Derivative at a point, Standard differentiation Questions in English

(A) Given $y = \sinh^{-1}\left(\frac{1-x}{1+x}\right)$.
Using the chain rule,$\frac{dy}{dx} = \frac{d}{du}(\sinh^{-1}(u)) \cdot \frac{du}{dx}$,where $u = \frac{1-x}{1+x}$.
The derivative of $\sinh^{-1}(u)$ is $\frac{1}{\sqrt{u^2+1}}$.
Now,$\frac{du}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.
Substituting these into the chain rule:
$\frac{dy}{dx} = \frac{1}{\sqrt{(\frac{1-x}{1+x})^2 + 1}} \cdot \frac{-2}{(1+x)^2}$.
$\frac{dy}{dx} = \frac{1}{\sqrt{\frac{(1-x)^2 + (1+x)^2}{(1+x)^2}}} \cdot \frac{-2}{(1+x)^2}$.
$\frac{dy}{dx} = \frac{|1+x|}{\sqrt{1-2x+x^2+1+2x+x^2}} \cdot \frac{-2}{(1+x)^2}$.
$\frac{dy}{dx} = \frac{|1+x|}{\sqrt{2+2x^2}} \cdot \frac{-2}{(1+x)^2} = \frac{|1+x|}{\sqrt{2}\sqrt{1+x^2}} \cdot \frac{-2}{(1+x)^2}$.
Since $\frac{|1+x|}{(1+x)^2} = \frac{1}{|1+x|}$,we get:
$\frac{dy}{dx} = \frac{-\sqrt{2}}{|1+x|\sqrt{1+x^2}}$.

(B) Given $y = \tan^{-1} \left\{ \frac{ax - b}{bx + a} \right\}$.
We can rewrite the expression inside the inverse tangent function by dividing the numerator and denominator by $a$:
$y = \tan^{-1} \left\{ \frac{x - \frac{b}{a}}{1 + \frac{b}{a}x} \right\}$.
Using the identity $\tan^{-1} \left( \frac{A - B}{1 + AB} \right) = \tan^{-1} A - \tan^{-1} B$,we get:
$y = \tan^{-1} x - \tan^{-1} \left( \frac{b}{a} \right)$.
Now,differentiating with respect to $x$:
$y' = \frac{d}{dx} (\tan^{-1} x) - \frac{d}{dx} (\tan^{-1} \frac{b}{a})$.
Since $\tan^{-1} \left( \frac{b}{a} \right)$ is a constant,its derivative is $0$.
Therefore,$y' = \frac{1}{1 + x^2} - 0 = \frac{1}{1 + x^2}$.

(B) Given $y = (\log_{\cot x} \tan x)(\log_{\tan x} \cot x) + \tan^{-1}\left(\frac{4x}{4-x^2}\right)$.
Since $\log_{\cot x} \tan x = \frac{1}{\log_{\tan x} \cot x}$,the product is $1$.
So,$y = 1 + \tan^{-1}\left(\frac{4x}{4-x^2}\right)$.
Let $x = 2 \tan \theta$,then $\frac{4x}{4-x^2} = \frac{8 \tan \theta}{4 - 4 \tan^2 \theta} = 2 \left(\frac{2 \tan \theta}{1 - \tan^2 \theta}\right) = 2 \tan(2\theta)$.
Alternatively,differentiate directly: $\frac{dy}{dx} = \frac{d}{dx} \tan^{-1}\left(\frac{4x}{4-x^2}\right)$.
Using the chain rule: $\frac{dy}{dx} = \frac{1}{1 + \left(\frac{4x}{4-x^2}\right)^2} \cdot \frac{d}{dx} \left(\frac{4x}{4-x^2}\right)$.
$= \frac{(4-x^2)^2}{(4-x^2)^2 + 16x^2} \cdot \frac{4(4-x^2) - 4x(-2x)}{(4-x^2)^2}$.
$= \frac{16 - 8x^2 + x^4 + 16x^2}{1} \text{ (denominator)} = 16 + 8x^2 + x^4 = (4+x^2)^2$.
$= \frac{16 - 4x^2 + 8x^2}{(4+x^2)^2} = \frac{16 + 4x^2}{(4+x^2)^2} = \frac{4(4+x^2)}{(4+x^2)^2} = \frac{4}{4+x^2}$.

(C) To find the derivative of the product $(1+x^2) \tan^{-1}(x)$,we use the product rule: $\frac{d}{dx}(u \cdot v) = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}$.
Let $u = (1+x^2)$ and $v = \tan^{-1}(x)$.
Then,$\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = \frac{1}{1+x^2}$.
Applying the product rule:
$\frac{d}{dx} \left\{ (1+x^2) \tan^{-1}(x) \right\} = (1+x^2) \cdot \frac{d}{dx}(\tan^{-1}(x)) + \tan^{-1}(x) \cdot \frac{d}{dx}(1+x^2)$
$= (1+x^2) \cdot \frac{1}{1+x^2} + \tan^{-1}(x) \cdot (2x)$
$= 1 + 2x \tan^{-1}(x)$.
Thus,the correct option is $C$.

(B) Given,$y=\frac{1}{4} \log \left(\frac{1+x}{1-x}\right)-\frac{1}{2} \tan ^{-1} x$.
Using the property $\log \left(\frac{1+x}{1-x}\right) = 2 \tanh ^{-1} x$,we get:
$y = \frac{1}{4} (2 \tanh ^{-1} x) - \frac{1}{2} \tan ^{-1} x = \frac{1}{2} \tanh ^{-1} x - \frac{1}{2} \tan ^{-1} x$.
On differentiating with respect to $x$,we get:
$\frac{d y}{d x} = \frac{1}{2} \frac{d}{d x}(\tanh ^{-1} x) - \frac{1}{2} \frac{d}{d x}(\tan ^{-1} x)$.
Since $\frac{d}{d x}(\tanh ^{-1} x) = \frac{1}{1-x^2}$ and $\frac{d}{d x}(\tan ^{-1} x) = \frac{1}{1+x^2}$,we have:
$\frac{d y}{d x} = \frac{1}{2} \left(\frac{1}{1-x^2}\right) - \frac{1}{2} \left(\frac{1}{1+x^2}\right)$.
$\frac{d y}{d x} = \frac{1}{2} \left(\frac{1+x^2 - (1-x^2)}{(1-x^2)(1+x^2)}\right) = \frac{1}{2} \left(\frac{2x^2}{1-x^4}\right) = \frac{x^2}{1-x^4}$.

