(1-3x^2)\frac{dy}{dx} + y(-6x) = 3-3x^2$.
Differentiating again with respect to $x$:
$(1-3x^2)\frac{d^2y}{dx^2} + (-6x)\frac{dy}{dx} - 6(x\frac{dy}{dx} + y) = -6x$.
$(1-3x^2)\frac{d^2y}{dx^2} - 6x\frac{dy}{dx} - 6x\frac{dy}{dx} - 6y = -6x$.
$(1-3x^2)\frac{d^2y}{dx^2} - 12x\frac{dy}{dx} = 6y - 6x = 6(y-x)$.

(C) Given $y = \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}}$.
Using the trigonometric identities $1 - \cos x = 2 \sin^2(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$,we get:
$y = \tan^{-1} \sqrt{\frac{2 \sin^2(x/2)}{2 \cos^2(x/2)}}$
$y = \tan^{-1} \sqrt{\tan^2(x/2)}$
$y = \tan^{-1}(\tan(x/2)) = \frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.
Again,differentiating with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{2}) = 0$.
Thus,the values are $\frac{1}{2}$ and $0$.

(D) Given,$y = \cos^{-1} \left( \frac{a \cos x - b \sin x}{\sqrt{a^2 + b^2}} \right)$.
Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \theta$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \theta$,where $\theta = \tan^{-1} \left( \frac{b}{a} \right)$.
Then,$y = \cos^{-1} (\cos \theta \cos x - \sin \theta \sin x)$.
Using the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get:
$y = \cos^{-1} (\cos(x + \theta))$.
$y = x + \theta$.
Since $\theta = \tan^{-1} \left( \frac{b}{a} \right)$ is a constant,differentiating with respect to $x$ gives:
$\frac{dy}{dx} = \frac{d}{dx} (x + \theta) = 1 + 0 = 1$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx} (1) = 0$.

(C) Given,$y=e^{\sin ^{-1} x}$.
Differentiating with respect to $x$:
$y_1 = \frac{d}{dx} e^{\sin ^{-1} x} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$
$y_1 = \frac{y}{\sqrt{1-x^2}}$
$y_1 \sqrt{1-x^2} = y$
Differentiating both sides with respect to $x$ using the product rule:
$y_2 \sqrt{1-x^2} + y_1 \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = y_1$
$y_2 \sqrt{1-x^2} - \frac{x y_1}{\sqrt{1-x^2}} = y_1$
Multiplying both sides by $\sqrt{1-x^2}$:
$y_2 (1-x^2) - x y_1 = y_1 \sqrt{1-x^2}$
Since $y_1 \sqrt{1-x^2} = y$,we have:
$(1-x^2) y_2 - x y_1 = y$
Thus,option $(c)$ is correct.

(B) Given: $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$ for $x>0$.
Differentiating both sides with respect to $x$:
$-\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} \frac{d y}{d x} = 2 \cdot \frac{1}{\left(\frac{x}{2}\right)} \cdot \frac{1}{2}$
$-\frac{1}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x} = \frac{2}{x}$
$x \frac{d y}{d x} = -2 \sqrt{b^2-y^2} \quad \dots (i)$
Differentiating again with respect to $x$:
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = -2 \cdot \frac{1}{2} (b^2-y^2)^{-1/2} (-2y) \frac{d y}{d x}$
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{2y}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x}$
From equation $(i)$,we have $\sqrt{b^2-y^2} = -\frac{x}{2} \frac{d y}{d x}$. Substituting this:
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{2y}{-\frac{x}{2} \frac{d y}{d x}} \cdot \frac{d y}{d x} = -\frac{4y}{x}$
Multiplying by $x$:
$x^2 \frac{d^2 y}{d x^2} + x \frac{d y}{d x} = -4y$.

