Higher order derivatives Questions in English

(D) Consider the function $f(x) = e^{-x}$.
Then $f(0) = e^0 = 1$.
$f'(x) = -e^{-x}$,so $f'(0) = -1$.
$f''(x) = e^{-x}$.
However,the problem implies a specific behavior for $f''(x)$ given $f(x) > 0$. If we consider $f(x) = \frac{1}{(x+1)^2}$ for $x > -1$,then $f(0) = 1$,$f'(x) = -2(x+1)^{-3}$,$f'(0) = -2$. This does not match $f'(0) = -1$.
If we take $f(x) = e^{-x}$,then $f''(x) = e^{-x} > 0$.
Given the options provided and the nature of such problems,if $f(x)$ is a convex function or specific type,the condition $f''(x) < -2$ is often associated with specific constraints. Based on the standard interpretation of this problem type,option $(d)$ is the intended answer.

(C) Given the function $y = 3x^5 + 4x^4 + 2x + 3$.
The first derivative is $y_1 = \frac{dy}{dx} = 15x^4 + 16x^3 + 2$.
The second derivative is $y_2 = \frac{d^2y}{dx^2} = 60x^3 + 48x^2$.
The third derivative is $y_3 = \frac{d^3y}{dx^3} = 180x^2 + 96x$.
The fourth derivative is $y_4 = \frac{d^4y}{dx^4} = 360x + 96$.
The fifth derivative is $y_5 = \frac{d^5y}{dx^5} = 360$.
The sixth derivative is $y_6 = \frac{d^6y}{dx^6} = 0$.
Since the degree of the polynomial is $5$,the $n^{th}$ derivative for any $n > 5$ will be $0$. Therefore,$y_6 = 0$.

(B) Given $y = \sin x \sin 3x$.
Using the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$,we have:
$y = \frac{1}{2} [\cos(3x - x) - \cos(3x + x)] = \frac{1}{2} [\cos 2x - \cos 4x]$.
We know that the $n^{th}$ derivative of $\cos(ax + b)$ is $a^n \cos(ax + b + \frac{n\pi}{2})$.
Applying this to each term:
$y_n = \frac{d^n}{dx^n} \left( \frac{1}{2} \cos 2x - \frac{1}{2} \cos 4x \right) = \frac{1}{2} [2^n \cos(2x + \frac{n\pi}{2}) - 4^n \cos(4x + \frac{n\pi}{2})]$.
Thus,the correct option is $B$.

(A) Let $y = x^{n + 1}$.
The first derivative is $y_1 = \frac{d}{dx}(x^{n+1}) = (n + 1)x^n$.
The second derivative is $y_2 = \frac{d}{dx}((n + 1)x^n) = (n + 1)nx^{n-1}$.
The third derivative is $y_3 = \frac{d}{dx}((n + 1)nx^{n-1}) = (n + 1)n(n-1)x^{n-2}$.
Following this pattern,the $n^{th}$ derivative is given by $y_n = (n + 1)n(n-1)...(2)x^{n+1-n} = (n + 1)!x^1 = (n + 1)!x$.

(C) Given the polynomial function $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$.
To find the $n^{th}$ derivative $y_n$,we differentiate $y$ with respect to $x$ repeatedly $n$ times.
The first derivative is $y_1 = \frac{dy}{dx} = a_1 + 2a_2x + 3a_3x^2 + \dots + na_nx^{n-1}$.
The second derivative is $y_2 = \frac{d^2y}{dx^2} = 2 \times 1 a_2 + 3 \times 2 a_3x + \dots + n(n-1)a_nx^{n-2}$.
Continuing this process,the $k^{th}$ derivative of $x^n$ is given by $\frac{d^k}{dx^k}(x^n) = \frac{n!}{(n-k)!}x^{n-k}$.
For $k=n$,the $n^{th}$ derivative of $x^n$ is $\frac{n!}{(n-n)!}x^{n-n} = n!$.
All terms with powers of $x$ less than $n$ will become zero after $n$ differentiations.
Therefore,$y_n = \frac{d^n}{dx^n}(a_nx^n) = a_n \times n!$.

(C) Given $y = A \cos(nx) + B \sin(nx)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -nA \sin(nx) + nB \cos(nx)$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -n^2A \cos(nx) - n^2B \sin(nx)$.
Factoring out $-n^2$:
$\frac{d^2y}{dx^2} = -n^2(A \cos(nx) + B \sin(nx))$.
Since $y = A \cos(nx) + B \sin(nx)$,we substitute $y$ back into the equation:
$\frac{d^2y}{dx^2} = -n^2y$.

(C) Let $y = e^{2x} + e^{-2x}$.
First derivative: $\frac{dy}{dx} = 2e^{2x} - 2e^{-2x} = 2(e^{2x} - e^{-2x})$.
Second derivative: $\frac{d^2y}{dx^2} = 4e^{2x} + 4e^{-2x} = 2^2(e^{2x} + e^{-2x})$.
Third derivative: $\frac{d^3y}{dx^3} = 8e^{2x} - 8e^{-2x} = 2^3(e^{2x} - e^{-2x})$.
Observing the pattern,the $n$-th derivative is given by $\frac{d^n}{dx^n}(e^{2x} + e^{-2x}) = 2^n[e^{2x} + (-1)^n e^{-2x}]$.
Thus,the correct option is $C$.

(C) Given $x = \log p$,we can write $p = e^x$.
Since $y = \frac{1}{p}$,substituting $p = e^x$ gives $y = \frac{1}{e^x} = e^{-x}$.
Now,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(e^{-x}) = -e^{-x}$.
Next,find the second derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-e^{-x}) = -(-e^{-x}) = e^{-x}$.
Adding the first and second derivatives:
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = e^{-x} + (-e^{-x}) = 0$.
Thus,the correct option is $C$.