(B) Given $y = \frac{ax+b}{cx+d}$.
Rearranging for $x$:
$y(cx+d) = ax+b$
$cxy + dy = ax + b$
$x(cy - a) = b - dy$
$x = \frac{dy-b}{a-cy}$
Differentiating with respect to $y$:
$\frac{dx}{dy} = \frac{(a-cy)(d) - (dy-b)(-c)}{(a-cy)^2}$
$\frac{dx}{dy} = \frac{ad - cdy + cdy - bc}{(a-cy)^2} = \frac{ad-bc}{c^2y^2 - 2acy + a^2}$
Comparing this with $\frac{ad-bc}{Py^2+Qy+R}$,we get:
$P = c^2, Q = -2ac, R = a^2$
Therefore,$P+Q+R = c^2 - 2ac + a^2 = (a-c)^2$.

(B) Given,$f(t) = \frac{1 + \operatorname{cosec} t}{1 - \operatorname{cosec} t}$.
Taking the natural logarithm on both sides,$\ln f(t) = \ln(1 + \operatorname{cosec} t) - \ln(1 - \operatorname{cosec} t)$.
Differentiating with respect to $t$:
$\frac{f^{\prime}(t)}{f(t)} = \frac{-\operatorname{cosec} t \cot t}{1 + \operatorname{cosec} t} - \frac{\operatorname{cosec} t \cot t}{1 - \operatorname{cosec} t}$.
$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{1}{1 + \operatorname{cosec} t} + \frac{1}{1 - \operatorname{cosec} t} \right)$.
$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{1 - \operatorname{cosec} t + 1 + \operatorname{cosec} t}{1 - \operatorname{cosec}^2 t} \right)$.
$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{2}{-\cot^2 t} \right) = \frac{2 \operatorname{cosec} t \cot t}{\cot^2 t} = \frac{2 \operatorname{cosec} t}{\cot t} = 2 \sec t \operatorname{cosec} t$.
Since $2 \sec t \operatorname{cosec} t = \frac{2}{\sin t \cos t} = \frac{4}{2 \sin t \cos t} = \frac{4}{\sin 2t} = 4 \operatorname{cosec} 2t$.
Thus,$g(t) = 4 \operatorname{cosec} 2t$.

(B) Given equation is $3 f(x)-2 f\left(\frac{1}{x}\right)=x$ ...$(i)$
Replacing $x$ with $\frac{1}{x}$ in equation $(i)$,we get:
$3 f\left(\frac{1}{x}\right)-2 f(x)=\frac{1}{x}$ ...(ii)
To eliminate $f\left(\frac{1}{x}\right)$,multiply equation $(i)$ by $3$ and equation (ii) by $2$:
$9 f(x)-6 f\left(\frac{1}{x}\right)=3 x$ ...(iii)
$6 f\left(\frac{1}{x}\right)-4 f(x)=\frac{2}{x}$ ...(iv)
Adding equation (iii) and (iv):
$5 f(x) = 3 x + \frac{2}{x} \Rightarrow f(x) = \frac{3}{5} x + \frac{2}{5 x}$
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{3}{5} - \frac{2}{5 x^2}$
Now,substitute $x=2$:
$f^{\prime}(2) = \frac{3}{5} - \frac{2}{5(2^2)} = \frac{3}{5} - \frac{2}{20} = \frac{3}{5} - \frac{1}{10} = \frac{6-1}{10} = \frac{5}{10} = \frac{1}{2}$.

(B) Given $y = (x-1)(x+2)(x^2+5)(x^4+8)$.
Using the product rule for differentiation $\frac{d}{dx}(uvwz) = u'vwz + uv'wz + uvw'z + uvwz'$,we get:
$\frac{dy}{dx} = (1)(x+2)(x^2+5)(x^4+8) + (x-1)(1)(x^2+5)(x^4+8) + (x-1)(x+2)(2x)(x^4+8) + (x-1)(x+2)(x^2+5)(4x^3)$.
Now,evaluate the limit as $x \rightarrow -1$:
For $x = -1$:
Term $1$: $(1)(-1+2)((-1)^2+5)((-1)^4+8) = (1)(1)(6)(9) = 54$.
Term $2$: $(-1-1)(1)((-1)^2+5)((-1)^4+8) = (-2)(1)(6)(9) = -108$.
Term $3$: $(-1-1)(-1+2)(2(-1))((-1)^4+8) = (-2)(1)(-2)(9) = 36$.
Term $4$: $(-1-1)(-1+2)((-1)^2+5)(4(-1)^3) = (-2)(1)(6)(-4) = 48$.
Summing these values: $54 - 108 + 36 + 48 = 30$.

(A) Given the function $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$.
We can define $f(x)$ in different intervals:
$f(x) = \begin{cases} -4x & x \leq -5 \\ -2x+10 & -5 < x \leq -4 \\ 18 & -4 < x \leq 4 \\ 2x+10 & 4 < x \leq 5 \\ 4x & x > 5 \end{cases}$
Now,the derivative $f^{\prime}(x)$ is:
$f^{\prime}(x) = \begin{cases} -4 & x < -5 \\ -2 & -5 < x < -4 \\ 0 & -4 < x < 4 \\ 2 & 4 < x < 5 \\ 4 & x > 5 \end{cases}$
Evaluating the required values:
$f^{\prime}(1) = 0$ (since $-4 < 1 < 4$)
$f^{\prime}(-6) = -4$ (since $-6 < -5$)
$f^{\prime}(-1) = 0$ (since $-4 < -1 < 4$)
$f^{\prime}(6) = 4$ (since $6 > 5$)
Substituting these into the expression:
$\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)} = \frac{0-(-4)}{0+4} = \frac{4}{4} = 1$.

