Derivative at a point, Standard differentiation Questions in English

Let $f(x) = \frac{p x^{2}+q x+r}{x}$.
We can simplify the function by dividing each term in the numerator by $x$:
$f(x) = \frac{p x^{2}}{x} + \frac{q x}{x} + \frac{r}{x} = p x + q + r x^{-1}$.
Now,differentiate with respect to $x$:
$f'(x) = \frac{d}{dx}(p x) + \frac{d}{dx}(q) + \frac{d}{dx}(r x^{-1})$.
Using the power rule $\frac{d}{dx}(x^n) = n x^{n-1}$:
$f'(x) = p(1) + 0 + r(-1)x^{-2}$.
Therefore,$f'(x) = p - \frac{r}{x^{2}}$.

(N/A) Let $f(x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$.
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the derivative of trigonometric functions $\frac{d}{dx}(\cos x) = -\sin x$:
$f'(x) = \frac{d}{dx}(ax^{-4}) - \frac{d}{dx}(bx^{-2}) + \frac{d}{dx}(\cos x)$
$f'(x) = a(-4x^{-5}) - b(-2x^{-3}) - \sin x$
$f'(x) = -\frac{4a}{x^{5}} + \frac{2b}{x^{3}} - \sin x$.

(N/A) Let $f(x) = 4 \sqrt{x} - 2$.
$f'(x) = \frac{d}{dx}(4 \sqrt{x} - 2)$
$= \frac{d}{dx}(4 \sqrt{x}) - \frac{d}{dx}(2)$
$= 4 \frac{d}{dx}(x^{1/2}) - 0$
$= 4 \left( \frac{1}{2} x^{1/2 - 1} \right)$
$= 2 x^{-1/2} = \frac{2}{\sqrt{x}}$.

(N/A) Let $f(x) = (ax + b)^n (cx + d)^m$.
Using the product rule for differentiation,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$.
Let $u(x) = (ax + b)^n$ and $v(x) = (cx + d)^m$.
Using the chain rule,$u'(x) = n(ax + b)^{n-1} \cdot \frac{d}{dx}(ax + b) = n(ax + b)^{n-1} \cdot a = na(ax + b)^{n-1}$.
Similarly,$v'(x) = m(cx + d)^{m-1} \cdot \frac{d}{dx}(cx + d) = m(cx + d)^{m-1} \cdot c = mc(cx + d)^{m-1}$.
Now,substitute these into the product rule formula:
$f'(x) = (ax + b)^n \cdot [mc(cx + d)^{m-1}] + (cx + d)^m \cdot [na(ax + b)^{n-1}]$.
Factor out the common terms $(ax + b)^{n-1}$ and $(cx + d)^{m-1}$:
$f'(x) = (ax + b)^{n-1} (cx + d)^{m-1} [mc(ax + b) + na(cx + d)]$.
Thus,the derivative is $(ax + b)^{n-1} (cx + d)^{m-1} [mc(ax + b) + na(cx + d)]$.

(N/A) Let $f(x) = \operatorname{cosec} x \cot x$.
Using the product rule for differentiation,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$.
Here,$u(x) = \operatorname{cosec} x$ and $v(x) = \cot x$.
We know that $\frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x$ and $\frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x$.
Applying the product rule:
$f'(x) = \operatorname{cosec} x \frac{d}{dx}(\cot x) + \cot x \frac{d}{dx}(\operatorname{cosec} x)$
$f'(x) = \operatorname{cosec} x (-\operatorname{cosec}^2 x) + \cot x (-\operatorname{cosec} x \cot x)$
$f'(x) = -\operatorname{cosec}^3 x - \operatorname{cosec} x \cot^2 x$
$f'(x) = -\operatorname{cosec} x (\operatorname{cosec}^2 x + \cot^2 x)$.