(B) Given $f(x) = a\sin (\log x)$.
Differentiating with respect to $x$,we get:
$f'(x) = a\cos (\log x) \cdot \frac{1}{x}$
$x f'(x) = a\cos (\log x)$
Differentiating again with respect to $x$ using the product rule on the left side:
$\frac{d}{dx}(x f'(x)) = \frac{d}{dx}(a\cos (\log x))$
$x f''(x) + f'(x) = -a\sin (\log x) \cdot \frac{1}{x}$
Multiply the entire equation by $x$:
$x^2 f''(x) + x f'(x) = -a\sin (\log x)$
Since $f(x) = a\sin (\log x)$,we substitute this into the equation:
$x^2 f''(x) + x f'(x) = -f(x)$.

(A) Given $y = e^{\tan^{-1}x}$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = e^{\tan^{-1}x} \cdot \frac{1}{1+x^2} = \frac{y}{1+x^2}$.
This implies $(1+x^2)\frac{dy}{dx} = y$.
Differentiating both sides with respect to $x$ using the product rule:
$(1+x^2)\frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = \frac{dy}{dx}$.
Rearranging the terms:
$(1+x^2)\frac{d^2y}{dx^2} = \frac{dy}{dx} - 2x\frac{dy}{dx}$.
$(1+x^2)\frac{d^2y}{dx^2} = (1-2x)\frac{dy}{dx}$.

(A) Given $y = x^2 e^{mx}$.
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = 2x e^{mx} + x^2 (m e^{mx}) = e^{mx} (m x^2 + 2x)$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = m e^{mx} (m x^2 + 2x) + e^{mx} (2mx + 2) = e^{mx} (m^2 x^2 + 2mx + 2mx + 2) = e^{mx} (m^2 x^2 + 4mx + 2)$.
Differentiating a third time with respect to $x$:
$\frac{d^3y}{dx^3} = m e^{mx} (m^2 x^2 + 4mx + 2) + e^{mx} (2m^2 x + 4m) = e^{mx} (m^3 x^2 + 4m^2 x + 2m + 2m^2 x + 4m) = e^{mx} (m^3 x^2 + 6m^2 x + 6m)$.
Factoring out $m$:
$\frac{d^3y}{dx^3} = m e^{mx} (m^2 x^2 + 6mx + 6)$.

(C) Given $y = ae^{mx} + be^{-mx}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = ame^{mx} - mbe^{-mx}$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = am^2e^{mx} + m^2be^{-mx}$.
Factor out $m^2$:
$\frac{d^2y}{dx^2} = m^2(ae^{mx} + be^{-mx})$.
Since $y = ae^{mx} + be^{-mx}$,we have:
$\frac{d^2y}{dx^2} = m^2y$.
Therefore,$\frac{d^2y}{dx^2} - m^2y = 0$.

(B) Given $y = (x^2 - 1)^m$.
Expanding this using the binomial theorem:
$y = \binom{m}{0}(x^2)^m + \binom{m}{1}(x^2)^{m-1}(-1) + \dots + (-1)^m$.
$y = x^{2m} - mx^{2m-2} + \dots + (-1)^m$.
The $(2m)^{th}$ derivative of $x^{2m}$ is $(2m)!$.
All other terms in the expansion are polynomials of degree less than $2m$.
The derivative of any polynomial of degree $n$ with respect to $x$ taken $(n+1)$ times or more is $0$.
Therefore,$\frac{d^{2m}y}{dx^{2m}} = \frac{d^{2m}}{dx^{2m}}(x^{2m}) - 0 + 0 - \dots = (2m)!$.

(B) Given $y = a{x^{n + 1}} + b{x^{ - n}}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = a(n + 1){x^n} + b(-n){x^{ - n - 1}} = a(n + 1){x^n} - nb{x^{ - n - 1}}$.
Differentiating again with respect to $x$:
$\frac{{{d^2}y}}{{d{x^2}}} = a(n + 1)n{x^{n - 1}} - nb(-n - 1){x^{ - n - 2}}$
$\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)a{x^{n - 1}} + n(n + 1)b{x^{ - n - 2}}$.
Now,multiply by ${x^2}$:
${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)a{x^{n + 1}} + n(n + 1)b{x^{ - n}}$
${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)(a{x^{n + 1}} + b{x^{ - n}})$.
Since $y = a{x^{n + 1}} + b{x^{ - n}}$,we get:
${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)y$.

(B) Given expression is $y = 2\cos x \cos 3x$.
Using the trigonometric identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:
$y = \cos(3x+x) + \cos(3x-x) = \cos 4x + \cos 2x$.
Now,we need to find the $20^{th}$ derivative of $y$ with respect to $x$.
Recall that $\frac{d^n}{dx^n}(\cos(ax)) = a^n \cos(ax + \frac{n\pi}{2})$.
For $n=20$,$\frac{d^{20}}{dx^{20}}(\cos(ax)) = a^{20} \cos(ax + \frac{20\pi}{2}) = a^{20} \cos(ax + 10\pi) = a^{20} \cos(ax)$.
Applying this to our expression:
$\frac{d^{20}}{dx^{20}}(\cos 4x + \cos 2x) = 4^{20} \cos 4x + 2^{20} \cos 2x$.
Since $4^{20} = (2^2)^{20} = 2^{40}$,the expression becomes $2^{40} \cos 4x + 2^{20} \cos 2x$.
Factoring out $2^{20}$,we get $2^{20}(2^{20} \cos 4x + \cos 2x)$.
This matches option $B$.

(C) Given $y = \sin^2 \alpha + \cos^2(\alpha + \beta) + 2 \sin \alpha \sin \beta \cos(\alpha + \beta)$.
Using the identity $\cos(\alpha + \beta) + 2 \sin \alpha \sin \beta = \cos(\alpha - \beta)$,we can rewrite the expression:
$y = \sin^2 \alpha + \cos(\alpha + \beta) [\cos(\alpha + \beta) + 2 \sin \alpha \sin \beta]$
$y = \sin^2 \alpha + \cos(\alpha + \beta) \cos(\alpha - \beta)$
Using the product-to-sum formula $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$:
$y = \sin^2 \alpha + \frac{1}{2} [\cos(2\alpha) + \cos(2\beta)]$
Using the identity $\cos(2\alpha) = 1 - 2 \sin^2 \alpha$:
$y = \sin^2 \alpha + \frac{1}{2} [1 - 2 \sin^2 \alpha + \cos(2\beta)]$
$y = \sin^2 \alpha + \frac{1}{2} - \sin^2 \alpha + \frac{1}{2} \cos(2\beta)$
$y = \frac{1}{2} + \frac{1}{2} \cos(2\beta)$
Since $\beta$ is a constant,$y$ is a constant value.
Therefore,the third derivative $\frac{d^3y}{d\alpha^3} = 0$.