(B) Given $f(t) = \frac{t}{2} + \frac{1}{4} \log(2t - 1)$.
First,find the derivative $f^{\prime}(t)$:
$f^{\prime}(t) = \frac{d}{dt} \left( \frac{t}{2} + \frac{1}{4} \log(2t - 1) \right) = \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2t - 1} \cdot 2 = \frac{1}{2} + \frac{1}{2(2t - 1)}$.
Now,substitute $t$ with $\frac{t+1}{2t+1}$ in $f^{\prime}(t)$:
$f^{\prime}\left(\frac{t+1}{2t+1}\right) = \frac{1}{2} + \frac{1}{2\left(2\left(\frac{t+1}{2t+1}\right) - 1\right)}$.
Simplify the denominator term:
$2\left(\frac{t+1}{2t+1}\right) - 1 = \frac{2t + 2 - (2t + 1)}{2t + 1} = \frac{1}{2t + 1}$.
Substitute this back into the expression:
$f^{\prime}\left(\frac{t+1}{2t+1}\right) = \frac{1}{2} + \frac{1}{2\left(\frac{1}{2t+1}\right)} = \frac{1}{2} + \frac{2t+1}{2} = \frac{1 + 2t + 1}{2} = \frac{2t + 2}{2} = t + 1$.

(C) Given that $y = x + \frac{1}{x}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 1 - \frac{1}{x^2}$.
This can be written as $y' = \frac{x^2 - 1}{x^2}$.
Multiplying both sides by $x^2$:
$x^2 y' = x^2 - 1$.
Rearranging the terms:
$x^2 y' - x^2 + 1 = 0$.
We know that $y = x + \frac{1}{x}$,so $x y = x(x + \frac{1}{x}) = x^2 + 1$.
Substituting $x^2 = x y - 1$ into the equation $x^2 y' - x^2 + 1 = 0$:
$x^2 y' - (x y - 1) + 1 = 0$.
$x^2 y' - x y + 1 + 1 = 0$.
$x^2 y' - x y + 2 = 0$.

(A) Given equation is $2 f(x)-3 f\left(\frac{1}{x}\right)=x+1 \quad ...(i)$
Replace $x$ with $\frac{1}{x}$ in equation $(i)$:
$2 f\left(\frac{1}{x}\right)-3 f(x)=\frac{1}{x}+1 \quad ...(ii)$
Multiply equation $(i)$ by $2$ and equation $(ii)$ by $3$:
$4 f(x)-6 f\left(\frac{1}{x}\right)=2 x+2 \quad ...(iii)$
$6 f\left(\frac{1}{x}\right)-9 f(x)=\frac{3}{x}+3 \quad ...(iv)$
Adding equation $(iii)$ and $(iv)$:
$(4 f(x)-9 f(x)) = 2 x + \frac{3}{x} + 5$
$-5 f(x) = 2 x + \frac{3}{x} + 5$
$f(x) = -\frac{2}{5} x - \frac{3}{5 x} - 1$
Differentiating with respect to $x$:
$f^{\prime}(x) = -\frac{2}{5} + \frac{3}{5 x^2}$
Now,substitute $x = \sqrt{3}$:
$f^{\prime}(\sqrt{3}) = -\frac{2}{5} + \frac{3}{5(\sqrt{3})^2}$
$f^{\prime}(\sqrt{3}) = -\frac{2}{5} + \frac{3}{5 \times 3}$
$f^{\prime}(\sqrt{3}) = -\frac{2}{5} + \frac{1}{5} = -\frac{1}{5}$

(A) Given,$y = \sqrt{\frac{1+\tan x}{1-\tan x}} = \sqrt{\tan(\frac{\pi}{4}+x)}$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan(\frac{\pi}{4}+x)}} \cdot \sec^2(\frac{\pi}{4}+x) \cdot \frac{d}{dx}(\frac{\pi}{4}+x)$.
Since $\frac{d}{dx}(\frac{\pi}{4}+x) = 1$,we have:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{\tan(\frac{\pi}{4}+x)}} \cdot \sec^2(\frac{\pi}{4}+x)$.
Substituting $\sqrt{\tan(\frac{\pi}{4}+x)} = \sqrt{\frac{1+\tan x}{1-\tan x}}$,we get:
$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \sec^2(\frac{\pi}{4}+x)$.
Thus,option $A$ is correct.

(B) Given $f(x) = 3 e^{x^2}$.
First,find the derivative $f^{\prime}(x)$ using the chain rule:
$f^{\prime}(x) = 3 \cdot e^{x^2} \cdot \frac{d}{dx}(x^2) = 3 e^{x^2} \cdot 2x = 6x e^{x^2}$.
Now,calculate the terms:
$2x f(x) = 2x(3 e^{x^2}) = 6x e^{x^2}$.
$f(0) = 3 e^{0^2} = 3(1) = 3$.
$f^{\prime}(0) = 6(0) e^{0^2} = 0$.
Substitute these into the expression $f^{\prime}(x) - 2x f(x) + \frac{1}{3} f(0) - f^{\prime}(0)$:
$= 6x e^{x^2} - 6x e^{x^2} + \frac{1}{3}(3) - 0$
$= 0 + 1 - 0 = 1$.

(C) Given the formula $T = 2 \pi \sqrt{\frac{L}{g}}$.
Taking the natural logarithm on both sides: $\ln T = \ln(2 \pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g$.
Differentiating both sides,we get $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.
The relative error in $T$ is $\frac{dT}{T}$.
The percentage error in $L$ is $\frac{dL}{L} \times 100$.
According to the problem,$\frac{dT}{T} = k \times (\frac{dL}{L} \times 100)$.
Comparing this with $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$,we have $k \times 100 = \frac{1}{2}$.
Therefore,$k = \frac{1}{200}$.
Thus,$\frac{1}{k} = 200$.