(D) Given $y=a \sin x+(5+2 x) \cos x$.
Differentiating with respect to $x$ using the product rule:
$y^{\prime} = a \cos x + (5+2x)(-\sin x) + (2)\cos x$
$y^{\prime} = a \cos x - (5+2x)\sin x + 2 \cos x$.
Differentiating again with respect to $x$:
$y^{\prime \prime} = -a \sin x - [(5+2x)\cos x + (2)\sin x] - 2 \sin x$
$y^{\prime \prime} = -a \sin x - (5+2x)\cos x - 2 \sin x - 2 \sin x$
$y^{\prime \prime} = -(a \sin x + (5+2x)\cos x) - 4 \sin x$
Since $y = a \sin x + (5+2x)\cos x$,we substitute $y$ into the equation:
$y^{\prime \prime} = -y - 4 \sin x$
Therefore,$y^{\prime \prime} + y = -4 \sin x$.

(D) Given,$y = \frac{\sinh^{-1} x}{\sqrt{1+x^2}}$
$\sqrt{1+x^2} y = \sinh^{-1} x$
On differentiating with respect to $x$,we get:
$\sqrt{1+x^2} y_1 + y \cdot \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{1}{\sqrt{1+x^2}}$
Multiplying both sides by $\sqrt{1+x^2}$,we get:
$(1+x^2) y_1 + xy = 1$
Again,on differentiating with respect to $x$,we get:
$(1+x^2) y_2 + y_1(2x) + x y_1 + y = 0$
$(1+x^2) y_2 + 3xy_1 + y = 0$

(B) Given $y = \frac{\alpha x + \beta}{\gamma x + \delta}$.
First derivative: $y_1 = \frac{\alpha(\gamma x + \delta) - \gamma(\alpha x + \beta)}{(\gamma x + \delta)^2} = \frac{\alpha \delta - \gamma \beta}{(\gamma x + \delta)^2}$.
Second derivative: $y_2 = \frac{d}{dx} [(\alpha \delta - \gamma \beta)(\gamma x + \delta)^{-2}] = -2(\alpha \delta - \gamma \beta)(\gamma x + \delta)^{-3} \cdot \gamma = \frac{-2 \gamma(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^3}$.
Third derivative: $y_3 = \frac{d}{dx} [-2 \gamma(\alpha \delta - \gamma \beta)(\gamma x + \delta)^{-3}] = (-2 \gamma)(\alpha \delta - \gamma \beta) \cdot (-3)(\gamma x + \delta)^{-4} \cdot \gamma = \frac{6 \gamma^2(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^4}$.
Now,calculate $2 y_1 y_3 = 2 \cdot \left( \frac{\alpha \delta - \gamma \beta}{(\gamma x + \delta)^2} \right) \cdot \left( \frac{6 \gamma^2(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^4} \right) = \frac{12 \gamma^2(\alpha \delta - \gamma \beta)^2}{(\gamma x + \delta)^6}$.
Also,$3 y_2^2 = 3 \cdot \left( \frac{-2 \gamma(\alpha \delta - \gamma \beta)}{(\gamma x + \delta)^3} \right)^2 = 3 \cdot \frac{4 \gamma^2(\alpha \delta - \gamma \beta)^2}{(\gamma x + \delta)^6} = \frac{12 \gamma^2(\alpha \delta - \gamma \beta)^2}{(\gamma x + \delta)^6}$.
Thus,$2 y_1 y_3 = 3 y_2^2$.

(A) Given $y = \sin^{-1} x$.
Differentiating with respect to $x$,we get:
$y_1 = \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.
This implies $\sqrt{1 - x^2} y_1 = 1$.
Squaring both sides,we get $(1 - x^2) y_1^2 = 1$.
Differentiating again with respect to $x$:
$(1 - x^2) \cdot 2 y_1 y_2 + y_1^2 \cdot (-2x) = 0$.
Dividing by $2 y_1$ (assuming $y_1 \neq 0$):
$(1 - x^2) y_2 - x y_1 = 0$.