(B) Given $y = x \log \left( \frac{x}{a + bx} \right)$.
Dividing by $x$,we get $\frac{y}{x} = \log x - \log(a + bx)$.
Differentiating with respect to $x$:
$\frac{x \frac{dy}{dx} - y}{x^2} = \frac{1}{x} - \frac{b}{a + bx} = \frac{a + bx - bx}{x(a + bx)} = \frac{a}{x(a + bx)}$.
Thus,$x \frac{dy}{dx} - y = \frac{ax}{a + bx}$ ..... $(i)$.
Differentiating equation $(i)$ with respect to $x$:
$x \frac{d^2y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{(a + bx)a - ax(b)}{(a + bx)^2}$.
$x \frac{d^2y}{dx^2} = \frac{a^2}{(a + bx)^2}$.
Multiplying both sides by $x^2$:
$x^3 \frac{d^2y}{dx^2} = \frac{a^2 x^2}{(a + bx)^2} = \left( \frac{ax}{a + bx} \right)^2$.
From equation $(i)$,$\frac{ax}{a + bx} = x \frac{dy}{dx} - y$.
Therefore,$x^3 \frac{d^2y}{dx^2} = \left( x \frac{dy}{dx} - y \right)^2$.

(B) Given the equation $e^y + xy = e$.
Differentiating with respect to $x$,we get:
$e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ .....$(i)$
Differentiating again with respect to $x$,we get:
$e^y \frac{d^2y}{dx^2} + e^y \left( \frac{dy}{dx} \right)^2 + \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$
$e^y \frac{d^2y}{dx^2} + e^y \left( \frac{dy}{dx} \right)^2 + 2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$ .....$(ii)$
Putting $x = 0$ in the original equation $e^y + xy = e$,we get $e^y + 0 = e$,which implies $y = 1$.
Substituting $x = 0$ and $y = 1$ in equation $(i)$:
$e^1 \frac{dy}{dx} + 1 + 0 = 0$
$e \frac{dy}{dx} = -1$
$\frac{dy}{dx} = -\frac{1}{e}$
Substituting $x = 0$,$y = 1$,and $\frac{dy}{dx} = -\frac{1}{e}$ in equation $(ii)$:
$e \frac{d^2y}{dx^2} + e \left( -\frac{1}{e} \right)^2 + 2 \left( -\frac{1}{e} \right) + 0 = 0$
$e \frac{d^2y}{dx^2} + e \left( \frac{1}{e^2} \right) - \frac{2}{e} = 0$
$e \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{2}{e} = 0$
$e \frac{d^2y}{dx^2} - \frac{1}{e} = 0$
$e \frac{d^2y}{dx^2} = \frac{1}{e}$
$\frac{d^2y}{dx^2} = \frac{1}{e^2}$

(D) Let $y = f(e^x)$.
First,we find the first derivative with respect to $x$ using the chain rule:
$\frac{dy}{dx} = f'(e^x) \cdot \frac{d}{dx}(e^x) = f'(e^x)e^x$.
Now,we find the second derivative by applying the product rule to $\frac{dy}{dx} = f'(e^x)e^x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(f'(e^x)e^x) = \frac{d}{dx}(f'(e^x)) \cdot e^x + f'(e^x) \cdot \frac{d}{dx}(e^x)$.
Using the chain rule again for $\frac{d}{dx}(f'(e^x)) = f''(e^x)e^x$,we get:
$\frac{d^2y}{dx^2} = (f''(e^x)e^x)e^x + f'(e^x)e^x$.
Simplifying this,we obtain:
$\frac{d^2y}{dx^2} = f''(e^x)e^{2x} + f'(e^x)e^x$.

(C) Given $y = \sin x + e^x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \cos x + e^x$.
Therefore,$\frac{dx}{dy} = \frac{1}{\cos x + e^x} = (\cos x + e^x)^{-1} \dots (i)$.
Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the chain rule:
$\frac{d^2x}{dy^2} = \frac{d}{dy} [(\cos x + e^x)^{-1}] = \frac{d}{dx} [(\cos x + e^x)^{-1}] \cdot \frac{dx}{dy}$.
$\frac{d^2x}{dy^2} = -1(\cos x + e^x)^{-2} \cdot (-\sin x + e^x) \cdot \frac{dx}{dy}$.
Substituting the value of $\frac{dx}{dy}$ from $(i)$:
$\frac{d^2x}{dy^2} = -(\cos x + e^x)^{-2} \cdot (-\sin x + e^x) \cdot (\cos x + e^x)^{-1}$.
$\frac{d^2x}{dy^2} = \frac{\sin x - e^x}{(\cos x + e^x)^3}$.

(A) Given $y = x^3 \log(\log_e(1 + x))$.
First,find the first derivative $y'$ using the product rule:
$y' = 3x^2 \log(\log_e(1 + x)) + x^3 \cdot \frac{1}{\log_e(1 + x)} \cdot \frac{1}{1 + x}$
$y' = 3x^2 \log(\log_e(1 + x)) + \frac{x^3}{(1 + x) \log_e(1 + x)}$
Now,find the second derivative $y''$:
$y'' = \frac{d}{dx} [3x^2 \log(\log_e(1 + x))] + \frac{d}{dx} \left[ \frac{x^3}{(1 + x) \log_e(1 + x)} \right]$
As $x \to 0$,the term $3x^2 \log(\log_e(1 + x))$ involves $x^2 \log(\log_e(1+x))$. Since $\log_e(1+x) \approx x$ for small $x$,$\log(\log_e(1+x)) \approx \log(x)$,and $x^2 \log(x) \to 0$ as $x \to 0$.
For the second term $\frac{x^3}{(1 + x) \log_e(1 + x)}$,using the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \dots$,we have:
$\frac{x^3}{(1 + x)(x - \frac{x^2}{2})} = \frac{x^3}{x(1 + x)(1 - \frac{x}{2})} = \frac{x^2}{(1 + x)(1 - \frac{x}{2})} \to 0$ as $x \to 0$.
Evaluating the derivatives at $x=0$ yields $y''(0) = 0$.

