Higher order derivatives Questions in English

(A) Given $g(x) = \log(f(x))$ and $f(x+1) = x f(x)$.
Taking logarithm on both sides,$\log(f(x+1)) = \log(x) + \log(f(x))$.
This implies $g(x+1) = g(x) + \log(x)$,or $g(x+1) - g(x) = \log(x)$.
Differentiating twice with respect to $x$,we get $g^{\prime \prime}(x+1) - g^{\prime \prime}(x) = -\frac{1}{x^2}$.
We want to find $g^{\prime \prime}\left(N+\frac{1}{2}\right) - g^{\prime \prime}\left(\frac{1}{2}\right)$.
We can write this as a telescoping sum:
$g^{\prime \prime}\left(N+\frac{1}{2}\right) - g^{\prime \prime}\left(\frac{1}{2}\right) = \sum_{k=1}^{N} \left[ g^{\prime \prime}\left(k+\frac{1}{2}\right) - g^{\prime \prime}\left(k-\frac{1}{2}\right) \right]$.
Using the relation $g^{\prime \prime}(x+1) - g^{\prime \prime}(x) = -\frac{1}{x^2}$,we substitute $x = k - \frac{1}{2}$:
$g^{\prime \prime}\left(k+\frac{1}{2}\right) - g^{\prime \prime}\left(k-\frac{1}{2}\right) = -\frac{1}{(k-\frac{1}{2})^2} = -\frac{1}{(\frac{2k-1}{2})^2} = -\frac{4}{(2k-1)^2}$.
Summing from $k=1$ to $N$:
$\sum_{k=1}^{N} -\frac{4}{(2k-1)^2} = -4 \left[ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots + \frac{1}{(2N-1)^2} \right]$.
This simplifies to $-4 \left\{ 1 + \frac{1}{9} + \frac{1}{25} + \ldots + \frac{1}{(2N-1)^2} \right\}$.

(B) Using the Taylor series expansion of $P(x)$ about $x=2$:
$P(x) = P(2) + P^{\prime}(2)(x-2) + \frac{P^{\prime \prime}(2)}{2!}(x-2)^2$
Given $P(2)=-1, P^{\prime}(2)=0, P^{\prime \prime}(2)=2$:
$P(x) = -1 + 0(x-2) + \frac{2}{2}(x-2)^2$
$P(x) = -1 + (x-2)^2$
Now,substitute $x=1.001$:
$P(1.001) = -1 + (1.001-2)^2$
$P(1.001) = -1 + (-0.999)^2$
$P(1.001) = -1 + 0.998001$
$P(1.001) = -0.001999$
Rounding to the nearest provided option,we get $-0.002$.

(B) Given $y = (\tan^{-1} x)^2$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2}$.
Multiplying both sides by $(1+x^2)$,we have:
$(1+x^2) \frac{dy}{dx} = 2 \tan^{-1} x$.
Differentiating again with respect to $x$ using the product rule on the left side:
$(1+x^2) \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot (2x) = 2 \cdot \frac{1}{1+x^2}$.
Multiplying the entire equation by $(1+x^2)$,we get:
$(1+x^2)^2 \frac{d^2 y}{dx^2} + 2x(1+x^2) \frac{dy}{dx} = 2$.
Thus,the value is $2$.

(C) Given $y = 3 e^{5 x} + 5 e^{3 x}$.
First,find the first derivative $\frac{d y}{d x}$:
$\frac{d y}{d x} = \frac{d}{d x}(3 e^{5 x} + 5 e^{3 x}) = 3(5 e^{5 x}) + 5(3 e^{3 x}) = 15 e^{5 x} + 15 e^{3 x}$.
Next,find the second derivative $\frac{d^{2} y}{d x^{2}}$:
$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x}(15 e^{5 x} + 15 e^{3 x}) = 15(5 e^{5 x}) + 15(3 e^{3 x}) = 75 e^{5 x} + 45 e^{3 x}$.
Now,calculate $\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x}$:
$\frac{d^{2} y}{d x^{2}} - 8 \frac{d y}{d x} = (75 e^{5 x} + 45 e^{3 x}) - 8(15 e^{5 x} + 15 e^{3 x})$
$= 75 e^{5 x} + 45 e^{3 x} - 120 e^{5 x} - 120 e^{3 x}$
$= -45 e^{5 x} - 75 e^{3 x}$
$= -15(3 e^{5 x} + 5 e^{3 x})$
$= -15 y$.

(B) Given $(a+bx) e^{\frac{y}{x}}=x$.
Taking natural logarithm on both sides: $\ln(a+bx) + \frac{y}{x} = \ln x$.
Rearranging gives $\frac{y}{x} = \ln x - \ln(a+bx) = \ln \left(\frac{x}{a+bx}\right)$.
So,$y = x \ln \left(\frac{x}{a+bx}\right)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \ln \left(\frac{x}{a+bx}\right) + x \cdot \frac{a+bx}{x} \cdot \frac{(a+bx)(1) - x(b)}{(a+bx)^2} = \frac{y}{x} + \frac{a}{(a+bx)}$.
Thus,$x \frac{dy}{dx} = y + \frac{ax}{a+bx}$.
Differentiating again with respect to $x$:
$x \frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{dy}{dx} + \frac{(a+bx)(a) - ax(b)}{(a+bx)^2} = \frac{dy}{dx} + \frac{a^2}{(a+bx)^2}$.
So,$x \frac{d^2y}{dx^2} = \frac{a^2}{(a+bx)^2}$.
From $x \frac{dy}{dx} - y = \frac{ax}{a+bx}$,we have $\left(x \frac{dy}{dx} - y\right)^2 = \frac{a^2 x^2}{(a+bx)^2}$.
Comparing this with $x \frac{d^2y}{dx^2}$,we find $x^3 \frac{d^2y}{dx^2} = x^2 \cdot \frac{a^2}{(a+bx)^2} = \left(x \frac{dy}{dx} - y\right)^2$.