(D) Given that $|f(x) - f(y)| \leq 2|x - y|^{\frac{3}{2}}$.
Dividing both sides by $|x - y|$ (where $x \neq y$),we get $\left|\frac{f(x) - f(y)}{x - y}\right| \leq 2|x - y|^{\frac{1}{2}}$.
Taking the limit as $x \rightarrow y$,the left side becomes the definition of the derivative $|f'(y)|$.
Thus,$|f'(y)| \leq 2 \lim_{x \rightarrow y} |x - y|^{\frac{1}{2}} = 0$.
Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which implies $f'(y) = 0$ for all $y \in R$.
This means $f(x)$ is a constant function,$f(x) = c$.
Given $f(0) = 1$,we have $c = 1$,so $f(x) = 1$.
Therefore,$\int_0^1 f^2(x) dx = \int_0^1 (1)^2 dx = \int_0^1 1 dx = [x]_0^1 = 1 - 0 = 1$.

(A) Given $f(x) = x(x+3)(x-2)$.
Expanding the expression: $f(x) = x(x^2 + x - 6) = x^3 + x^2 - 6x$.
Differentiating with respect to $x$: $f^{\prime}(x) = 3x^2 + 2x - 6$.
We are given $f^{\prime}(c) = 10$,so $3c^2 + 2c - 6 = 10$.
Rearranging the equation: $3c^2 + 2c - 16 = 0$.
Factoring the quadratic equation: $3c^2 + 8c - 6c - 16 = 0$.
$c(3c + 8) - 2(3c + 8) = 0$.
$(3c + 8)(c - 2) = 0$.
Thus,$c = -\frac{8}{3}$ or $c = 2$.
Since we require $c \in (-1, 4)$,we reject $c = -\frac{8}{3}$ as it lies outside the interval.
Therefore,$c = 2$ is the correct value.

(A) Given,$f(x) = \int_0^x [(a+1)(t+1)^2 - (a-1)(t^2+t+1)] dt$.
By the Fundamental Theorem of Calculus,$f^{\prime}(x) = (a+1)(x+1)^2 - (a-1)(x^2+x+1)$.
Expanding the terms: $f^{\prime}(x) = (a+1)(x^2+2x+1) - (a-1)(x^2+x+1)$.
$f^{\prime}(x) = (a+1)x^2 + 2(a+1)x + (a+1) - (a-1)x^2 - (a-1)x - (a-1)$.
$f^{\prime}(x) = (a+1-a+1)x^2 + (2a+2-a+1)x + (a+1-a+1)$.
$f^{\prime}(x) = 2x^2 + (a+3)x + 2 = 0$.
For the quadratic equation $Ax^2 + Bx + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be $0$.
Here,$A = 2$,$B = (a+3)$,and $C = 2$.
$D = (a+3)^2 - 4(2)(2) = 0$.
$(a+3)^2 - 16 = 0$.
$(a+3)^2 = 16$.
$a+3 = \pm 4$.
Case $1$: $a+3 = 4 \Rightarrow a = 1$.
Case $2$: $a+3 = -4 \Rightarrow a = -7$.
Since the question asks for a positive value of '$a$',the answer is $1$.

(B) Given $f^{\prime}(x)=a \cos x+b \sin x$ ... $(i)$
Integrating both sides with respect to $x$:
$f(x) = \int (a \cos x + b \sin x) dx = a \sin x - b \cos x + C$ ... $(ii)$
Given $f^{\prime}(0) = 4$:
$f^{\prime}(0) = a \cos(0) + b \sin(0) = a(1) + b(0) = a = 4$.
Given $f(0) = 3$:
$f(0) = a \sin(0) - b \cos(0) + C = 0 - b(1) + C = -b + C = 3$.
Given $f\left(\frac{\pi}{2}\right) = 5$:
$f\left(\frac{\pi}{2}\right) = a \sin\left(\frac{\pi}{2}\right) - b \cos\left(\frac{\pi}{2}\right) + C = a(1) - b(0) + C = a + C = 5$.
Since $a = 4$,we have $4 + C = 5 \Rightarrow C = 1$.
Substituting $C = 1$ into $-b + C = 3$:
$-b + 1 = 3 \Rightarrow -b = 2 \Rightarrow b = -2$.
Substituting $a = 4, b = -2, C = 1$ into equation $(ii)$:
$f(x) = 4 \sin x - (-2) \cos x + 1 = 4 \sin x + 2 \cos x + 1$.

(C) Given,$f(x) = \begin{cases} \frac{x^2-16}{x-4} & \text{if } x > 4 \\ 2x & \text{if } x \leq 4 \end{cases}$
For $x > 4$,$f(x) = \frac{(x-4)(x+4)}{x-4} = x+4$.
For $x \leq 4$,$f(x) = 2x$.
Now,differentiating with respect to $x$:
$f^{\prime}(x) = \begin{cases} \frac{d}{dx}(x+4) = 1 & \text{if } x > 4 \\ \frac{d}{dx}(2x) = 2 & \text{if } x < 4 \end{cases}$
Therefore,$f^{\prime}(4^{+}) = \lim_{h \to 0} f^{\prime}(4+h) = 1$ and $f^{\prime}(4^{-}) = \lim_{h \to 0} f^{\prime}(4-h) = 2$.
Thus,$f^{\prime}(4^{-}) + f^{\prime}(4^{+}) = 2 + 1 = 3$.

(A) For $x \neq 0$,we find the derivative $f^{\prime}(x)$ using the product rule and chain rule:
$f^{\prime}(x) = \frac{d}{dx} \left( x^2 \sin \frac{1}{x} \right) = 2x \sin \frac{1}{x} + x^2 \left( \cos \frac{1}{x} \right) \left( -\frac{1}{x^2} \right)$
$f^{\prime}(x) = 2x \sin \frac{1}{x} - \cos \frac{1}{x}$
Now,we evaluate the limit as $x \rightarrow 0$:
$\lim_{x \rightarrow 0} f^{\prime}(x) = \lim_{x \rightarrow 0} \left( 2x \sin \frac{1}{x} - \cos \frac{1}{x} \right)$
$= \lim_{x \rightarrow 0} (2x \sin \frac{1}{x}) - \lim_{x \rightarrow 0} \cos \frac{1}{x}$
Since $\lim_{x \rightarrow 0} 2x \sin \frac{1}{x} = 0$ (by the squeeze theorem) and $\lim_{x \rightarrow 0} \cos \frac{1}{x}$ does not exist because the function oscillates between $-1$ and $1$ as $x \rightarrow 0$,the overall limit does not exist.