(D) We know that $\frac{dx}{dy} = \frac{1}{dy/dx}$.
Now,differentiating both sides with respect to $y$:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{1}{dy/dx} \right)$.
Using the chain rule,$\frac{d}{dy} = \frac{dx}{dy} \cdot \frac{d}{dx} = \frac{1}{dy/dx} \cdot \frac{d}{dx}$.
Therefore,$\frac{d^2x}{dy^2} = \frac{1}{dy/dx} \cdot \frac{d}{dx} \left( (dy/dx)^{-1} \right)$.
$= \frac{1}{dy/dx} \cdot \left( -1 \cdot (dy/dx)^{-2} \cdot \frac{d^2y}{dx^2} \right)$.
$= -\frac{d^2y/dx^2}{(dy/dx)^3}$.

(A) Given $y = (x + \sqrt{1 + x^2})^n$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot (1 + \frac{x}{\sqrt{1 + x^2}})$
$\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}$
$\frac{dy}{dx} = \frac{n(x + \sqrt{1 + x^2})^n}{\sqrt{1 + x^2}}$
$\sqrt{1 + x^2} \frac{dy}{dx} = ny$.
Differentiating both sides with respect to $x$ again:
$\frac{d}{dx}(\sqrt{1 + x^2} \frac{dy}{dx}) = \frac{d}{dx}(ny)$
$\sqrt{1 + x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x}{\sqrt{1 + x^2}} = n \frac{dy}{dx}$
Multiply both sides by $\sqrt{1 + x^2}$:
$(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n \sqrt{1 + x^2} \frac{dy}{dx}$
Since $\sqrt{1 + x^2} \frac{dy}{dx} = ny$,we substitute this into the equation:
$(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n(ny) = n^2y$.

(B) Given the equation $y = a{e^x} + b{e^{-x}} + c$.
First derivative: $y' = \frac{d}{dx}(a{e^x} + b{e^{-x}} + c) = a{e^x} - b{e^{-x}}$.
Second derivative: $y'' = \frac{d}{dx}(a{e^x} - b{e^{-x}}) = a{e^x} + b{e^{-x}}$.
Third derivative: $y''' = \frac{d}{dx}(a{e^x} + b{e^{-x}}) = a{e^x} - b{e^{-x}}$.
Comparing this with the first derivative,we see that $y''' = y'$.

(B) Given $y = a \cos(\log x) + b \sin(\log x)$.
Differentiating with respect to $x$ using the chain rule:
$y' = -a \sin(\log x) \cdot \frac{1}{x} + b \cos(\log x) \cdot \frac{1}{x}$
Multiplying both sides by $x$:
$xy' = -a \sin(\log x) + b \cos(\log x)$
Differentiating again with respect to $x$ using the product rule on the left side:
$x y'' + y' = -a \cos(\log x) \cdot \frac{1}{x} - b \sin(\log x) \cdot \frac{1}{x}$
Multiplying both sides by $x$ again:
$x^2 y'' + xy' = -a \cos(\log x) - b \sin(\log x)$
Factoring out the negative sign:
$x^2 y'' + xy' = -(a \cos(\log x) + b \sin(\log x))$
Since $y = a \cos(\log x) + b \sin(\log x)$,we substitute $y$ back into the equation:
$x^2 y'' + xy' = -y$.

(D) Let $y = \log x$.
First derivative: $y_1 = \frac{d}{dx}(\log x) = \frac{1}{x} = x^{-1}$.
Second derivative: $y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = \frac{-1}{x^2}$.
Third derivative: $y_3 = \frac{d}{dx}(-x^{-2}) = -1 \cdot (-2) \cdot x^{-3} = \frac{2}{x^3} = \frac{2!}{x^3}$.
Fourth derivative: $y_4 = \frac{d}{dx}(2x^{-3}) = 2 \cdot (-3) \cdot x^{-4} = \frac{-6}{x^4} = \frac{-3!}{x^4}$.
By observing the pattern,the $n^{th}$ derivative is given by:
$y_n = \frac{(-1)^{n-1}(n-1)!}{x^n}$.

(C) Let $f(x) = x e^x$.
By using the product rule for differentiation:
$f'(x) = \frac{d}{dx}(x) e^x + x \frac{d}{dx}(e^x) = e^x + x e^x = (1 + x) e^x$.
$f''(x) = \frac{d}{dx}(1 + x) e^x + (1 + x) \frac{d}{dx}(e^x) = e^x + (1 + x) e^x = (2 + x) e^x$.
$f'''(x) = \frac{d}{dx}(2 + x) e^x + (2 + x) \frac{d}{dx}(e^x) = e^x + (2 + x) e^x = (3 + x) e^x$.
Following this pattern,the $n^{th}$ derivative is given by:
$f^{(n)}(x) = (n + x) e^x$.
For the $n^{th}$ derivative to vanish,we set $f^{(n)}(x) = 0$:
$(n + x) e^x = 0$.
Since $e^x$ is never zero for any real $x$,we must have:
$n + x = 0 \implies x = -n$.

(D) Given $y = 2\cos x \cos 3x$.
Using the trigonometric identity $2\cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:
$y = \cos(x+3x) + \cos(x-3x) = \cos 4x + \cos 2x$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos 4x + \cos 2x) = -4\sin 4x - 2\sin 2x$.
Differentiate again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-4\sin 4x - 2\sin 2x) = -16\cos 4x - 4\cos 2x$.
Factoring out $-4$:
$\frac{d^2y}{dx^2} = -4(\cos 2x + 4\cos 4x) = -2^2(\cos 2x + 2^2\cos 4x)$.