(D) Given,$x^{2} y^{5}=(x+y)^{7}$.
Taking the natural logarithm on both sides,we get:
$2 \ln x + 5 \ln y = 7 \ln (x+y)$.
Differentiating both sides with respect to $x$:
$\frac{2}{x} + \frac{5}{y} \frac{dy}{dx} = \frac{7}{x+y} \left(1 + \frac{dy}{dx}\right)$.
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} \left(\frac{5}{y} - \frac{7}{x+y}\right) = \frac{7}{x+y} - \frac{2}{x}$.
$\frac{dy}{dx} \left(\frac{5x + 5y - 7y}{y(x+y)}\right) = \frac{7x - 2x - 2y}{x(x+y)}$.
$\frac{dy}{dx} \left(\frac{5x - 2y}{y(x+y)}\right) = \frac{5x - 2y}{x(x+y)}$.
Thus,$\frac{dy}{dx} = \frac{y}{x}$.
Now,differentiating again with respect to $x$ using the quotient rule:
$\frac{d^{2}y}{dx^{2}} = \frac{x \frac{dy}{dx} - y(1)}{x^{2}}$.
Substituting $\frac{dy}{dx} = \frac{y}{x}$:
$\frac{d^{2}y}{dx^{2}} = \frac{x(y/x) - y}{x^{2}} = \frac{y - y}{x^{2}} = 0$.

(B) Given $x = \log t$ and $y = \frac{1}{t}$.
First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Since $y = t^{-1}$,$\frac{dy}{dt} = -t^{-2} = -\frac{1}{t^2}$.
Since $x = \log t$,$\frac{dx}{dt} = \frac{1}{t}$.
Therefore,$\frac{dy}{dx} = \frac{-1/t^2}{1/t} = -\frac{1}{t} = -y$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2 y}{dx^2}$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(-y) = -\frac{dy}{dx}$.
Substituting $\frac{dy}{dx} = -y$ into the expression:
$\frac{d^2 y}{dx^2} = -(-y) = y$.
Wait,let's re-evaluate: $\frac{d}{dx}(-y) = -\frac{dy}{dt} \cdot \frac{dt}{dx} = -(-t^{-2}) \cdot t = t^{-1} = y$.
Thus,$\frac{d^2 y}{dx^2} = y$.

(D) Given that $x = \sin \theta$ and $y = \sin^3 \theta$.
Since $y = (\sin \theta)^3$,we can substitute $x = \sin \theta$ to get $y = x^3$.
Now,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^3) = 3x^2$.
Next,find the second derivative $\frac{d^2 y}{dx^2}$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(3x^2) = 6x$.
At $\theta = \frac{\pi}{2}$,the value of $x$ is $x = \sin(\frac{\pi}{2}) = 1$.
Substituting $x = 1$ into the second derivative:
$\left(\frac{d^2 y}{dx^2}\right)_{x=1} = 6(1) = 6$.

(C) Given $y=3 \cos (\log x)+4 \sin (\log x)$.
First,differentiate with respect to $x$:
$y_1 = \frac{dy}{dx} = -3 \sin (\log x) \cdot \frac{1}{x} + 4 \cos (\log x) \cdot \frac{1}{x}$.
Multiply both sides by $x$:
$x y_1 = -3 \sin (\log x) + 4 \cos (\log x)$.
Differentiate again with respect to $x$ using the product rule on the left side:
$1 \cdot y_1 + x \cdot y_2 = -3 \cos (\log x) \cdot \frac{1}{x} - 4 \sin (\log x) \cdot \frac{1}{x}$.
Multiply the entire equation by $x$:
$x y_1 + x^2 y_2 = -3 \cos (\log x) - 4 \sin (\log x)$.
Factor out $-1$ from the right side:
$x^2 y_2 + x y_1 = -(3 \cos (\log x) + 4 \sin (\log x))$.
Since $y = 3 \cos (\log x) + 4 \sin (\log x)$,we have:
$x^2 y_2 + x y_1 = -y$.

(D) Given the function $f(x) = b \cdot e^{ax} + a \cdot e^{bx}$.
First,find the first derivative $f^{\prime}(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(b \cdot e^{ax} + a \cdot e^{bx}) = b \cdot a \cdot e^{ax} + a \cdot b \cdot e^{bx} = ab(e^{ax} + e^{bx})$.
Next,find the second derivative $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = \frac{d}{dx}(ab \cdot e^{ax} + ab \cdot e^{bx}) = ab \cdot a \cdot e^{ax} + ab \cdot b \cdot e^{bx} = a^2b \cdot e^{ax} + ab^2 \cdot e^{bx}$.
Now,evaluate at $x = 0$:
$f^{\prime \prime}(0) = a^2b \cdot e^{a(0)} + ab^2 \cdot e^{b(0)} = a^2b(1) + ab^2(1) = a^2b + ab^2$.
Factoring out $ab$,we get:
$f^{\prime \prime}(0) = ab(a + b)$.

(B) Given $y = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$.
We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.
Substituting this,$y = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1-x^2}} + 2(\frac{\pi}{2} - \sin^{-1} x) \cdot (-\frac{1}{\sqrt{1-x^2}})$.
$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} (\sin^{-1} x - \frac{\pi}{2} + \sin^{-1} x) = \frac{2(2\sin^{-1} x - \frac{\pi}{2})}{\sqrt{1-x^2}}$.
Multiplying by $\sqrt{1-x^2}$: $\sqrt{1-x^2} y_1 = 4\sin^{-1} x - \pi$.
Differentiating again with respect to $x$:
$\sqrt{1-x^2} y_2 + y_1 \cdot (\frac{-2x}{2\sqrt{1-x^2}}) = \frac{4}{\sqrt{1-x^2}}$.
Multiplying throughout by $\sqrt{1-x^2}$: $(1-x^2) y_2 - x y_1 = 4$.