(B) Given $y = \sqrt{\sin x + y}$. Squaring both sides,we get $y^2 = \sin x + y$.
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = \cos x + \frac{dy}{dx} \implies \frac{dy}{dx}(2y - 1) = \cos x \implies \frac{dy}{dx} = \frac{\cos x}{2y - 1}$.
Differentiating again with respect to $x$ using the quotient rule:
$\frac{d^2 y}{dx^2} = \frac{(2y - 1)(-\sin x) - \cos x (2 \frac{dy}{dx})}{(2y - 1)^2}$.
Substituting $\frac{dy}{dx} = \frac{\cos x}{2y - 1}$:
$\frac{d^2 y}{dx^2} = \frac{-(2y - 1)\sin x - \frac{2 \cos^2 x}{2y - 1}}{(2y - 1)^2}$.
At the point $(\pi, 1)$,we have $x = \pi$ and $y = 1$:
$\sin \pi = 0$ and $\cos \pi = -1$.
$\left(\frac{d^2 y}{dx^2}\right)_{(\pi, 1)} = \frac{-(2(1) - 1)(0) - \frac{2(-1)^2}{2(1) - 1}}{(2(1) - 1)^2} = \frac{0 - \frac{2(1)}{1}}{1^2} = -2$.

(A) Given the infinite geometric series $y = 1 + x + x^2 + x^3 + . . . + \infty$ for $|x| < 1$.
Using the sum formula for an infinite geometric series,$y = \frac{1}{1 - x}$.
Differentiating with respect to $x$,we get $y^{\prime} = \frac{d}{dx}(1 - x)^{-1} = -1(1 - x)^{-2}(-1) = \frac{1}{(1 - x)^2}$.
Now,differentiating $y^{\prime}$ with respect to $x$ to find $y^{\prime \prime}$,we get $y^{\prime \prime} = \frac{d}{dx}(1 - x)^{-2} = -2(1 - x)^{-3}(-1) = \frac{2}{(1 - x)^3}$.
We can rewrite this as $y^{\prime \prime} = 2 \times \frac{1}{(1 - x)} \times \frac{1}{(1 - x)^2}$.
Since $y = \frac{1}{1 - x}$ and $y^{\prime} = \frac{1}{(1 - x)^2}$,we have $y^{\prime \prime} = 2 y y^{\prime}$.

(A) Given $x^x y^y=e^e$. Taking natural logarithm on both sides:
$\ln(x^x y^y) = \ln(e^e)$
$x \ln x + y \ln y = e$
Differentiating with respect to $x$:
$\frac{d}{dx}(x \ln x) + \frac{d}{dx}(y \ln y) = \frac{d}{dx}(e)$
$(1 + \ln x) + (1 + \ln y) \frac{dy}{dx} = 0$
At point $(e, e)$,$\ln e = 1$:
$(1 + 1) + (1 + 1) \left(\frac{dy}{dx}\right)_{(e, e)} = 0$
$2 + 2 \left(\frac{dy}{dx}\right)_{(e, e)} = 0 \Rightarrow \left(\frac{dy}{dx}\right)_{(e, e)} = -1$
Differentiating $(1 + \ln x) + (1 + \ln y) y' = 0$ again with respect to $x$:
$\frac{1}{x} + \frac{1}{y} y' \cdot y' + (1 + \ln y) y'' = 0$
$\frac{1}{x} + \frac{(y')^2}{y} + (1 + \ln y) y'' = 0$
At $(e, e)$ where $y' = -1$:
$\frac{1}{e} + \frac{(-1)^2}{e} + (1 + \ln e) y'' = 0$
$\frac{1}{e} + \frac{1}{e} + 2 y'' = 0$
$\frac{2}{e} + 2 y'' = 0 \Rightarrow y'' = -\frac{1}{e}$
Since $\left(\frac{dy}{dx}\right)_{(e, e)} = -1$,we can write $y'' = \frac{-1}{e} = \frac{1}{e} \left(\frac{dy}{dx}\right)_{(e, e)}$.