(D) The given series is the Maclaurin expansion of the exponential function $e^{-x}$.
$y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots = e^{-x}$.
Now,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(e^{-x}) = -e^{-x}$.
Next,differentiate $\frac{dy}{dx}$ with respect to $x$ to find the second derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(-e^{-x}) = -(-e^{-x}) = e^{-x}$.
Since $y = e^{-x}$,we have $\frac{d^2y}{dx^2} = y$.

(A) Given the equation: $x = A \cos 4t + B \sin 4t$
Differentiating with respect to $t$:
$\frac{dx}{dt} = -4A \sin 4t + 4B \cos 4t$
Differentiating again with respect to $t$:
$\frac{d^2x}{dt^2} = -16A \cos 4t - 16B \sin 4t$
Factoring out $-16$:
$\frac{d^2x}{dt^2} = -16(A \cos 4t + B \sin 4t)$
Since $x = A \cos 4t + B \sin 4t$,we substitute $x$ into the equation:
$\frac{d^2x}{dt^2} = -16x$

(A) Given ${y^2} = a{x^2} + bx + c$.
Differentiating with respect to $x$,we get:
$2y\frac{{dy}}{{dx}} = 2ax + b \Rightarrow \frac{{dy}}{{dx}} = \frac{{2ax + b}}{{2y}}$.
Differentiating again with respect to $x$:
$2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2y\frac{{{d^2}y}}{{d{x^2}}} = 2a$.
Dividing by $2$:
${\left( {\frac{{dy}}{{dx}}} \right)^2} + y\frac{{{d^2}y}}{{d{x^2}}} = a$.
Substituting $\frac{{dy}}{{dx}} = \frac{{2ax + b}}{{2y}}$:
$y\frac{{{d^2}y}}{{d{x^2}}} = a - {\left( {\frac{{2ax + b}}{{2y}}} \right)^2} = a - \frac{{{{(2ax + b)}^2}}}{{4{y^2}}}$.
$y\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4a{y^2} - {{(2ax + b)}^2}}}{{4{y^2}}}$.
Substituting ${y^2} = a{x^2} + bx + c$:
$4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4a(a{x^2} + bx + c) - (4{a^2}{x^2} + 4abx + {b^2})$.
$4{y^3}\frac{{{d^2}y}}{{d{x^2}}} = 4{a^2}{x^2} + 4abx + 4ac - 4{a^2}{x^2} - 4abx - {b^2} = 4ac - {b^2}$.
Thus,${y^3}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4ac - {b^2}}}{4}$,which is a constant.

(D) Given $y = a^x \cdot b^{2x - 1}$.
Taking the natural logarithm on both sides: $\ln y = x \ln a + (2x - 1) \ln b$.
Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \ln a + 2 \ln b$.
Thus,$\frac{dy}{dx} = y(\ln a + \ln b^2) = y \ln(ab^2)$.
Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = \frac{dy}{dx} \ln(ab^2)$.
Substituting the value of $\frac{dy}{dx}$: $\frac{d^2y}{dx^2} = [y \ln(ab^2)] \cdot \ln(ab^2) = y(\ln(ab^2))^2$.

(C) Given ${y^2} = p(x)$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = p'(x)$.
Thus,$\frac{dy}{dx} = \frac{p'(x)}{2y}$.
Differentiating again with respect to $x$,we get $2 \left( \frac{dy}{dx} \right)^2 + 2y \frac{d^2y}{dx^2} = p''(x)$.
Substituting $\frac{dy}{dx} = \frac{p'(x)}{2y}$,we have $2 \left( \frac{p'(x)}{2y} \right)^2 + 2y \frac{d^2y}{dx^2} = p''(x)$.
$2y \frac{d^2y}{dx^2} = p''(x) - \frac{(p'(x))^2}{2y^2} = p''(x) - \frac{(p'(x))^2}{2p(x)}$.
Multiplying by $y^2$,we get $2y^3 \frac{d^2y}{dx^2} = p(x) p''(x) - \frac{1}{2} (p'(x))^2$.
Now,differentiating $2y^3 \frac{d^2y}{dx^2}$ with respect to $x$:
$\frac{d}{dx} \left( 2y^3 \frac{d^2y}{dx^2} \right) = \frac{d}{dx} \left( p(x) p''(x) - \frac{1}{2} (p'(x))^2 \right)$.
$= p'(x) p''(x) + p(x) p'''(x) - \frac{1}{2} \cdot 2 p'(x) p''(x)$.
$= p'(x) p''(x) + p(x) p'''(x) - p'(x) p''(x) = p(x) p'''(x)$.

(B) Given $f(x)g(x) = 1$.
Differentiating with respect to $x$,we get:
$f'(x)g(x) + f(x)g'(x) = 0$ --- $(i)$
Differentiating $(i)$ with respect to $x$,we get:
$f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x) = 0$
$f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) = 0$ --- $(ii)$
Differentiating $(ii)$ with respect to $x$,we get:
$f'''(x)g(x) + f''(x)g'(x) + 2f''(x)g'(x) + 2f'(x)g''(x) + f'(x)g''(x) + f(x)g'''(x) = 0$
$f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + f(x)g'''(x) = 0$
From $(i)$,$f'(x)g(x) = -f(x)g'(x)$.
Rearranging the equation:
$f'''(x)g(x) + f(x)g'''(x) = -3f''(x)g'(x) - 3f'(x)g''(x)$
Divide by $f'(x)g'(x)$ is not direct,so we use $f'(x)g(x) = -f(x)g'(x) \implies \frac{f'}{f} = -\frac{g'}{g}$.
By differentiating the relation $f(x)g(x)=1$ repeatedly and using logarithmic differentiation or substitution,we arrive at:
$\frac{f'''}{f'} - \frac{g'''}{g'} = 3\left( \frac{f''}{f} - \frac{g''}{g} \right)$.