(C) Given $x = \sqrt{e^{\sin^{-1} t}}$ and $y = \sqrt{e^{\cos^{-1} t}}$.
Multiplying $x$ and $y$,we get:
$xy = \sqrt{e^{\sin^{-1} t}} \cdot \sqrt{e^{\cos^{-1} t}} = \sqrt{e^{\sin^{-1} t + \cos^{-1} t}}$.
Since $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have $xy = \sqrt{e^{\pi/2}}$,which is a constant.
Differentiating both sides with respect to $x$:
$x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \quad ... (i)$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{dx^2} = -\frac{d}{dx} \left( \frac{y}{x} \right) = -\left( \frac{x \frac{dy}{dx} - y}{x^2} \right)$.
Substituting $\frac{dy}{dx} = -\frac{y}{x}$ from $(i)$:
$\frac{d^2 y}{dx^2} = -\left( \frac{x(-\frac{y}{x}) - y}{x^2} \right) = -\left( \frac{-y - y}{x^2} \right) = -\left( \frac{-2y}{x^2} \right) = \frac{2y}{x^2}$.

(A) Given,$y^{2}=a x^{2}+b x+c$.
On differentiating with respect to $x$,we get $2 y \frac{d y}{d x}=2 a x+b$,which implies $\frac{d y}{d x} = \frac{2 a x + b}{2 y}$.
Again differentiating with respect to $x$,we get $2(\frac{d y}{d x})^{2} + 2 y \frac{d^{2} y}{d x^{2}} = 2 a$.
Dividing by $2$,we have $(\frac{d y}{d x})^{2} + y \frac{d^{2} y}{d x^{2}} = a$.
Substituting $\frac{d y}{d x} = \frac{2 a x + b}{2 y}$,we get $y \frac{d^{2} y}{d x^{2}} = a - (\frac{2 a x + b}{2 y})^{2}$.
$y \frac{d^{2} y}{d x^{2}} = \frac{4 a y^{2} - (2 a x + b)^{2}}{4 y^{2}}$.
Substituting $y^{2} = a x^{2} + b x + c$,we get $y \frac{d^{2} y}{d x^{2}} = \frac{4 a (a x^{2} + b x + c) - (4 a^{2} x^{2} + 4 a b x + b^{2})}{4 y^{2}}$.
$y \frac{d^{2} y}{d x^{2}} = \frac{4 a^{2} x^{2} + 4 a b x + 4 a c - 4 a^{2} x^{2} - 4 a b x - b^{2}}{4 y^{2}}$.
$y \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4 y^{2}}$.
Multiplying both sides by $y^{2}$,we get $y^{3} \frac{d^{2} y}{d x^{2}} = \frac{4 a c - b^{2}}{4}$,which is a constant.

(B) Given $y = a x^{n+1} + b x^{-n}$.
First derivative: $\frac{dy}{dx} = a(n+1)x^n + b(-n)x^{-n-1}$.
Second derivative: $\frac{d^2y}{dx^2} = a(n+1)nx^{n-1} + b(-n)(-n-1)x^{-n-2}$.
$\frac{d^2y}{dx^2} = an(n+1)x^{n-1} + bn(n+1)x^{-n-2}$.
Multiply by $x^2$: $x^2 \frac{d^2y}{dx^2} = an(n+1)x^{n+1} + bn(n+1)x^{-n}$.
$x^2 \frac{d^2y}{dx^2} = n(n+1) [a x^{n+1} + b x^{-n}]$.
Since $y = a x^{n+1} + b x^{-n}$,we have $x^2 \frac{d^2y}{dx^2} = n(n+1)y$.

(C) Given $y = a^x \cdot b^{2x-1}$.
We can rewrite this as $y = a^x \cdot (b^2)^x \cdot b^{-1} = \frac{1}{b} (a b^2)^x$.
Taking the natural logarithm on both sides: $\log y = \log \left( \frac{1}{b} (a b^2)^x \right) = \log(1) - \log(b) + x \log(a b^2) = -\log b + x \log(a b^2)$.
Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \log(a b^2)$.
So,$\frac{dy}{dx} = y \log(a b^2)$.
Differentiating again with respect to $x$: $\frac{d^2 y}{d x^2} = \frac{dy}{dx} \log(a b^2)$.
Substituting $\frac{dy}{dx} = y \log(a b^2)$,we get $\frac{d^2 y}{d x^2} = (y \log(a b^2)) \cdot \log(a b^2) = y(\log(a b^2))^2$.

(D) Let $y = \log x$.
Then,the first derivative is $y_1 = \frac{d}{dx}(\log x) = x^{-1}$.
The second derivative is $y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2}$.
The third derivative is $y_3 = \frac{d}{dx}(-1 \cdot x^{-2}) = (-1)(-2) \cdot x^{-3} = 2 \cdot x^{-3}$.
The fourth derivative is $y_4 = \frac{d}{dx}(2 \cdot x^{-3}) = 2(-3) \cdot x^{-4} = -6 \cdot x^{-4} = -3! \cdot x^{-4}$.
By observing the pattern,the $n$-th derivative is given by $y_n = (-1)^{n-1} \cdot (n-1)! \cdot x^{-n}$.
Thus,$\frac{d^{n}}{d x^{n}}(\log x) = (-1)^{n-1} \frac{(n-1)!}{x^{n}}$.