(B) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.
First derivative $y_1 = \frac{d}{dx}(e^{\sqrt{x}} + e^{-\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - e^{-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}$.
Multiplying by $2\sqrt{x}$,we get $2\sqrt{x} y_1 = e^{\sqrt{x}} - e^{-\sqrt{x}}$.
Differentiating both sides with respect to $x$:
$2\sqrt{x} y_2 + 2 y_1 \cdot \frac{1}{2\sqrt{x}} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$.
$2\sqrt{x} y_2 + \frac{y_1}{\sqrt{x}} = \frac{e^{\sqrt{x}} + e^{-\sqrt{x}}}{2\sqrt{x}} = \frac{y}{2\sqrt{x}}$.
Multiplying the entire equation by $2\sqrt{x}$:
$4 x y_2 + 2 y_1 = y$.

(B) Given $y = \frac{\log x}{x}$.
First,find the first derivative $\frac{dy}{dx}$ using the quotient rule:
$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2}$.
Now,find the second derivative $\frac{d^2 y}{dx^2}$ by differentiating $\frac{dy}{dx} = \frac{1 - \log x}{x^2}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{x^2 \cdot \frac{d}{dx}(1 - \log x) - (1 - \log x) \cdot \frac{d}{dx}(x^2)}{(x^2)^2}$
$\frac{d^2 y}{dx^2} = \frac{x^2 \cdot (-\frac{1}{x}) - (1 - \log x) \cdot (2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4} = \frac{-3 + 2 \log x}{x^3}$.
At $x = 1$:
$\frac{d^2 y}{dx^2} = \frac{-3 + 2 \log 1}{1^3} = \frac{-3 + 2(0)}{1} = -3$.

(B) $f(x) = \sin x \sin 2x \sin 3x$
$= \frac{1}{2} (2 \sin x \sin 2x) \sin 3x$
$= \frac{1}{2} (\cos x - \cos 3x) \sin 3x$
$= \frac{1}{2} (\cos x \sin 3x - \cos 3x \sin 3x)$
$= \frac{1}{4} (2 \sin 3x \cos x) - \frac{1}{4} (2 \sin 3x \cos 3x)$
$= \frac{1}{4} (\sin 4x + \sin 2x) - \frac{1}{4} \sin 6x$
$f'(x) = \frac{1}{4} (4 \cos 4x + 2 \cos 2x) - \frac{6}{4} \cos 6x = \cos 4x + \frac{1}{2} \cos 2x - \frac{3}{2} \cos 6x$
$f''(x) = -4 \sin 4x - \sin 2x + 9 \sin 6x$
Comparing with $f''(x) = a \sin bx + c \sin dx + e \sin kx$,we get:
$a = -4, b = 4, c = -1, d = 2, e = 9, k = 6$
$(a+c+e) - (b+d+k) = (-4 - 1 + 9) - (4 + 2 + 6) = 4 - 12 = -8$

(B) Let $p(x) = ax^3 + bx^2 + cx + d$ be a polynomial of degree $3$.
Then,the derivatives are:
$p^{\prime}(x) = 3ax^2 + 2bx + c$
$p^{\prime \prime}(x) = 6ax + 2b$
$p^{\prime \prime \prime}(x) = 6a$
Given $p^{\prime \prime \prime}(1) = 6$,we have $6a = 6$,which implies $a = 1$.
Given $p^{\prime \prime}(1) = 0$,we substitute $a = 1$ into $p^{\prime \prime}(1) = 6a(1) + 2b = 0$:
$6(1) + 2b = 0 \Rightarrow 2b = -6 \Rightarrow b = -3$.
Now,we need to find $p^{\prime \prime}(0)$:
$p^{\prime \prime}(0) = 6a(0) + 2b = 2b$.
Substituting $b = -3$,we get $p^{\prime \prime}(0) = 2(-3) = -6$.