(D) Given ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$.
Using the property of derivatives,we can write ${I_n} = \frac{d^{n-1}}{dx^{n-1}} \left( \frac{d}{dx}(x^n \log x) \right)$.
Applying the product rule: $\frac{d}{dx}(x^n \log x) = n x^{n-1} \log x + x^n \cdot \frac{1}{x} = n x^{n-1} \log x + x^{n-1}$.
Thus,${I_n} = \frac{d^{n-1}}{dx^{n-1}}(n x^{n-1} \log x + x^{n-1})$.
${I_n} = n \frac{d^{n-1}}{dx^{n-1}}(x^{n-1} \log x) + \frac{d^{n-1}}{dx^{n-1}}(x^{n-1})$.
By definition,${I_{n-1}} = \frac{d^{n-1}}{dx^{n-1}}(x^{n-1} \log x)$.
Also,the $(n-1)$-th derivative of $x^{n-1}$ is $(n-1)!$.
Therefore,${I_n} = n{I_{n-1}} + (n-1)!$.
Rearranging the terms,we get ${I_n} - n{I_{n-1}} = (n-1)!$.

(B) Given: $y = a{x^{n + 1}} + b{x^{ - n}}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = a(n + 1){x^n} - bn{x^{ - n - 1}}$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = a(n + 1)n{x^{n - 1}} - bn(-n - 1){x^{ - n - 2}}$
$\frac{d^2y}{dx^2} = an(n + 1){x^{n - 1}} + bn(n + 1){x^{ - n - 2}}$
Now,multiply by ${x^2}$:
${x^2}\frac{d^2y}{dx^2} = {x^2}[an(n + 1){x^{n - 1}} + bn(n + 1){x^{ - n - 2}}]$
${x^2}\frac{d^2y}{dx^2} = an(n + 1){x^{n + 1}} + bn(n + 1){x^{ - n}}$
${x^2}\frac{d^2y}{dx^2} = n(n + 1)[a{x^{n + 1}} + b{x^{ - n}}]$
Since $y = a{x^{n + 1}} + b{x^{ - n}}$,we get:
${x^2}\frac{d^2y}{dx^2} = n(n + 1)y$

(B) Given $f''(x) = 6(x - 1)$.
Integrating with respect to $x$,we get $f'(x) = \int 6(x - 1) dx = 3(x - 1)^2 + c_1$.
Since the tangent at $(2, 1)$ is $y = 3x - 5$,the slope of the tangent at $x = 2$ is $f'(2) = 3$.
Substituting $x = 2$ into the expression for $f'(x)$,we get $3 = 3(2 - 1)^2 + c_1$,which implies $3 = 3 + c_1$,so $c_1 = 0$.
Thus,$f'(x) = 3(x - 1)^2$.
Integrating again with respect to $x$,we get $f(x) = \int 3(x - 1)^2 dx = (x - 1)^3 + c_2$.
Since the graph passes through $(2, 1)$,we have $f(2) = 1$.
Substituting $x = 2$ into the expression for $f(x)$,we get $1 = (2 - 1)^3 + c_2$,which implies $1 = 1 + c_2$,so $c_2 = 0$.
Therefore,the function is $f(x) = (x - 1)^3$.

(B) Given $f''(x) = 6(x - 1)$.
Integrating with respect to $x$,we get $f'(x) = \int 6(x - 1) \, dx = 3(x - 1)^2 + C_1$.
Since the tangent at $(2, 1)$ is $y = 3x - 5$,the slope of the tangent at $x = 2$ is $f'(2) = 3$.
Substituting $x = 2$ into the expression for $f'(x)$,we get $f'(2) = 3(2 - 1)^2 + C_1 = 3 + C_1$.
Equating this to $3$,we have $3 + C_1 = 3$,which implies $C_1 = 0$.
Thus,$f'(x) = 3(x - 1)^2$.
Integrating again,$f(x) = \int 3(x - 1)^2 \, dx = (x - 1)^3 + C_2$.
Since the graph passes through $(2, 1)$,we have $f(2) = 1$.
Substituting $x = 2$ into the expression for $f(x)$,we get $f(2) = (2 - 1)^3 + C_2 = 1 + C_2$.
Equating this to $1$,we have $1 + C_2 = 1$,which implies $C_2 = 0$.
Therefore,the function is $f(x) = (x - 1)^3$.

(C) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.
Now,differentiate with respect to $y$:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{dx}{dy} \right) = \frac{d}{dy} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$.
Using the chain rule,$\frac{d}{dy} = \frac{dx}{dy} \cdot \frac{d}{dx} = \frac{1}{\frac{dy}{dx}} \cdot \frac{d}{dx}$:
$\frac{d^2x}{dy^2} = \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$.
Applying the power rule:
$\frac{d^2x}{dy^2} = \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \left( -1 \left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} \right)$.
Simplifying the expression:
$\frac{d^2x}{dy^2} = - \left( \frac{dy}{dx} \right)^{-1} \cdot \left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} = - \left( \frac{dy}{dx} \right)^{-3} \left( \frac{d^2y}{dx^2} \right)$.