Let $f(x) = \frac{\cos x}{1+\sin x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$f'(x) = \frac{(1+\sin x) \frac{d}{dx}(\cos x) - (\cos x) \frac{d}{dx}(1+\sin x)}{(1+\sin x)^2}$
$f'(x) = \frac{(1+\sin x)(-\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2}$
$f'(x) = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2}$
Since $\sin^2 x + \cos^2 x = 1$,we have:
$f'(x) = \frac{-\sin x - 1}{(1+\sin x)^2}$
$f'(x) = \frac{-(1+\sin x)}{(1+\sin x)^2}$
$f'(x) = \frac{-1}{1+\sin x}$

Let $f(x) = \frac{\sin x+\cos x}{\sin x-\cos x}$.
Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{(\sin x - \cos x) \frac{d}{dx}(\sin x + \cos x) - (\sin x + \cos x) \frac{d}{dx}(\sin x - \cos x)}{(\sin x - \cos x)^2}$
$f'(x) = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-(\sin^2 x + \cos^2 x - 2\sin x \cos x) - (\sin^2 x + \cos^2 x + 2\sin x \cos x)}{(\sin x - \cos x)^2}$
Since $\sin^2 x + \cos^2 x = 1$:
$f'(x) = \frac{-(1 - 2\sin x \cos x) - (1 + 2\sin x \cos x)}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-1 + 2\sin x \cos x - 1 - 2\sin x \cos x}{(\sin x - \cos x)^2}$
$f'(x) = \frac{-2}{(\sin x - \cos x)^2}$

Let $f(x) = \frac{\sec x - 1}{\sec x + 1}$.
We can simplify the function using trigonometric identities:
$f(x) = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1} = \frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} = \tan^2(x/2)$.
Now,differentiate $f(x) = \tan^2(x/2)$ with respect to $x$ using the chain rule:
$f'(x) = 2 \tan(x/2) \cdot \sec^2(x/2) \cdot \frac{1}{2} = \tan(x/2) \sec^2(x/2)$.
Alternatively,using the quotient rule on $\frac{1 - \cos x}{1 + \cos x}$:
$f'(x) = \frac{(1 + \cos x)(\sin x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2}$
$f'(x) = \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{(1 + \cos x)^2}$
$f'(x) = \frac{2 \sin x}{(1 + \cos x)^2}$.

(N/A) Let $y = \sin^{n} x$.
Using the chain rule for differentiation,we treat $\sin^{n} x$ as a composite function $f(g(x))$,where $f(u) = u^{n}$ and $g(x) = \sin x$.
According to the chain rule,$\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.
Here,$\frac{df}{du} = n u^{n-1} = n(\sin x)^{n-1}$ and $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$.
Therefore,$\frac{dy}{dx} = n(\sin x)^{n-1} \cdot \cos x$.
Hence,$\frac{d}{dx}(\sin^{n} x) = n \sin^{n-1} x \cos x$.

Let $f(x) = \frac{a+b \sin x}{c+d \cos x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$f'(x) = \frac{(c+d \cos x) \frac{d}{dx}(a+b \sin x) - (a+b \sin x) \frac{d}{dx}(c+d \cos x)}{(c+d \cos x)^2}$
$f'(x) = \frac{(c+d \cos x)(b \cos x) - (a+b \sin x)(-d \sin x)}{(c+d \cos x)^2}$
$f'(x) = \frac{bc \cos x + bd \cos^2 x + ad \sin x + bd \sin^2 x}{(c+d \cos x)^2}$
Using the identity $\sin^2 x + \cos^2 x = 1$:
$f'(x) = \frac{bc \cos x + ad \sin x + bd(\cos^2 x + \sin^2 x)}{(c+d \cos x)^2}$
$f'(x) = \frac{bc \cos x + ad \sin x + bd}{(c+d \cos x)^2}$

Let $f(x) = \frac{\sin (x+a)}{\cos x}$.
Using the quotient rule,$\frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = \sin (x+a)$ and $v = \cos x$.
We know that $\frac{d}{dx} [\sin (x+a)] = \cos (x+a)$ and $\frac{d}{dx} [\cos x] = -\sin x$.
Substituting these into the quotient rule formula:
$f'(x) = \frac{\cos x \cdot \cos (x+a) - \sin (x+a) \cdot (-\sin x)}{\cos^2 x}$
$f'(x) = \frac{\cos x \cos (x+a) + \sin x \sin (x+a)}{\cos^2 x}$
Using the trigonometric identity $\cos A \cos B + \sin A \sin B = \cos (A - B)$,where $A = x+a$ and $B = x$:
$f'(x) = \frac{\cos ((x+a) - x)}{\cos^2 x}$
$f'(x) = \frac{\cos a}{\cos^2 x}$