(D) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.
Now,differentiating both sides with respect to $y$ using the chain rule:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{\frac{dy}{dx}} \right) = \frac{d}{dx} \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{dx}{dy}$.
Using the quotient rule or power rule:
$\frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right) = -\left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^2}$.
Substituting this back:
$\frac{d^2x}{dy^2} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^2} \cdot \frac{1}{\frac{dy}{dx}} = -\frac{\frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}$.
Given $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at point $P$:
$\left( \frac{d^2x}{dy^2} \right)_P = -\frac{-3}{(4)^3} = \frac{3}{64}$.
Thus,option $D$ is correct.

(B) Given $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$.
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = -2a^2 \cos x \sin x + 2b^2 \sin x \cos x = (b^2 - a^2) \sin 2x$.
Thus,$y \frac{dy}{dx} = \frac{b^2 - a^2}{2} \sin 2x$.
Differentiating again with respect to $x$:
$y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = (b^2 - a^2) \cos 2x$.
From the first derivative,$\frac{dy}{dx} = \frac{(b^2 - a^2) \sin 2x}{2y}$.
Substituting this into the second derivative equation:
$y \frac{d^2 y}{dx^2} + \frac{(b^2 - a^2)^2 \sin^2 2x}{4y^2} = (b^2 - a^2) \cos 2x$.
Multiplying by $y^2$:
$y^3 \frac{d^2 y}{dx^2} + \frac{(b^2 - a^2)^2 \sin^2 2x}{4} = y^2 (b^2 - a^2) \cos 2x$.
Using $y^2 = a^2 \cos^2 x + b^2 \sin^2 x$,we find that $y^3 (\frac{d^2 y}{dx^2} + y) = a^2 b^2$.
Therefore,$\frac{d^2 y}{dx^2} + y = \frac{a^2 b^2}{y^3} = \frac{1}{y} \left( \frac{ab}{y} \right)^2$.

(A) Given,$x = \sin \theta$ and $y = \cos(p \theta)$.
$\frac{dx}{d\theta} = \cos \theta$ and $\frac{dy}{d\theta} = -p \sin(p \theta)$.
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-p \sin(p \theta)}{\cos \theta}$.
Since $\sin(p \theta) = \sqrt{1 - y^2}$ and $\cos \theta = \sqrt{1 - x^2}$,we have $\frac{dy}{dx} = -p \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}$.
$\sqrt{1 - x^2} y_1 = -p \sqrt{1 - y^2}$.
Squaring both sides: $(1 - x^2) y_1^2 = p^2 (1 - y^2)$.
Differentiating with respect to $x$:
$(1 - x^2) \cdot 2 y_1 y_2 + y_1^2 (-2x) = p^2 (-2y y_1)$.
Dividing by $2 y_1$ (assuming $y_1 \neq 0$):
$(1 - x^2) y_2 - x y_1 = -p^2 y$.
Therefore,$(1 - x^2) y_2 = x y_1 - p^2 y$.
Thus,option $A$ is correct.

(A) Given the equation $y^2 = a x^2 + 2 x + c$.
Differentiating both sides with respect to $x$,we get:
$2 y \frac{d y}{d x} = 2 a x + 2$
$y \frac{d y}{d x} = a x + 1$
Differentiating again with respect to $x$ using the product rule:
$y \frac{d^2 y}{d x^2} + (\frac{d y}{d x})^2 = a$
From the first derivative,$\frac{d y}{d x} = \frac{a x + 1}{y}$.
Substituting this into the second derivative equation:
$y \frac{d^2 y}{d x^2} + (\frac{a x + 1}{y})^2 = a$
$y \frac{d^2 y}{d x^2} = a - \frac{(a x + 1)^2}{y^2}$
$y \frac{d^2 y}{d x^2} = \frac{a y^2 - (a x + 1)^2}{y^2}$
Substitute $y^2 = a x^2 + 2 x + c$:
$y \frac{d^2 y}{d x^2} = \frac{a(a x^2 + 2 x + c) - (a^2 x^2 + 2 a x + 1)}{y^2}$
$y \frac{d^2 y}{d x^2} = \frac{a^2 x^2 + 2 a x + a c - a^2 x^2 - 2 a x - 1}{y^2}$
$y \frac{d^2 y}{d x^2} = \frac{a c - 1}{y^2}$
Multiplying both sides by $y^2$:
$y^3 \frac{d^2 y}{d x^2} = a c - 1$.