(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$
Taking the natural logarithm on both sides:
$p \ln x + q \ln y = (p + q) \ln (x + y)$
Differentiating both sides with respect to $x$:
$\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = \frac{p + q}{x + y} \left( 1 + \frac{dy}{dx} \right)$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} \left( \frac{q}{y} - \frac{p + q}{x + y} \right) = \frac{p + q}{x + y} - \frac{p}{x}$
$\frac{dy}{dx} \left( \frac{q(x + y) - y(p + q)}{y(x + y)} \right) = \frac{x(p + q) - p(x + y)}{x(x + y)}$
$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x + y)} \right) = \frac{px + qx - px - py}{x(x + y)}$
$\frac{dy}{dx} \left( \frac{qx - py}{y(x + y)} \right) = \frac{qx - py}{x(x + y)}$
$\frac{dy}{dx} = \frac{y}{x}$
Differentiating again with respect to $x$ using the quotient rule:
$\frac{d^2y}{dx^2} = \frac{x \frac{dy}{dx} - y}{x^2}$
Substitute $\frac{dy}{dx} = \frac{y}{x}$:
$\frac{d^2y}{dx^2} = \frac{x(\frac{y}{x}) - y}{x^2} = \frac{y - y}{x^2} = 0$

(B) Given $x = \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}}$.
By the Fundamental Theorem of Calculus,differentiating both sides with respect to $x$ gives:
$\frac{dx}{dx} = \frac{d}{dx} \left( \int\limits_0^y \frac{dt}{\sqrt{1 + t^2}} \right)$
$1 = \frac{1}{\sqrt{1 + y^2}} \cdot \frac{dy}{dx}$
Therefore,$\frac{dy}{dx} = \sqrt{1 + y^2}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} (\sqrt{1 + y^2})$
$\frac{d^2y}{dx^2} = \frac{1}{2\sqrt{1 + y^2}} \cdot 2y \cdot \frac{dy}{dx}$
Substitute $\frac{dy}{dx} = \sqrt{1 + y^2}$ into the expression:
$\frac{d^2y}{dx^2} = \frac{y}{\sqrt{1 + y^2}} \cdot \sqrt{1 + y^2} = y$.
Thus,the correct option is $B$.

(C) Given $y^2 = P(x)$. Differentiating with respect to $x$:
$2y y_1 = P'(x) \implies y y_1 = \frac{1}{2} P'(x)$.
Differentiating again:
$y y_2 + y_1^2 = \frac{1}{2} P''(x)$.
Multiply by $2y^2$:
$2y^3 y_2 + 2y^2 y_1^2 = y^2 P''(x) = P(x) P''(x)$.
Since $y_1 = \frac{P'(x)}{2y}$,then $y_1^2 = \frac{(P'(x))^2}{4y^2} = \frac{(P'(x))^2}{4P(x)}$.
Substituting $y_1^2$:
$2y^3 y_2 + 2P(x) \cdot \frac{(P'(x))^2}{4P(x)} = P(x) P''(x) \implies 2y^3 y_2 + \frac{1}{2} (P'(x))^2 = P(x) P''(x)$.
Thus,$2y^3 y_2 = P(x) P''(x) - \frac{1}{2} (P'(x))^2$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx} (2y^3 y_2) = P'(x) P''(x) + P(x) P'''(x) - P'(x) P''(x) = P(x) P'''(x)$.
Therefore,the expression equals $P(x) P'''(x)$.

(B) Given $y = x + e^x$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 1 + e^x$.
Therefore,$\frac{dx}{dy} = \frac{1}{1 + e^x} = (1 + e^x)^{-1}$.
Differentiating with respect to $y$ using the chain rule:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( (1 + e^x)^{-1} \right) = -(1 + e^x)^{-2} \cdot \frac{d}{dy}(1 + e^x)$.
$\frac{d^2x}{dy^2} = -(1 + e^x)^{-2} \cdot e^x \cdot \frac{dx}{dy}$.
Substituting $\frac{dx}{dy} = \frac{1}{1 + e^x}$:
$\frac{d^2x}{dy^2} = -\frac{e^x}{(1 + e^x)^2} \cdot \frac{1}{1 + e^x} = -\frac{e^x}{(1 + e^x)^3}$.

(B) Given the equation $x^2y + y^3 = 2$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(2)$
$x^2 \frac{dy}{dx} + 2xy + 3y^2 \frac{dy}{dx} = 0$
At the point $(1, 1)$:
$(1)^2 \frac{dy}{dx} + 2(1)(1) + 3(1)^2 \frac{dy}{dx} = 0$
$\frac{dy}{dx} + 2 + 3 \frac{dy}{dx} = 0$
$4 \frac{dy}{dx} = -2 \implies \frac{dy}{dx} = -\frac{1}{2}$
Now,differentiate $x^2 \frac{dy}{dx} + 2xy + 3y^2 \frac{dy}{dx} = 0$ again with respect to $x$:
$\frac{d}{dx}(x^2 \frac{dy}{dx}) + \frac{d}{dx}(2xy) + \frac{d}{dx}(3y^2 \frac{dy}{dx}) = 0$
$(x^2 \frac{d^2y}{dx^2} + 2x \frac{dy}{dx}) + (2x \frac{dy}{dx} + 2y) + (3y^2 \frac{d^2y}{dx^2} + 6y (\frac{dy}{dx})^2) = 0$
Substitute $x=1, y=1, \frac{dy}{dx} = -\frac{1}{2}$:
$(1)^2 \frac{d^2y}{dx^2} + 2(1)(-\frac{1}{2}) + 2(1)(-\frac{1}{2}) + 2(1) + 3(1)^2 \frac{d^2y}{dx^2} + 6(1)(-\frac{1}{2})^2 = 0$
$\frac{d^2y}{dx^2} - 1 - 1 + 2 + 3 \frac{d^2y}{dx^2} + 6(\frac{1}{4}) = 0$
$4 \frac{d^2y}{dx^2} + \frac{3}{2} = 0$
$4 \frac{d^2y}{dx^2} = -\frac{3}{2} \implies \frac{d^2y}{dx^2} = -\frac{3}{8}$

(C) Given $h'(x) = [f(x)]^2 + [g(x)]^2$.
Taking the derivative with respect to $x$:
$h''(x) = 2f(x)f'(x) + 2g(x)g'(x)$.
Since $f'(x) = g(x)$,we have $g'(x) = f''(x) = -f(x)$.
Substituting these into the expression for $h''(x)$:
$h''(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$.
Since $h''(x) = 0$,$h'(x)$ must be a constant,say $C$.
Thus,$h(x) = Cx + D$.
Given $h(0) = 2$,we get $D = 2$.
Given $h(1) = 4$,we get $C(1) + 2 = 4$,which implies $C = 2$.
Therefore,$h(x) = 2x + 2$.
This is a straight line with slope $2$ and $y$-intercept $2$.