Let $f(x) = x^{4}(5 \sin x - 3 \cos x)$.
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = x^{4} \frac{d}{dx}(5 \sin x - 3 \cos x) + (5 \sin x - 3 \cos x) \frac{d}{dx}(x^{4})$
$f'(x) = x^{4}(5 \cos x - 3(-\sin x)) + (5 \sin x - 3 \cos x)(4x^{3})$
$f'(x) = x^{4}(5 \cos x + 3 \sin x) + 4x^{3}(5 \sin x - 3 \cos x)$
$f'(x) = 5x^{4} \cos x + 3x^{4} \sin x + 20x^{3} \sin x - 12x^{3} \cos x$
Factoring out $x^{3}$:
$f'(x) = x^{3}(5x \cos x + 3x \sin x + 20 \sin x - 12 \cos x)$

(N/A) Let $f(x) = (x^{2}+1) \cos x$.
Using the product rule for differentiation,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (x^{2}+1) \frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(x^{2}+1)$
$f'(x) = (x^{2}+1)(-\sin x) + \cos x(2x)$
$f'(x) = -x^{2} \sin x - \sin x + 2x \cos x$
Therefore,the derivative is $2x \cos x - (x^{2}+1) \sin x$.

Let $f(x) = (ax^{2} + \sin x)(p + q \cos x)$.
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$.
Here,$u(x) = ax^{2} + \sin x$ and $v(x) = p + q \cos x$.
$f'(x) = (ax^{2} + \sin x) \frac{d}{dx}(p + q \cos x) + (p + q \cos x) \frac{d}{dx}(ax^{2} + \sin x)$.
$f'(x) = (ax^{2} + \sin x)(-q \sin x) + (p + q \cos x)(2ax + \cos x)$.
$f'(x) = -q \sin x(ax^{2} + \sin x) + (p + q \cos x)(2ax + \cos x)$.

Let $f(x) = \frac{4x + 5 \sin x}{3x + 7 \cos x}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$f'(x) = \frac{(3x + 7 \cos x) \frac{d}{dx}(4x + 5 \sin x) - (4x + 5 \sin x) \frac{d}{dx}(3x + 7 \cos x)}{(3x + 7 \cos x)^2}$
$f'(x) = \frac{(3x + 7 \cos x)(4 + 5 \cos x) - (4x + 5 \sin x)(3 - 7 \sin x)}{(3x + 7 \cos x)^2}$
Expanding the numerator:
$= \frac{(12x + 15x \cos x + 28 \cos x + 35 \cos^2 x) - (12x - 28x \sin x + 15 \sin x - 35 \sin^2 x)}{(3x + 7 \cos x)^2}$
$= \frac{12x + 15x \cos x + 28 \cos x + 35 \cos^2 x - 12x + 28x \sin x - 15 \sin x + 35 \sin^2 x}{(3x + 7 \cos x)^2}$
Since $\cos^2 x + \sin^2 x = 1$,we have $35(\cos^2 x + \sin^2 x) = 35$:
$= \frac{35 + 15x \cos x + 28 \cos x + 28x \sin x - 15 \sin x}{(3x + 7 \cos x)^2}$

Let $f(x) = \frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}$.
Using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^{2}}$,where $u = x^{2} \cos \left(\frac{\pi}{4}\right)$ and $v = \sin x$:
$f^{\prime}(x) = \cos \left(\frac{\pi}{4}\right) \left[ \frac{\sin x \cdot \frac{d}{dx}(x^{2}) - x^{2} \cdot \frac{d}{dx}(\sin x)}{\sin^{2} x} \right]$
$f^{\prime}(x) = \cos \left(\frac{\pi}{4}\right) \left[ \frac{\sin x(2x) - x^{2}(\cos x)}{\sin^{2} x} \right]$
$f^{\prime}(x) = \frac{x \cos \left(\frac{\pi}{4}\right) (2 \sin x - x \cos x)}{\sin^{2} x}$

Let $f(x) = \frac{x}{1+\tan x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = x$ and $v = 1 + \tan x$.
$\frac{du}{dx} = 1$ and $\frac{dv}{dx} = \sec^2 x$.
Substituting these into the quotient rule formula:
$f'(x) = \frac{(1 + \tan x)(1) - (x)(\sec^2 x)}{(1 + \tan x)^2}$.
Thus,$f'(x) = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}$.