(D) Given: $y = a \sin x + (5 + 2x) \cos x$ . . . . . . $(i)$
Differentiating with respect to $x$,we get:
$y' = a \cos x + 2 \cos x - (5 + 2x) \sin x$
$y' = (a + 2) \cos x - (5 + 2x) \sin x$
Again differentiating with respect to $x$,we get:
$y'' = -(a + 2) \sin x - 2 \sin x - (5 + 2x) \cos x$
$y'' = -(a + 4) \sin x - (5 + 2x) \cos x$ . . . . . . $(ii)$
Adding $(i)$ and $(ii)$,we get:
$y'' + y = [-(a + 4) \sin x - (5 + 2x) \cos x] + [a \sin x + (5 + 2x) \cos x]$
$y'' + y = -a \sin x - 4 \sin x - (5 + 2x) \cos x + a \sin x + (5 + 2x) \cos x$
$y'' + y = -4 \sin x$.

(C) Given the equation $a y^4 = (x + b)^5$.
Taking the natural logarithm on both sides: $\ln(a) + 4 \ln(y) = 5 \ln(x + b)$.
Differentiating with respect to $x$: $\frac{4}{y} \cdot \frac{dy}{dx} = \frac{5}{x + b}$.
Thus,$\frac{dy}{dx} = \frac{5y}{4(x + b)}$.
Differentiating again with respect to $x$ using the quotient rule: $\frac{d^2y}{dx^2} = \frac{5}{4} \cdot \frac{(x + b) \frac{dy}{dx} - y}{(x + b)^2}$.
Substitute $\frac{dy}{dx} = \frac{5y}{4(x + b)}$ into the expression: $\frac{d^2y}{dx^2} = \frac{5}{4} \cdot \frac{(x + b) \cdot \frac{5y}{4(x + b)} - y}{(x + b)^2} = \frac{5}{4} \cdot \frac{\frac{5y}{4} - y}{(x + b)^2} = \frac{5}{4} \cdot \frac{y}{4(x + b)^2} = \frac{5y}{16(x + b)^2}$.
Now,calculate the ratio: $\frac{y \cdot (\frac{d^2y}{dx^2})}{(\frac{dy}{dx})^2} = \frac{y \cdot \frac{5y}{16(x + b)^2}}{(\frac{5y}{4(x + b)})^2} = \frac{\frac{5y^2}{16(x + b)^2}}{\frac{25y^2}{16(x + b)^2}} = \frac{5}{25} = \frac{1}{5}$.

(D) Given $y = a \cos(\log x) + b \sin(\log x)$.
Differentiating with respect to $x$:
$y_1 = \frac{dy}{dx} = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(\log x) \cdot \frac{1}{x}$
$x y_1 = -a \sin(\log x) + b \cos(\log x)$.
Differentiating again with respect to $x$:
$x y_2 + y_1 = -a \cos(\log x) \cdot \frac{1}{x} - b \sin(\log x) \cdot \frac{1}{x}$
Multiply by $x$:
$x^2 y_2 + x y_1 = -[a \cos(\log x) + b \sin(\log x)]$
Since $y = a \cos(\log x) + b \sin(\log x)$,we have:
$x^2 y_2 + x y_1 = -y$.