(C) Given that,$f(2)=4$ and $f^{\prime}(2)=1$.
We need to evaluate the limit:
$L = \lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$
Adding and subtracting $2f(2)$ in the numerator:
$L = \lim _{x \rightarrow 2} \frac{x f(2)-2 f(2)+2 f(2)-2 f(x)}{x-2}$
$L = \lim _{x \rightarrow 2} \left[ \frac{f(2)(x-2)}{x-2} - 2 \frac{f(x)-f(2)}{x-2} \right]$
$L = f(2) - 2 \lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}$
$L = f(2) - 2 f^{\prime}(2)$
Substituting the given values:
$L = 4 - 2(1) = 4 - 2 = 2$.

(B) Given that $f(x) = \frac{x - 2}{x^2 - 3x + 2} = \frac{x - 2}{(x - 2)(x - 1)} = \frac{1}{x - 1}$ for $x \neq 1, 2$.
We need to evaluate $\lim_{x \rightarrow 2} \frac{f(x) - f(2)}{x - 2}$.
Given $f(2) = 1$.
Substituting the values,we get $\lim_{x \rightarrow 2} \frac{\frac{1}{x - 1} - 1}{x - 2}$.
$= \lim_{x \rightarrow 2} \frac{\frac{1 - (x - 1)}{x - 1}}{x - 2} = \lim_{x \rightarrow 2} \frac{2 - x}{(x - 1)(x - 2)}$.
$= \lim_{x \rightarrow 2} \frac{-(x - 2)}{(x - 1)(x - 2)} = \lim_{x \rightarrow 2} \frac{-1}{x - 1}$.
$= \frac{-1}{2 - 1} = -1$.

(D) Given $f(x) = \tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$.
Let $x = \cos 2\theta$,then $1+x = 2\cos^2\theta$ and $1-x = 2\sin^2\theta$.
$f(x) = \tan^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right) = \tan^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right) = \tan^{-1}\left(\tan(\frac{\pi}{4}-\theta)\right) = \frac{\pi}{4}-\theta$.
Since $x = \cos 2\theta$,$\theta = \frac{1}{2}\cos^{-1}x$,so $f(x) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x$.
The limit is $\lim_{x \rightarrow \frac{1}{2}} \frac{2[f(x)-f(\frac{1}{2})]}{2x-1} = \lim_{x \rightarrow \frac{1}{2}} \frac{f(x)-f(\frac{1}{2})}{x-\frac{1}{2}} = f'(\frac{1}{2})$.
$f'(x) = -\frac{1}{2} \left(-\frac{1}{\sqrt{1-x^2}}\right) = \frac{1}{2\sqrt{1-x^2}}$.
$f'(\frac{1}{2}) = \frac{1}{2\sqrt{1-(\frac{1}{2})^2}} = \frac{1}{2\sqrt{3/4}} = \frac{1}{2(\frac{\sqrt{3}}{2})} = \frac{1}{\sqrt{3}}$.

(D) The given expression is the definition of the derivative of $f(x)$ at $x = 1$,i.e.,$f'(1)$.
Given $f(x) = x \tan^{-1} x$.
Using the product rule,$f'(x) = \frac{d}{dx}(x) \cdot \tan^{-1} x + x \cdot \frac{d}{dx}(\tan^{-1} x)$.
$f'(x) = 1 \cdot \tan^{-1} x + x \cdot \frac{1}{1 + x^2} = \tan^{-1} x + \frac{x}{1 + x^2}$.
Now,evaluate at $x = 1$:
$f'(1) = \tan^{-1}(1) + \frac{1}{1 + 1^2} = \frac{\pi}{4} + \frac{1}{2}$.
$f'(1) = \frac{\pi}{4} + \frac{2}{4} = \frac{\pi + 2}{4}$.

(C) The function is defined as $f(x) = \begin{cases} x-5 & \text{for } x \leq 1 \\ 4x^2-9 & \text{for } 1 < x < 2 \\ 3x+4 & \text{for } x \geq 2 \end{cases}$.
To find $f^{\prime}(2^{+})$,we use the definition of the right-hand derivative at $x = 2$:
$f^{\prime}(2^{+}) = \lim_{x \rightarrow 2^{+}} \frac{f(x) - f(2)}{x-2}$
For $x \geq 2$,$f(x) = 3x+4$. Thus,$f(2) = 3(2) + 4 = 10$.
$f^{\prime}(2^{+}) = \lim_{x \rightarrow 2^{+}} \frac{(3x+4) - 10}{x-2}$
$f^{\prime}(2^{+}) = \lim_{x \rightarrow 2^{+}} \frac{3x-6}{x-2}$
$f^{\prime}(2^{+}) = \lim_{x \rightarrow 2^{+}} \frac{3(x-2)}{x-2}$
$f^{\prime}(2^{+}) = 3$.

(D) Given $f(x) = \sum_{p=1}^7 p^2 \sin^{-1}\left(\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)\right)$.
Let $\cos \alpha = \frac{4}{5}$ and $\sin \alpha = \frac{3}{5}$.
Then the expression becomes $\sin^{-1}(\sin(px) \cos \alpha - \cos(px) \sin \alpha) = \sin^{-1}(\sin(px - \alpha)) = px - \alpha$.
Thus,$f(x) = \sum_{p=1}^7 p^2(px - \alpha) = \sum_{p=1}^7 (p^3 x - p^2 \alpha) = x \sum_{p=1}^7 p^3 - \alpha \sum_{p=1}^7 p^2$.
Taking the derivative with respect to $x$:
$\frac{df}{dx} = \frac{d}{dx} \left( x \sum_{p=1}^7 p^3 - \alpha \sum_{p=1}^7 p^2 \right) = \sum_{p=1}^7 p^3$.
The sum of cubes is given by $\left(\frac{n(n+1)}{2}\right)^2$ for $n=7$.
$\frac{df}{dx} = \left(\frac{7(8)}{2}\right)^2 = (28)^2 = 784$.