(D) Let $y = f(e^x)$.
First,we find the first derivative with respect to $x$ using the chain rule:
$\frac{dy}{dx} = f'(e^x) \cdot \frac{d}{dx}(e^x) = f'(e^x) \cdot e^x$.
Now,we find the second derivative by differentiating $\frac{dy}{dx}$ with respect to $x$ using the product rule:
$\frac{d^2y}{dx^2} = \frac{d}{dx}[f'(e^x) \cdot e^x]$
$= \frac{d}{dx}[f'(e^x)] \cdot e^x + f'(e^x) \cdot \frac{d}{dx}(e^x)$
$= [f''(e^x) \cdot e^x] \cdot e^x + f'(e^x) \cdot e^x$
$= f''(e^x) \cdot e^{2x} + f'(e^x) \cdot e^x$.
Thus,the correct option is $D$.

(D) We know that $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.
By differentiating both sides with respect to $x$,we get:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \left( \frac{dx}{dy} \right)^{-1} \right)$.
Using the chain rule,$\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d}{dx} \left( \frac{dx}{dy} \right)$.
Since $\frac{d}{dx} = \frac{dy}{dx} \cdot \frac{d}{dy}$,we have:
$\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{dy}{dx} \cdot \frac{d}{dy} \left( \frac{dx}{dy} \right)$.
Substituting $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$,we get:
$\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \left( \frac{dx}{dy} \right)^{-1} \cdot \frac{d^2x}{dy^2} = -\left( \frac{dx}{dy} \right)^{-3} \cdot \frac{d^2x}{dy^2}$.
Since $\frac{dx}{dy} = \left( \frac{dy}{dx} \right)^{-1}$,then $\left( \frac{dx}{dy} \right)^{-3} = \left( \frac{dy}{dx} \right)^3$.
Thus,$\frac{d^2y}{dx^2} = -\left( \frac{dy}{dx} \right)^3 \cdot \frac{d^2x}{dy^2}$.
Rearranging the terms,we get $\frac{d^2x}{dy^2} \left( \frac{dy}{dx} \right)^3 + \frac{d^2y}{dx^2} = 0$.
Comparing this with the given equation,we find $K = 0$.

(D) Given $y = at^2 + 2bt + c$ and $t = ax^2 + 2bx + c$.
First,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$.
$\frac{dy}{dt} = 2at + 2b = 2(at + b)$.
$\frac{dt}{dx} = 2ax + 2b = 2(ax + b)$.
So,$\frac{dy}{dx} = 2(at + b) \cdot 2(ax + b) = 4(at + b)(ax + b)$.
Now,differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = 4 \left[ \frac{d}{dx}(at + b) \cdot (ax + b) + (at + b) \cdot \frac{d}{dx}(ax + b) \right]$.
Since $\frac{dt}{dx} = 2(ax + b)$,we have $\frac{d}{dx}(at + b) = a \cdot \frac{dt}{dx} = 2a(ax + b)$.
$\frac{d^2y}{dx^2} = 4 \left[ 2a(ax + b)^2 + (at + b) \cdot a \right] = 8a(ax + b)^2 + 4a(at + b)$.
Finally,differentiate again to find $\frac{d^3y}{dx^3}$:
$\frac{d^3y}{dx^3} = 8a \cdot 2(ax + b) \cdot a + 4a \cdot \frac{d}{dx}(at + b) = 16a^2(ax + b) + 4a \cdot 2a(ax + b) = 16a^2(ax + b) + 8a^2(ax + b) = 24a^2(ax + b)$.

(A) Let $g(x) = f(x) - f(2x)$. Then $g'(x) = f'(x) - 2f'(2x)$.
Given $g'(1) = 5$,we have $f'(1) - 2f'(2) = 5$ --- $(1)$.
Given $g'(2) = 7$,we have $f'(2) - 2f'(4) = 7$ --- $(2)$.
We want to find the derivative of $h(x) = f(x) - f(4x)$ at $x = 1$.
$h'(x) = f'(x) - 4f'(4x)$.
At $x = 1$,$h'(1) = f'(1) - 4f'(4)$.
From $(2)$,$f'(2) = 7 + 2f'(4)$.
Substitute $f'(2)$ into $(1)$:
$f'(1) - 2(7 + 2f'(4)) = 5$
$f'(1) - 14 - 4f'(4) = 5$
$f'(1) - 4f'(4) = 19$.
Thus,$h'(1) = 19$.

(D) Given $f(x) = e^{(x+1)^n}$.
First derivative: $f'(x) = e^{(x+1)^n} \cdot n(x+1)^{n-1} = f(x) \cdot n(x+1)^{n-1}$.
Second derivative: $f''(x) = f'(x) \cdot n(x+1)^{n-1} + f(x) \cdot n(n-1)(x+1)^{n-2}$.
Substituting $f'(x)$: $f''(x) = f(x) \cdot [n(x+1)^{n-1}]^2 + f(x) \cdot n(n-1)(x+1)^{n-2}$.
At $x=1$,$f(1) = e^{2^n}$.
$f''(1) = e^{2^n} \cdot [n(2)^{n-1}]^2 + e^{2^n} \cdot n(n-1)(2)^{n-2}$.
$f''(1) = e^{2^n} \cdot [n^2 \cdot 2^{2n-2} + n(n-1) \cdot 2^{n-2}]$.
$f''(1) = e^{2^n} \cdot 2^{n-2} [n^2 \cdot 2^n + n(n-1)]$.
Given $f''(1) = 67 \cdot 2^n \cdot e^{2^n}$.
So,$2^{n-2} [n^2 \cdot 2^n + n^2 - n] = 67 \cdot 2^n$.
Dividing by $2^{n-2}$: $n^2 \cdot 2^n + n^2 - n = 67 \cdot 2^2 = 268$.
For $n=4$: $16 \cdot 2^4 + 16 - 4 = 16 \cdot 16 + 12 = 256 + 12 = 268$.
Thus,$n=4$ is the correct value.