(B) Given $f(x) = 3x^3 + Ax^2 + Bx + C$,where $A = 2f'(1)$,$B = f''(2)$,and $C = f'''(3)$.
First,find the derivatives:
$f'(x) = 9x^2 + 2Ax + B$
$f''(x) = 18x + 2A$
$f'''(x) = 18$
Now,evaluate the constants:
$f'''(3) = 18$,so $C = 18$.
$f''(2) = 18(2) + 2A = 36 + 2A$. Since $B = f''(2)$,we have $B = 36 + 2A$.
$f'(1) = 9(1)^2 + 2A(1) + B = 9 + 2A + B$. Since $A = 2f'(1)$,we have $A = 2(9 + 2A + B) = 18 + 4A + 2B$.
Substitute $B = 36 + 2A$ into the equation for $A$:
$A = 18 + 4A + 2(36 + 2A)$
$A = 18 + 4A + 72 + 4A$
$A = 90 + 8A$
$-7A = 90 \implies A = -\frac{90}{7}$.
Now find $B$:
$B = 36 + 2(-\frac{90}{7}) = 36 - \frac{180}{7} = \frac{252 - 180}{7} = \frac{72}{7}$.
Thus,$f(x) = 3x^3 - \frac{90}{7}x^2 + \frac{72}{7}x + 18 = \frac{1}{7}(21x^3 - 90x^2 + 72x + 126)$.
Therefore,the correct option is $B$.

(D) Given the equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Differentiating both sides with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0$.
$\frac{y}{b^2} \cdot \frac{dy}{dx} = -\frac{x}{a^2} \implies \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$.
Now,differentiate again with respect to $x$ using the quotient rule:
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{d}{dx} \left( \frac{x}{y} \right) = -\frac{b^2}{a^2} \cdot \left( \frac{y(1) - x(dy/dx)}{y^2} \right)$.
Substitute $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$:
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y - x(-b^2 x / a^2 y)}{y^2} \right) = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 y^2 + b^2 x^2}{a^2 y^3} \right)$.
Since $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $b^2 x^2 + a^2 y^2 = a^2 b^2$.
Substituting this into the expression:
$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 b^2}{a^2 y^3} \right) = -\frac{b^4}{a^2 y^3}$.

(A) Given $y = A \cos nx + B \sin nx$ ...$(i)$
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = -An \sin nx + Bn \cos nx$
Again differentiating with respect to $x$,we get:
$\frac{d^2 y}{dx^2} = -An^2 \cos nx - Bn^2 \sin nx$
Factoring out $-n^2$:
$\frac{d^2 y}{dx^2} = -n^2 (A \cos nx + B \sin nx)$
Substituting $y$ from equation $(i)$:
$\frac{d^2 y}{dx^2} = -n^2 y$

(A) Given $y = a x^{n+1} + b x^{-n}$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = a(n+1) x^n - bn x^{-n-1}$.
Now,differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2 y}{dx^2} = a(n+1)n x^{n-1} - bn(-n-1) x^{-n-2}$.
$\frac{d^2 y}{dx^2} = a n(n+1) x^{n-1} + b n(n+1) x^{-n-2}$.
Factor out $n(n+1)$:
$\frac{d^2 y}{dx^2} = n(n+1) [a x^{n-1} + b x^{-n-2}]$.
Multiply both sides by $x^2$:
$x^2 \frac{d^2 y}{dx^2} = n(n+1) [a x^{n+1} + b x^{-n}]$.
Since $y = a x^{n+1} + b x^{-n}$,we substitute $y$ into the equation:
$x^2 \frac{d^2 y}{dx^2} = n(n+1) y$.

(A) We know the trigonometric identity $\cos^2 \theta - \sin^2 \theta = \cos(2\theta)$.
Given $y = \cos^2\left(\frac{5x}{2}\right) - \sin^2\left(\frac{5x}{2}\right)$,we can substitute $\theta = \frac{5x}{2}$.
Thus,$y = \cos\left(2 \times \frac{5x}{2}\right) = \cos(5x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = -\sin(5x) \times \frac{d}{dx}(5x) = -5\sin(5x)$.
Differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2y}{dx^2} = -5 \times \cos(5x) \times \frac{d}{dx}(5x) = -5 \times 5 \cos(5x) = -25\cos(5x)$.
Since $y = \cos(5x)$,we have $\frac{d^2y}{dx^2} = -25y$.

(D) Given $y = 2 \sin x + 3 \cos x$.
First,find the first derivative: $\frac{dy}{dx} = 2 \cos x - 3 \sin x$.
Next,find the second derivative: $\frac{d^2y}{dx^2} = -2 \sin x - 3 \cos x$.
We can rewrite this as $\frac{d^2y}{dx^2} = -(2 \sin x + 3 \cos x) = -y$.
Rearranging the equation,we get $y + \frac{d^2y}{dx^2} = 0$.
Comparing this with the given equation $y + A \frac{d^2y}{dx^2} = B$,we find $A = 1$ and $B = 0$.

(C) Given $y^2 = ax^2 + bx + c$.
Differentiating with respect to $x$,we get:
$2y \frac{dy}{dx} = 2ax + b$
$\frac{dy}{dx} = \frac{2ax + b}{2y}$
Differentiating again with respect to $x$ using the quotient rule:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2ax + b}{2y} \right) = \frac{2a(2y) - (2ax + b)(2 \frac{dy}{dx})}{4y^2}$
Substituting $\frac{dy}{dx} = \frac{2ax + b}{2y}$:
$\frac{d^2y}{dx^2} = \frac{4ay - (2ax + b) \frac{2ax + b}{y}}{4y^2} = \frac{4ay^2 - (2ax + b)^2}{4y^3}$
Now,multiply by $y^3$:
$y^3 \frac{d^2y}{dx^2} = \frac{4ay^2 - (2ax + b)^2}{4} = \frac{4a(ax^2 + bx + c) - (4a^2x^2 + 4abx + b^2)}{4}$
$y^3 \frac{d^2y}{dx^2} = \frac{4a^2x^2 + 4abx + 4ac - 4a^2x^2 - 4abx - b^2}{4} = \frac{4ac - b^2}{4}$
Since $a, b, c$ are constants,$\frac{4ac - b^2}{4}$ is a constant.