(D) Given that $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$.
This implies that $f(x)$ can be written as $f(x)=(x-a)^2 g(x)$,where $g(a) \neq 0$.
First,we find the first derivative $f^{\prime}(x)$:
$f^{\prime}(x) = 2(x-a)g(x) + (x-a)^2 g^{\prime}(x)$.
Evaluating at $x=a$,we get $f^{\prime}(a) = 2(a-a)g(a) + (a-a)^2 g^{\prime}(a) = 0$.
Next,we find the second derivative $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = 2g(x) + 2(x-a)g^{\prime}(x) + 2(x-a)g^{\prime}(x) + (x-a)^2 g^{\prime \prime}(x) = 2g(x) + 4(x-a)g^{\prime}(x) + (x-a)^2 g^{\prime \prime}(x)$.
Evaluating at $x=a$,we get $f^{\prime \prime}(a) = 2g(a) + 4(a-a)g^{\prime}(a) + (a-a)^2 g^{\prime \prime}(a) = 2g(a)$.
Since $g(a) \neq 0$,it follows that $f^{\prime \prime}(a) \neq 0$.
Thus,$f(a)=0$,$f^{\prime}(a)=0$,and $f^{\prime \prime}(a) \neq 0$.

(C) Given,$y=e^{a \sin ^{-1} x}$.
On differentiating with respect to $x$,we get:
$y_1 = e^{a \sin ^{-1} x} \cdot \frac{a}{\sqrt{1-x^2}}$
$\Rightarrow y_1 \sqrt{1-x^2} = ay$
Squaring both sides:
$(1-x^2) y_1^2 = a^2 y^2$
Differentiating again with respect to $x$:
$(1-x^2) 2 y_1 y_2 - 2x y_1^2 = a^2 2y y_1$
Dividing by $2y_1$ (assuming $y_1 \neq 0$):
$(1-x^2) y_2 - x y_1 - a^2 y = 0$
Applying Leibnitz's theorem for the $n$-th derivative:
$[(1-x^2) y_{n+2} + n(-2x) y_{n+1} + \frac{n(n-1)}{2}(-2) y_n] - [x y_{n+1} + n(1) y_n] - a^2 y_n = 0$
$(1-x^2) y_{n+2} - 2nx y_{n+1} - n(n-1) y_n - x y_{n+1} - n y_n - a^2 y_n = 0$
$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} - (n^2 - n + n + a^2) y_n = 0$
$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} = (n^2 + a^2) y_n$

(B) Given that $f(x)=10 \cos x+(13+2 x) \sin x \quad ...(i)$
On differentiating with respect to $x$,we get
$f^{\prime}(x)=-10 \sin x+(13+2 x) \cos x+2 \sin x$
$f^{\prime}(x)=-8 \sin x+(13+2 x) \cos x$
Again differentiating with respect to $x$,we get
$f^{\prime \prime}(x)=-8 \cos x+(13+2 x)(-\sin x)+2 \cos x$
$f^{\prime \prime}(x)=-6 \cos x-(13+2 x) \sin x \quad ...(ii)$
On adding equations $(i)$ and $(ii)$,we get
$f^{\prime \prime}(x)+f(x) = [-6 \cos x-(13+2 x) \sin x] + [10 \cos x+(13+2 x) \sin x]$
$f^{\prime \prime}(x)+f(x) = 4 \cos x$

(C) Given $y = \sin^{-1} x$.
On differentiating both sides with respect to $x$,we get:
$\frac{d y}{d x} = \frac{1}{\sqrt{1-x^2}}$ ... $(i)$
Squaring both sides,we get:
$\left(\frac{d y}{d x}\right)^2 = \frac{1}{1-x^2}$
$(1-x^2) \left(\frac{d y}{d x}\right)^2 = 1$
Differentiating both sides with respect to $x$ using the product rule:
$(1-x^2) \cdot 2 \left(\frac{d y}{d x}\right) \cdot \frac{d^2 y}{d x^2} + \left(\frac{d y}{d x}\right)^2 \cdot (-2x) = 0$
Dividing by $2 \frac{d y}{d x}$ (assuming $\frac{d y}{d x} \neq 0$):
$(1-x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x} = 0$
Therefore,$(1-x^2) \frac{d^2 y}{d x^2} = x \frac{d y}{d x}$.