(A) Given $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$.
Let $x = \sin \theta$,then $\sqrt{1-x^2} = \cos \theta$.
So,$f(x) = \sqrt{\cos^{-1}(\cos \theta)} = \sqrt{\theta} = \sqrt{\sin^{-1} x}$.
Now,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(\sin^{-1} x)^{1/2} = \frac{1}{2}(\sin^{-1} x)^{-1/2} \cdot \frac{d}{dx}(\sin^{-1} x) = \frac{1}{2\sqrt{\sin^{-1} x}} \cdot \frac{1}{\sqrt{1-x^2}}$.
At $x = \frac{1}{2}$,$\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ and $\sqrt{1-(\frac{1}{2})^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
$f^{\prime}(\frac{1}{2}) = \frac{1}{2\sqrt{\pi/6}} \cdot \frac{1}{\sqrt{3}/2} = \frac{1}{\sqrt{\pi/6} \cdot \sqrt{3}} = \frac{1}{\sqrt{\pi/2}} = \sqrt{\frac{2}{\pi}}$.

(B) Given that $f(x)$ is an even function,so $f(-x) = f(x)$.
By differentiating both sides with respect to $x$ using the chain rule:
$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$
$-f^{\prime}(-x) = f^{\prime}(x)$,which implies $f^{\prime}(-x) = -f^{\prime}(x)$.
Thus,the first derivative $f^{\prime}(x)$ is an odd function.
Now,differentiating again:
$\frac{d}{dx}[f^{\prime}(-x)] = \frac{d}{dx}[-f^{\prime}(x)]$
$-f^{\prime \prime}(-x) = -f^{\prime \prime}(x)$,which implies $f^{\prime \prime}(-x) = f^{\prime \prime}(x)$.
Thus,the second derivative $f^{\prime \prime}(x)$ is an even function.
Similarly,the third derivative $f^{\prime \prime \prime}(x)$ will be an odd function.
Therefore,$f^{\prime}(x)$ is an odd function. Comparing with the options,option $B$ is correct.

(C) The function is defined as $f(x) = |x-1| + |x-2|$.
We analyze the derivative $f^{\prime}(x)$ in different intervals:
$1$. For $x < 1$,$f(x) = -(x-1) - (x-2) = -2x + 3$,so $f^{\prime}(x) = -2$.
Since $-2023 < 1$,$f^{\prime}(-2023) = -2$.
$2$. For $1 < x < 2$,$f(x) = (x-1) - (x-2) = 1$,so $f^{\prime}(x) = 0$.
Since $1 < \frac{2024}{2023} < 2$,$f^{\prime}\left(\frac{2024}{2023}\right) = 0$.
$3$. For $x > 2$,$f(x) = (x-1) + (x-2) = 2x - 3$,so $f^{\prime}(x) = 2$.
Since $2023 > 2$,$f^{\prime}(2023) = 2$.
Summing these values: $f^{\prime}(-2023) + f^{\prime}\left(\frac{2024}{2023}\right) + f^{\prime}(2023) = -2 + 0 + 2 = 0$.

(D) Given the equation: $y f(x+y) + \cos(mxy) = 1 + y f(x)$.
Rearranging the terms,we get: $y(f(x+y) - f(x)) = 1 - \cos(mxy)$.
Given $m=2$,the equation becomes: $y(f(x+y) - f(x)) = 1 - \cos(2xy)$.
Dividing by $y^2$ on both sides (assuming $y \neq 0$): $\frac{f(x+y) - f(x)}{y} = \frac{1 - \cos(2xy)}{y^2}$.
Using the identity $1 - \cos(2\theta) = 2 \sin^2(\theta)$,we have: $\frac{f(x+y) - f(x)}{y} = \frac{2 \sin^2(xy)}{y^2}$.
This can be written as: $\frac{f(x+y) - f(x)}{y} = 2 \left( \frac{\sin(xy)}{y} \right)^2$.
To find $f'(x)$,take the limit as $y \rightarrow 0$:
$f'(x) = \lim_{y \rightarrow 0} \frac{f(x+y) - f(x)}{y} = \lim_{y \rightarrow 0} 2 \left( \frac{\sin(xy)}{y} \right)^2$.
Multiply and divide by $x^2$: $f'(x) = 2x^2 \lim_{y \rightarrow 0} \left( \frac{\sin(xy)}{xy} \right)^2$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get: $f'(x) = 2x^2(1)^2 = 2x^2$.

(A) By the definition of the derivative,$f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.
Given the functional equation $f(x+y) = f(x) + f(y) - 3xy$,we substitute $y = h$ to get $f(x+h) = f(x) + f(h) - 3xh$.
Substituting this into the derivative formula:
$f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x) + f(h) - 3xh - f(x)}{h}$.
Simplifying the expression:
$f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(h) - 3xh}{h} = \lim _{h \rightarrow 0} \left( \frac{f(h)}{h} - 3x \right)$.
Given that $\lim _{h \rightarrow 0} \frac{f(h)}{h} = 7$,we have:
$f^{\prime}(x) = 7 - 3x$.

(A) Let $L = \lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$.
Since the limit is in the $\frac{0}{0}$ form,we can apply $L$'$H$ôpital's rule or manipulate the expression.
Adding and subtracting $a f(a)$ in the numerator:
$L = \lim _{x \rightarrow a} \frac{x f(a) - a f(a) + a f(a) - a f(x)}{x-a}$
$L = \lim _{x \rightarrow a} \left[ \frac{f(a)(x-a)}{x-a} - a \frac{f(x)-f(a)}{x-a} \right]$
$L = f(a) - a \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$
By the definition of the derivative,$\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} = f^{\prime}(a)$.
So,$L = f(a) - a f^{\prime}(a)$.
Given $f^{\prime}(a) = a f(a)$,we substitute this into the expression:
$L = f(a) - a(a f(a))$
$L = f(a) - a^2 f(a)$
$L = (1-a^2) f(a)$.