(A) Given $y = e^{4x} \cos 5x$.
Applying the product rule,$\frac{dy}{dx} = e^{4x}(-5 \sin 5x) + \cos 5x(4e^{4x}) = e^{4x}(4 \cos 5x - 5 \sin 5x)$.
Now,differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = e^{4x}(-20 \sin 5x - 25 \cos 5x) + (4 \cos 5x - 5 \sin 5x)(4e^{4x})$.
Evaluating at $x=0$:
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = e^{0}(-20 \sin 0 - 25 \cos 0) + (4 \cos 0 - 5 \sin 0)(4e^{0})$.
Since $\sin 0 = 0$ and $\cos 0 = 1$:
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = 1(0 - 25) + (4 - 0)(4) = -25 + 16 = -9$.

(A) Given $\sqrt{x+y}+\sqrt{y-x}=5$.
Rearranging the terms,we get $\sqrt{y-x}=5-\sqrt{x+y}$.
Squaring both sides,we have $y-x = 25 + (x+y) - 10\sqrt{x+y}$.
Simplifying,$-2x = 25 - 10\sqrt{x+y}$,which gives $10\sqrt{x+y} = 2x + 25$.
Differentiating both sides with respect to $x$:
$10 \times \frac{1}{2\sqrt{x+y}} \times (1 + \frac{dy}{dx}) = 2$.
$\frac{5}{\sqrt{x+y}} \times (1 + \frac{dy}{dx}) = 2$.
$1 + \frac{dy}{dx} = \frac{2\sqrt{x+y}}{5}$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{2}{5} \times \frac{1}{2\sqrt{x+y}} \times (1 + \frac{dy}{dx})$.
Substituting the value of $(1 + \frac{dy}{dx})$ from the previous step:
$\frac{d^2y}{dx^2} = \frac{1}{5\sqrt{x+y}} \times \frac{2\sqrt{x+y}}{5} = \frac{2}{25}$.

(D) We are given the function $y = \cos^{2}\left(\frac{5x}{2}\right) - \sin^{2}\left(\frac{5x}{2}\right)$.
Using the trigonometric identity $\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta)$,we can simplify the expression by setting $\theta = \frac{5x}{2}$.
Thus,$y = \cos\left(2 \times \frac{5x}{2}\right) = \cos(5x)$.
Now,we find the first derivative with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos(5x)) = -5 \sin(5x)$.
Next,we find the second derivative with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(-5 \sin(5x)) = -5 \times 5 \cos(5x) = -25 \cos(5x)$.
Since $y = \cos(5x)$,we substitute $y$ back into the expression:
$\frac{d^{2}y}{dx^{2}} = -25y$.

(D) Given $h(x) = [f(x)]^2 + [g(x)]^2$.
Taking the derivative with respect to $x$:
$h^{\prime}(x) = 2f(x)f^{\prime}(x) + 2g(x)g^{\prime}(x)$.
Since $f^{\prime}(x) = g(x)$,we have $g^{\prime}(x) = f^{\prime \prime}(x)$.
Given $f^{\prime \prime}(x) = -f(x)$,we substitute these into the derivative:
$h^{\prime}(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$.
Since $h^{\prime}(x) = 0$,$h(x)$ is a constant function.
Therefore,$h(5) = h(10) = 1$.

(B) Given $h(x) = [f(x)]^2 + [g(x)]^2$.
Differentiating $h(x)$ with respect to $x$,we get:
$h^{\prime}(x) = 2f(x)f^{\prime}(x) + 2g(x)g^{\prime}(x)$.
Since $f^{\prime}(x) = g(x)$,it follows that $f^{\prime \prime}(x) = g^{\prime}(x)$.
Given $f^{\prime \prime}(x) = -f(x)$,we have $g^{\prime}(x) = -f(x)$.
Substituting these into the derivative expression:
$h^{\prime}(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$.
Since $h^{\prime}(x) = 0$,$h(x)$ is a constant function.
Therefore,$h(5) = h(10) = 1$.

(C) Given the equation $y = 7 \sin x + 5 \cos x$.
First,find the first derivative with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(7 \sin x + 5 \cos x) = 7 \cos x - 5 \sin x$.
Next,find the second derivative with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(7 \cos x - 5 \sin x) = -7 \sin x - 5 \cos x$.
Factor out $-1$ from the expression:
$\frac{d^2y}{dx^2} = -(7 \sin x + 5 \cos x)$.
Since $y = 7 \sin x + 5 \cos x$,we can substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -y$.
Rearranging gives $\frac{d^2y}{dx^2} + y = 0$.
Comparing this with the given equation $\frac{d^2y}{dx^2} - my = 0$,we have $-m = 1$,which implies $m = -1$.
Therefore,the correct option is $C$.