(D) Given $y = \cos(\sin x)$.
First derivative $y_1 = \frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$.
Second derivative $y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}[-\sin(\sin x) \cdot \cos x]$.
Using the product rule: $y_2 = -[\cos(\sin x) \cdot \cos x \cdot \cos x + \sin(\sin x) \cdot (-\sin x)]$.
$y_2 = -\cos(\sin x) \cos^2 x + \sin(\sin x) \sin x$.
Now,substitute $y_1$ and $y_2$ into the expression $y_1 \sin x + y_2 \cos x$:
$= [-\sin(\sin x) \cos x] \sin x + [-\cos(\sin x) \cos^2 x + \sin(\sin x) \sin x] \cos x$.
$= -\sin(\sin x) \sin x \cos x - \cos(\sin x) \cos^3 x + \sin(\sin x) \sin x \cos x$.
$= -\cos(\sin x) \cos^3 x$.
Since $y = \cos(\sin x)$,the expression simplifies to $-y \cos^3 x$.

(C) Given $f(x) = \frac{x^2}{x+a}$.
Using the quotient rule,$f^{\prime}(x) = \frac{(x+a)(2x) - x^2(1)}{(x+a)^2} = \frac{2x^2 + 2ax - x^2}{(x+a)^2} = \frac{x^2 + 2ax}{(x+a)^2}$.
Now,find $f^{\prime \prime}(x)$ by differentiating $f^{\prime}(x) = \frac{x^2 + 2ax}{(x+a)^2}$:
$f^{\prime \prime}(x) = \frac{(x+a)^2(2x + 2a) - (x^2 + 2ax)(2)(x+a)}{(x+a)^4}$.
Simplify by canceling $(x+a)$:
$f^{\prime \prime}(x) = \frac{(x+a)(2x + 2a) - 2(x^2 + 2ax)}{(x+a)^3} = \frac{2x^2 + 4ax + 2a^2 - 2x^2 - 4ax}{(x+a)^3} = \frac{2a^2}{(x+a)^3}$.
Substitute $x = a$:
$f^{\prime \prime}(a) = \frac{2a^2}{(a+a)^3} = \frac{2a^2}{(2a)^3} = \frac{2a^2}{8a^3} = \frac{1}{4a}$.

(A) Given equation is $y = A \cos n x + B \sin n x$.
First,differentiate with respect to $x$:
$y_1 = \frac{dy}{dx} = -A n \sin n x + B n \cos n x$.
Now,differentiate again with respect to $x$ to find $y_2$:
$y_2 = \frac{d^2y}{dx^2} = -A n^2 \cos n x - B n^2 \sin n x$.
Factor out $-n^2$ from the expression:
$y_2 = -n^2 (A \cos n x + B \sin n x)$.
Since $y = A \cos n x + B \sin n x$,we substitute $y$ into the equation:
$y_2 = -n^2 y$.
Rearranging the terms,we get:
$y_2 + n^2 y = 0$.

(D) Given,$y=f(x)$ is a differentiable and bijective function.
We know that $\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} = \left(\frac{d y}{d x}\right)^{-1}$.
Differentiating both sides with respect to $y$:
$\frac{d^2 x}{d y^2} = \frac{d}{d y} \left[ \left(\frac{d y}{d x}\right)^{-1} \right]$.
Using the chain rule:
$\frac{d^2 x}{d y^2} = \frac{d}{d x} \left[ \left(\frac{d y}{d x}\right)^{-1} \right] \cdot \frac{d x}{d y}$.
$\frac{d^2 x}{d y^2} = -\left(\frac{d y}{d x}\right)^{-2} \cdot \frac{d^2 y}{d x^2} \cdot \frac{1}{\frac{d y}{d x}}$.
$\frac{d^2 x}{d y^2} = -\frac{\frac{d^2 y}{d x^2}}{\left(\frac{d y}{d x}\right)^3}$.
Multiplying both sides by $\left(\frac{d y}{d x}\right)^3$:
$\frac{d^2 x}{d y^2} \cdot \left(\frac{d y}{d x}\right)^3 = -\frac{d^2 y}{d x^2}$.
Therefore,$\frac{d^2 x}{d y^2} \left(\frac{d y}{d x}\right)^3 + \frac{d^2 y}{d x^2} = 0$.