(D) Let $y = \frac{x-1}{x-\sqrt{x}} e^{2x+1}$.
We can simplify the expression $\frac{x-1}{x-\sqrt{x}}$ as follows:
$\frac{x-1}{x-\sqrt{x}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + \frac{1}{\sqrt{x}} = 1 + x^{-1/2}$.
So,$y = (1 + x^{-1/2}) e^{2x+1}$.
Now,differentiate $y$ with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(1 + x^{-1/2}) \cdot e^{2x+1} + (1 + x^{-1/2}) \cdot \frac{d}{dx}(e^{2x+1})$.
$\frac{dy}{dx} = (- \frac{1}{2} x^{-3/2}) e^{2x+1} + (1 + x^{-1/2}) \cdot 2 e^{2x+1}$.
Factor out $e^{2x+1}$:
$\frac{dy}{dx} = e^{2x+1} [-\frac{1}{2} x^{-3/2} + 2 + 2x^{-1/2}]$.
We are given $\frac{dy}{dx} = \frac{x-1}{x-\sqrt{x}} e^{2x+1} f(x) = (1 + x^{-1/2}) e^{2x+1} f(x)$.
Therefore,$f(x) = \frac{-\frac{1}{2} x^{-3/2} + 2 + 2x^{-1/2}}{1 + x^{-1/2}}$.
Substitute $x = 4$:
$f(4) = \frac{-\frac{1}{2} (4)^{-3/2} + 2 + 2(4)^{-1/2}}{1 + (4)^{-1/2}} = \frac{-\frac{1}{2} (\frac{1}{8}) + 2 + 2(\frac{1}{2})}{1 + \frac{1}{2}} = \frac{-\frac{1}{16} + 2 + 1}{\frac{3}{2}} = \frac{3 - \frac{1}{16}}{\frac{3}{2}} = \frac{\frac{47}{16}}{\frac{3}{2}} = \frac{47}{16} \times \frac{2}{3} = \frac{47}{24}$.

(D) Given $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.
Using the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
Let $u = e^{2x} - e^{-2x}$ and $v = e^{3x} + e^{-3x}$.
Then $u^{\prime} = 2e^{2x} + 2e^{-2x}$ and $v^{\prime} = 3e^{3x} - 3e^{-3x}$.
At $x = 0$:
$u(0) = e^0 - e^0 = 1 - 1 = 0$.
$v(0) = e^0 + e^0 = 1 + 1 = 2$.
$u^{\prime}(0) = 2(1) + 2(1) = 4$.
$v^{\prime}(0) = 3(1) - 3(1) = 0$.
Now,$f^{\prime}(0) = \frac{u^{\prime}(0)v(0) - u(0)v^{\prime}(0)}{(v(0))^2}$.
$f^{\prime}(0) = \frac{(4)(2) - (0)(0)}{(2)^2} = \frac{8}{4} = 2$.

(B) Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$.
We need to evaluate the limit $L = \lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}$.
This expression is the definition of the derivative of the composite function $F(x) = f(g(h(x)))$ at $x=1$,i.e.,$F'(1)$.
First,find $F(x) = f(g(h(x))) = \sin(\cos(x^2))$.
Now,differentiate $F(x)$ with respect to $x$ using the chain rule:
$F'(x) = \cos(\cos(x^2)) \cdot \frac{d}{dx}(\cos(x^2)) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot \frac{d}{dx}(x^2)$.
$F'(x) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) = -2x \sin(x^2) \cos(\cos(x^2))$.
Now,evaluate at $x=1$:
$F'(1) = -2(1) \sin(1^2) \cos(\cos(1^2)) = -2 \sin 1 \cos(\cos 1)$.
Thus,the limit is $-2 \sin 1 \cos(\cos 1)$.
Therefore,option $B$ is correct.

(D) Given $f(x) = \frac{1+\sec x}{2(\sec x-1)}$.
Simplifying $f(x)$ by converting to $\cos x$:
$f(x) = \frac{1 + \frac{1}{\cos x}}{2(\frac{1}{\cos x} - 1)} = \frac{\frac{\cos x + 1}{\cos x}}{2(\frac{1 - \cos x}{\cos x})} = \frac{1 + \cos x}{2(1 - \cos x)}$.
Using the half-angle formulas $1 + \cos x = 2 \cos^2(\frac{x}{2})$ and $1 - \cos x = 2 \sin^2(\frac{x}{2})$:
$f(x) = \frac{2 \cos^2(\frac{x}{2})}{2(2 \sin^2(\frac{x}{2}))} = \frac{1}{2} \cot^2(\frac{x}{2})$.
Now,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{1}{2} \cdot 2 \cot(\frac{x}{2}) \cdot (-\operatorname{cosec}^2(\frac{x}{2})) \cdot \frac{1}{2} = -\frac{1}{2} \cot(\frac{x}{2}) \operatorname{cosec}^2(\frac{x}{2})$.
We know $f^{\prime}(x) = f(x) \cdot g(x)$,so $g(x) = \frac{f^{\prime}(x)}{f(x)}$.
$g(x) = \frac{-\frac{1}{2} \cot(\frac{x}{2}) \operatorname{cosec}^2(\frac{x}{2})}{\frac{1}{2} \cot^2(\frac{x}{2})} = -\frac{\operatorname{cosec}^2(\frac{x}{2})}{\cot(\frac{x}{2})} = -\frac{1}{\sin^2(\frac{x}{2})} \cdot \frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})} = -\frac{1}{\sin(\frac{x}{2}) \cos(\frac{x}{2})}$.
Multiplying numerator and denominator by $2$:
$g(x) = -\frac{2}{2 \sin(\frac{x}{2}) \cos(\frac{x}{2})} = -\frac{2}{\sin x} = -2 \operatorname{cosec} x$.

(D) Given $F(x)=f(x) g(x)$.
By the product rule,$F^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$.
We are given $G(x)=f^{\prime}(x) g^{\prime}(x)$.
Since $F^{\prime}(x)=G(x) H(x)$,we have $H(x)=\frac{F^{\prime}(x)}{G(x)}=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f^{\prime}(x) g^{\prime}(x)}=\frac{g(x)}{g^{\prime}(x)}+\frac{f(x)}{f^{\prime}(x)}$.
Also,$F^{\prime}(x)=F(x) K(x)$,so $K(x)=\frac{F^{\prime}(x)}{F(x)}=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f(x) g(x)}=\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}$.
Therefore,$H(x)+K(x)=\frac{f(x)}{f^{\prime}(x)}+\frac{g(x)}{g^{\prime}(x)}+\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}$.