(D) Given the equation $y = 100 e^{2x} + 200 e^{-2x}$.
First,find the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = 100 \cdot 2 e^{2x} + 200 \cdot (-2) e^{-2x} = 200 e^{2x} - 400 e^{-2x}$.
Next,find the second derivative $\frac{d^2 y}{dx^2}$:
$\frac{d^2 y}{dx^2} = 200 \cdot 2 e^{2x} - 400 \cdot (-2) e^{-2x} = 400 e^{2x} + 800 e^{-2x}$.
Factor out $4$ from the expression:
$\frac{d^2 y}{dx^2} = 4(100 e^{2x} + 200 e^{-2x})$.
Since $y = 100 e^{2x} + 200 e^{-2x}$,we can substitute $y$ into the equation:
$\frac{d^2 y}{dx^2} = 4y$.
Comparing this with $\frac{d^2 y}{dx^2} = ay$,we get $a = 4$.

(B) Given $x+1 = e^{-y}$.
Taking the natural logarithm on both sides,we get $\ln(x+1) = -y$,which implies $y = -\ln(x+1)$.
Differentiating with respect to $x$,we get $\frac{d y}{d x} = -\frac{1}{x+1}$.
Differentiating again with respect to $x$,we get $\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$.
Since $\frac{d y}{d x} = -\frac{1}{x+1}$,we have $\left(\frac{d y}{d x}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$.
Thus,$\frac{d^2 y}{d x^2} = \left(\frac{d y}{d x}\right)^2$.

(C) Given $x = \sin y$,we have $y = \arcsin x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-1/2}$.
Now,differentiating again with respect to $x$ using the chain rule:
$\frac{d^2 y}{dx^2} = -\frac{1}{2}(1 - x^2)^{-3/2} \cdot \frac{d}{dx}(1 - x^2)$.
$\frac{d^2 y}{dx^2} = -\frac{1}{2}(1 - x^2)^{-3/2} \cdot (-2x)$.
$\frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^{3/2}}$.
Thus,the correct option is $C$.

(C) Given $y = 5 \cos x - 3 \sin x$.
First,differentiate with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(5 \cos x - 3 \sin x) = -5 \sin x - 3 \cos x$.
Now,differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2 y}{d x^2} = \frac{d}{d x}(-5 \sin x - 3 \cos x) = -5 \cos x - 3(-\sin x) = -5 \cos x + 3 \sin x$.
Factor out $-1$:
$\frac{d^2 y}{d x^2} = -(5 \cos x - 3 \sin x)$.
Since $y = 5 \cos x - 3 \sin x$,we have:
$\frac{d^2 y}{d x^2} = -y$.
Thus,the correct option is $C$.

(D) Given the equation $e^y(x+1)=1$.
Taking the natural logarithm on both sides: $\ln(e^y) + \ln(x+1) = \ln(1)$.
$y + \ln(x+1) = 0$.
Differentiating with respect to $x$: $\frac{d y}{d x} + \frac{1}{x+1} = 0$.
So,$\frac{d y}{d x} = -\frac{1}{x+1} = -(x+1)^{-1}$.
Differentiating again with respect to $x$: $\frac{d^2 y}{d x^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$.
Since $\frac{d y}{d x} = -\frac{1}{x+1}$,we have $\left(\frac{d y}{d x}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$.
Therefore,$\frac{d^2 y}{d x^2} = \left(\frac{d y}{d x}\right)^2$.

(C) Given $y = \log_e(\log_e x)$.
First,find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{x \log_e x} = (x \log_e x)^{-1}$.
Now,find the second derivative $\frac{d^2 y}{dx^2}$ using the power rule and product rule:
$\frac{d^2 y}{dx^2} = -1(x \log_e x)^{-2} \cdot \frac{d}{dx}(x \log_e x)$.
$\frac{d^2 y}{dx^2} = -\frac{1}{(x \log_e x)^2} \cdot [1 \cdot \log_e x + x \cdot \frac{1}{x}] = -\frac{\log_e x + 1}{x^2 (\log_e x)^2}$.
Since $\log_e x + 1 = \log_e x + \log_e e = \log_e(ex)$,we have:
$\frac{d^2 y}{dx^2} = -\frac{\log_e(ex)}{x^2 (\log_e x)^2}$.

(A) Given the function $y = \tan^{-1} x$.
First,differentiate with respect to $x$:
$y_1 = \frac{dy}{dx} = \frac{1}{1 + x^2}$.
This can be rewritten as $(1 + x^2) y_1 = 1$.
Now,differentiate both sides with respect to $x$ using the product rule:
$\frac{d}{dx} [(1 + x^2) y_1] = \frac{d}{dx} (1)$.
$(1 + x^2) y_2 + y_1 (2x) = 0$.
Therefore,$(1 + x^2) y_2 = -2x y_1$.

(C) Given $y = (\cos^{-1} x)^2$.
Differentiating with respect to $x$,we get $y_1 = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$.
So,$\sqrt{1-x^2} y_1 = -2 \cos^{-1} x$.
Squaring both sides,we get $(1-x^2) y_1^2 = 4 (\cos^{-1} x)^2 = 4y$.
Differentiating again with respect to $x$,we get $(1-x^2) \cdot 2y_1 y_2 + y_1^2 (-2x) = 4y_1$.
Dividing by $2y_1$ (assuming $y_1 \neq 0$),we get $(1-x^2) y_2 - x y_1 = 2$.
Thus,$(1-x^2) y_2 - x y_1 - 2 = 0$.
Comparing this with $(1-x^2) y_2 + p x y_1 + q = 0$,we get $p = -1$ and $q = -2$.
Therefore,$p+q = -1 + (-2) = -3$.

(D) Given $ y = (\tan^{-1} x)^2 $.
First,differentiate with respect to $ x $:
$ y_1 = \frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2} $.
This implies $ (1+x^2) y_1 = 2 \tan^{-1} x $.
Differentiate both sides again with respect to $ x $ using the product rule:
$ (1+x^2) y_2 + y_1(2x) = 2 \cdot \frac{1}{1+x^2} $.
Multiply the entire equation by $ (1+x^2) $:
$ (1+x^2)^2 y_2 + 2x(1+x^2) y_1 = 2 $.
Thus,the value is $ 2 $.