(B) Given $y = \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \log \sqrt{1-x^2}$.
First,simplify the logarithmic term: $\log \sqrt{1-x^2} = \frac{1}{2} \log(1-x^2)$.
Now,differentiate $y$ with respect to $x$ using the quotient rule for the first term and the chain rule for the second term:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) + \frac{d}{dx} \left( \frac{1}{2} \log(1-x^2) \right)$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$ where $u = x \sin^{-1} x$ and $v = \sqrt{1-x^2}$:
$u' = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}}$ and $v' = \frac{-x}{\sqrt{1-x^2}}$.
$\frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) = \frac{(\sin^{-1} x + \frac{x}{\sqrt{1-x^2}})\sqrt{1-x^2} - (x \sin^{-1} x)(\frac{-x}{\sqrt{1-x^2}})}{1-x^2} = \frac{(1-x^2)\sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2} = \frac{\sin^{-1} x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}} + x}{1-x^2}$.
Now,differentiate the second term: $\frac{d}{dx} (\frac{1}{2} \log(1-x^2)) = \frac{1}{2} \cdot \frac{1}{1-x^2} \cdot (-2x) = \frac{-x}{1-x^2}$.
Adding both parts: $\frac{dy}{dx} = \frac{\sin^{-1} x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}} + x - x}{1-x^2} = \frac{\sin^{-1} x (1 + \frac{x^2}{\sqrt{1-x^2}})}{1-x^2} = \frac{\sin^{-1} x (\frac{\sqrt{1-x^2} + x^2}{\sqrt{1-x^2}})}{1-x^2}$.
Wait,simplifying the expression: $\frac{dy}{dx} = \frac{\sin^{-1} x (1-x^2 + x^2)}{\sqrt{1-x^2}(1-x^2)} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}}$.

(C) Let $y = \operatorname{cosech}^{-1}(\tan 2x)$.
Using the chain rule,$\frac{dy}{dx} = \frac{d}{dx} [\operatorname{cosech}^{-1}(\tan 2x)]$.
Recall the formula: $\frac{d}{dx} \operatorname{cosech}^{-1}(u) = \frac{-1}{|u| \sqrt{1+u^2}} \cdot \frac{du}{dx}$.
Here,$u = \tan 2x$,so $\frac{du}{dx} = 2 \sec^2 2x$.
Substituting these into the formula:
$\frac{dy}{dx} = \frac{-1}{|\tan 2x| \sqrt{1 + \tan^2 2x}} \cdot 2 \sec^2 2x$.
Since $1 + \tan^2 2x = \sec^2 2x$,we have $\sqrt{1 + \tan^2 2x} = |\sec 2x|$.
$\frac{dy}{dx} = \frac{-1}{|\tan 2x| \cdot |\sec 2x|} \cdot 2 \sec^2 2x$.
$\frac{dy}{dx} = \frac{-2 \sec^2 2x}{|\frac{\sin 2x}{\cos 2x}| \cdot |\frac{1}{\cos 2x}|} = \frac{-2 \sec^2 2x}{|\frac{\sin 2x}{\cos^2 2x}|} = \frac{-2 \sec^2 2x \cdot |\cos^2 2x|}{|\sin 2x|}$.
Since $\sec^2 2x \cdot \cos^2 2x = 1$,this simplifies to:
$\frac{dy}{dx} = \frac{-2}{|\sin 2x|} = -2 |\operatorname{cosec} 2x|$.

(B) Let $f(x) = \frac{x+5}{(x+1)^2(x+2)}$. Using partial fractions,we write:
$\frac{x+5}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$
$x+5 = A(x+1)(x+2) + B(x+2) + C(x+1)^2$
For $x = -1$: $-1+5 = B(-1+2) \Rightarrow B = 4$.
For $x = -2$: $-2+5 = C(-2+1)^2 \Rightarrow C = 3$.
Comparing coefficients of $x^2$: $0 = A + C \Rightarrow A = -C = -3$.
Thus,$f(x) = -\frac{3}{x+1} + \frac{4}{(x+1)^2} + \frac{3}{x+2}$.
Differentiating with respect to $x$:
$f'(x) = \frac{d}{dx} \left( -3(x+1)^{-1} + 4(x+1)^{-2} + 3(x+2)^{-1} \right)$
$f'(x) = 3(x+1)^{-2} - 8(x+1)^{-3} - 3(x+2)^{-2}$
$f'(x) = \frac{3}{(x+1)^2} - \frac{8}{(x+1)^3} - \frac{3}{(x+2)^2}$.

(B) Given $f(x) = \frac{1}{1 + \frac{1}{x}} = \frac{x}{x + 1}$.
Now,$g(x) = \frac{1}{1 + \frac{1}{f(x)}} = \frac{1}{1 + \frac{x + 1}{x}} = \frac{1}{\frac{x + x + 1}{x}} = \frac{x}{2x + 1}$.
To find $g^{\prime}(x)$,we use the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$g^{\prime}(x) = \frac{(1)(2x + 1) - (x)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x}{(2x + 1)^2} = \frac{1}{(2x + 1)^2}$.
Substituting $x = 2$:
$g^{\prime}(2) = \frac{1}{(2(2) + 1)^2} = \frac{1}{(5)^2} = \frac{1}{25}$.

(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.
Multiply and divide by $(x-1)$:
$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^4-1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^8-1)(x^8+1)}{(x-1)} = \frac{x^{16}-1}{x-1}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{d}{dx}\left(\frac{x^{16}-1}{x-1}\right) = \frac{(16x^{15})(x-1) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.
Comparing this with the given expression $\frac{15x^p - 16x^q + 1}{(x-1)^2}$,we get $p = 16$ and $q = 15$.
Thus,$(p, q) = (16, 15)$.