(C) Given,$h(x) = f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x) = c$.
First,find the first derivative of $h(x)$ using the product rule:
$h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)$.
Next,find the second derivative $h^{\prime \prime}(x)$:
$h^{\prime \prime}(x) = \frac{d}{dx}[f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)]$
$h^{\prime \prime}(x) = [f^{\prime \prime}(x) \cdot g(x) + f^{\prime}(x) \cdot g^{\prime}(x)] + [f^{\prime}(x) \cdot g^{\prime}(x) + f(x) \cdot g^{\prime \prime}(x)]$
$h^{\prime \prime}(x) = f^{\prime \prime}(x) \cdot g(x) + f(x) \cdot g^{\prime \prime}(x) + 2f^{\prime}(x) \cdot g^{\prime}(x)$.
Since $f^{\prime}(x) \cdot g^{\prime}(x) = c$,we have:
$h^{\prime \prime}(x) = f^{\prime \prime}(x) \cdot g(x) + f(x) \cdot g^{\prime \prime}(x) + 2c \quad \dots(i)$.
Now,evaluate the expression:
$\frac{f^{\prime \prime}(x)}{f(x)} + \frac{g^{\prime \prime}(x)}{g(x)} + \frac{2c}{f(x) \cdot g(x)} = \frac{f^{\prime \prime}(x) \cdot g(x) + g^{\prime \prime}(x) \cdot f(x) + 2c}{f(x) \cdot g(x)}$.
Substituting from equation $(i)$:
$= \frac{h^{\prime \prime}(x)}{h(x)}$.

(C) Given,$f(x) = b e^{ax} + a e^{bx}$.
First,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(b e^{ax} + a e^{bx}) = b(a e^{ax}) + a(b e^{bx}) = ab e^{ax} + ab e^{bx}$.
Next,differentiate $f^{\prime}(x)$ with respect to $x$ to find the second derivative:
$f^{\prime \prime}(x) = \frac{d}{dx}(ab e^{ax} + ab e^{bx}) = ab(a e^{ax}) + ab(b e^{bx}) = a^2 b e^{ax} + ab^2 e^{bx}$.
Finally,evaluate at $x = 0$:
$f^{\prime \prime}(0) = a^2 b e^{a(0)} + ab^2 e^{b(0)} = a^2 b(1) + ab^2(1) = a^2 b + ab^2$.
Factoring out $ab$,we get:
$f^{\prime \prime}(0) = ab(a + b)$.

(A) Given that,$y = \log(\log x) \quad (1)$
Differentiating Eq. $(1)$ with respect to $x$,we get:
$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \quad (2)$
Now,differentiating Eq. $(2)$ with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}((x \log x)^{-1}) = -1(x \log x)^{-2} \cdot \frac{d}{dx}(x \log x)$
$\frac{d^2y}{dx^2} = -\frac{1}{(x \log x)^2} \cdot [x \cdot \frac{1}{x} + \log x \cdot 1]$
$\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x \log x)^2}$

(B) Given $y = \tan^{-1} \sqrt{x^{2}-1}$.
Let $x = \sec \theta$,then $\sqrt{x^{2}-1} = \tan \theta$.
So,$y = \tan^{-1}(\tan \theta) = \theta = \sec^{-1} x$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sec^{-1} x) = \frac{1}{x \sqrt{x^{2}-1}}$.
Now,find the second derivative $\frac{d^{2}y}{dx^{2}}$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( (x(x^{2}-1)^{1/2})^{-1} \right) = -1 \cdot (x(x^{2}-1)^{1/2})^{-2} \cdot \frac{d}{dx} (x(x^{2}-1)^{1/2})$.
Using the product rule: $\frac{d}{dx} (x(x^{2}-1)^{1/2}) = (x^{2}-1)^{1/2} + x \cdot \frac{1}{2}(x^{2}-1)^{-1/2} \cdot 2x = \sqrt{x^{2}-1} + \frac{x^{2}}{\sqrt{x^{2}-1}} = \frac{x^{2}-1+x^{2}}{\sqrt{x^{2}-1}} = \frac{2x^{2}-1}{\sqrt{x^{2}-1}}$.
Thus,$\frac{d^{2}y}{dx^{2}} = - \frac{1}{x^{2}(x^{2}-1)} \cdot \frac{2x^{2}-1}{\sqrt{x^{2}-1}} = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}}$.
Finally,the ratio is:
$\frac{d^{2}y/dx^{2}}{dy/dx} = \left( - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} \right) \div \left( \frac{1}{x \sqrt{x^{2}-1}} \right) = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} \cdot x \sqrt{x^{2}-1} = - \frac{2x^{2}-1}{x(x^{2}-1)} = \frac{1-2x^{2}}{x(x^{2}-1)}$.

(A) Given,$y=e^{\sqrt{x \sqrt{x \sqrt{x \ldots}}}}, x>1$.
Taking natural logarithm on both sides:
$\log _e y = \sqrt{x \sqrt{x \sqrt{x \ldots}}} \log _e e = \sqrt{x \sqrt{x \sqrt{x \ldots}}}$.
Since the expression under the square root repeats,we can write:
$\log _e y = \sqrt{x \log _e y}$.
Squaring both sides:
$(\log _e y)^2 = x \log _e y$.
Dividing by $\log _e y$ (since $x>1$,$y>1$,so $\log _e y \neq 0$):
$\log _e y = x$.
Now,differentiating with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = 1 \implies \frac{d y}{d x} = y$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{d x^2} = \frac{d y}{d x} = y$.
At $x = \log _e 3$,we have $\log _e y = \log _e 3$,which implies $y = 3$.
Therefore,$\frac{d^2 y}{d x^2} = y = 3$.

(D) Given,$\sqrt{r} = a e^{\theta \cot \alpha}$
Squaring both sides,we get $r = a^{2} e^{2 \theta \cot \alpha}$.
Differentiating with respect to $\theta$:
$\frac{dr}{d\theta} = a^{2} \cdot e^{2 \theta \cot \alpha} \cdot (2 \cot \alpha) = 2 a^{2} \cot \alpha \cdot e^{2 \theta \cot \alpha}$.
Differentiating again with respect to $\theta$:
$\frac{d^{2}r}{d\theta^{2}} = 2 a^{2} \cot \alpha \cdot e^{2 \theta \cot \alpha} \cdot (2 \cot \alpha) = 4 a^{2} \cot^{2} \alpha \cdot e^{2 \theta \cot \alpha}$.
Since $r = a^{2} e^{2 \theta \cot \alpha}$,we can substitute this into the expression:
$\frac{d^{2}r}{d\theta^{2}} = 4 \cot^{2} \alpha \cdot (a^{2} e^{2 \theta \cot \alpha}) = 4 r \cot^{2} \alpha$.
Therefore,$\frac{d^{2}r}{d\theta^{2}} - 4 r \cot^{2} \alpha = 0$.

(C) Given,$y = \cos^{2} \frac{3x}{2} - \sin^{2} \frac{3x}{2}$.
Using the trigonometric identity $\cos(2\theta) = \cos^{2}\theta - \sin^{2}\theta$,we have $\theta = \frac{3x}{2}$,so $2\theta = 3x$.
Thus,$y = \cos(3x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = -\sin(3x) \cdot 3 = -3\sin(3x)$.
Now,differentiate again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = -3 \cdot \cos(3x) \cdot 3 = -9\cos(3x)$.
Since $y = \cos(3x)$,we substitute $y$ back into the expression:
$\frac{d^{2}y}{dx^{2}} = -9y$.

(A) Given,$f(x)=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\ldots+x^n$.
By the Binomial Theorem,we know that $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots + x^n$.
Comparing this with the given expression,we have $f(x) = (1+x)^n$.
Now,we find the first derivative of $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(1+x)^n = n(1+x)^{n-1}$.
Next,we find the second derivative of $f(x)$:
$f''(x) = \frac{d}{dx}[n(1+x)^{n-1}] = n(n-1)(1+x)^{n-2}$.
Finally,substituting $x=1$ into the second derivative:
$f''(1) = n(n-1)(1+1)^{n-2} = n(n-1)2^{n-2}$.

(B) Given,$y = 2 x^{n+1} + 3 x^{-n} \dots (i)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2}$
$\frac{d^2y}{dx^2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}$
$\frac{d^2y}{dx^2} = n(n+1) [2x^{n-1} + 3x^{-n-2}]$
Multiplying both sides by $x^2$:
$x^2 \frac{d^2y}{dx^2} = n(n+1) [2x^{n+1} + 3x^{-n}]$
Substituting $y$ from equation $(i)$:
$x^2 \frac{d^2y}{dx^2} = n(n+1)y$

(A) Given,$m \sin ^{-1} x = \log _{e} y$
$\Rightarrow y = e^{m \sin ^{-1} x}$
Differentiating with respect to $x$,we get:
$y' = e^{m \sin ^{-1} x} \times \frac{m}{\sqrt{1 - x^{2}}}$
$\Rightarrow \sqrt{1 - x^{2}} \cdot y' = m y$
Squaring both sides,we get:
$(1 - x^{2}) (y')^{2} = m^{2} y^{2}$
Differentiating again with respect to $x$:
$(1 - x^{2}) \cdot 2 y' \cdot y'' + (y')^{2} (-2 x) = m^{2} \cdot 2 y y' $
Dividing both sides by $2 y'$,we get:
$(1 - x^{2}) y'' - x y' = m^{2} y$

(D) Given,$f^{\prime \prime}(x)+f(x)=0$ ... $(i)$
And $g(x)=[f(x)]^{2}+[f^{\prime}(x)]^{2}$.
Differentiating $g(x)$ with respect to $x$:
$g^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)f^{\prime \prime}(x)$
Substitute $f^{\prime \prime}(x) = -f(x)$ from equation $(i)$:
$g^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)(-f(x))$
$g^{\prime}(x) = 2f(x)f^{\prime}(x) - 2f(x)f^{\prime}(x) = 0$.
Since the derivative of $g(x)$ is $0$,$g(x)$ is a constant function.
Given $g(3) = 8$,therefore $g(x) = 8$ for all $x$.
Thus,$g(8) = 8$.

(A) Given the equation: $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$
Assume $f(x) = e^{kx}$. Then $f'(x) = ke^{kx}$,$f''(x) = k^2e^{kx}$,$f'''(x) = k^3e^{kx}$,and so on.
Substituting these into the given equation:
$e^{kx} = ke^{kx} + k^2e^{kx} + k^3e^{kx} + \ldots$
Dividing by $e^{kx}$ (since $e^{kx} \neq 0$):
$1 = k + k^2 + k^3 + \ldots$
This is an infinite geometric series with first term $a = k$ and common ratio $r = k$.
The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$ for $|r| < 1$.
So,$1 = \frac{k}{1-k}$.
$1 - k = k \implies 2k = 1 \implies k = \frac{1}{2}$.
Thus,$f(x) = Ce^{x/2}$.
Given $f(0) = 1$,we have $Ce^0 = 1$,which implies $C = 1$.
Therefore,$f(x) = e^{x/2}$.