Let $f(x) = \frac{x}{\sin^{n} x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = x$ and $v = \sin^{n} x$.
$\frac{du}{dx} = 1$ and $\frac{dv}{dx} = n \sin^{n-1} x \cos x$ (by chain rule).
Substituting these into the quotient rule formula:
$f'(x) = \frac{\sin^{n} x (1) - x (n \sin^{n-1} x \cos x)}{(\sin^{n} x)^2}$.
$f'(x) = \frac{\sin^{n} x - n x \sin^{n-1} x \cos x}{\sin^{2n} x}$.
Factoring out $\sin^{n-1} x$ from the numerator:
$f'(x) = \frac{\sin^{n-1} x (\sin x - n x \cos x)}{\sin^{2n} x}$.
Simplifying the expression:
$f'(x) = \frac{\sin x - n x \cos x}{\sin^{n+1} x}$.

(C) Given the identity $f(x+y) = f(x) + f(y) + xy^2 + x^2y$.
Setting $x=0$ and $y=0$,we get $f(0) = f(0) + f(0) + 0 + 0$,which implies $f(0) = 0$.
By the definition of the derivative,$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}$.
Using the given identity,$f(x+h) = f(x) + f(h) + xh^2 + x^2h$.
Substituting this into the derivative formula:
$f'(x) = \lim_{h \rightarrow 0} \frac{f(x) + f(h) + xh^2 + x^2h - f(x)}{h} = \lim_{h \rightarrow 0} \left( \frac{f(h)}{h} + xh + x^2 \right)$.
Since $\lim_{h \rightarrow 0} \frac{f(h)}{h} = 1$ (given),we have $f'(x) = 1 + 0 + x^2 = 1 + x^2$.
Therefore,$f'(3) = 1 + (3)^2 = 1 + 9 = 10$.

(A) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.
By definition of the derivative at $x=0$,we have $f'(0) = \lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$.
Since $f(x+y)=f(x)f(y)$,setting $x=0, y=0$ gives $f(0)=f(0)^2$. Since $f(x) \neq 0$,we must have $f(0)=1$.
Substituting $f(0)=1$ into the derivative definition,we get $f'(0) = \lim_{h \rightarrow 0} \frac{f(h)-1}{h}$.
Given $f'(0)=3$,it follows that $\lim_{h \rightarrow 0} \frac{f(h)-1}{h} = 3$.

(D) Given the condition $|f(x) - f(y)| \leq |(x - y)^2|$.
Dividing both sides by $|x - y|$ (for $x \neq y$),we get $\left| \frac{f(x) - f(y)}{x - y} \right| \leq |x - y|$.
Taking the limit as $x \to y$,we have $\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| \leq \lim_{x \to y} |x - y|$.
This implies $|f'(y)| \leq 0$.
Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which means $f'(y) = 0$ for all $y \in R$.
$A$ function whose derivative is zero everywhere is a constant function,so $f(x) = C$.
Given $f(0) = 1$,we find $C = 1$.
Thus,$f(x) = 1$ for all $x \in R$.
Since $1 > 0$,the correct statement is $f(x) > 0, \forall x \in R$.

(B) Given $f^{\prime}(a) = 2$ and $f(a) = 4$.
We need to evaluate $L = \lim_{x \rightarrow a} \frac{x f(a) - a f(x)}{x - a}$.
Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim_{x \rightarrow a} \frac{\frac{d}{dx}(x f(a) - a f(x))}{\frac{d}{dx}(x - a)}$
$L = \lim_{x \rightarrow a} \frac{f(a) - a f^{\prime}(x)}{1}$
Substituting $x = a$:
$L = f(a) - a f^{\prime}(a)$
$L = 4 - a(2) = 4 - 2a$.

(A) Given $f(x) = x^{3} + x - 5$.
First,we find the derivative $f^{\prime}(x) = 3x^{2} + 1$.
Since $f^{\prime}(x) > 0$ for all $x \in R$,the function $f(x)$ is strictly increasing and therefore invertible.
Given $f(g(x)) = x$,by the chain rule,we have $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$,which implies $g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.
To find $g^{\prime}(63)$,we need to find $x$ such that $f(x) = 63$.
$x^{3} + x - 5 = 63 \Rightarrow x^{3} + x - 68 = 0$.
By inspection,for $x = 4$,$4^{3} + 4 - 5 = 64 + 4 - 5 = 63$. Thus,$g(63) = 4$.
Now,$g^{\prime}(63) = \frac{1}{f^{\prime}(g(63))} = \frac{1}{f^{\prime}(4)}$.
Since $f^{\prime}(x) = 3x^{2} + 1$,we have $f^{\prime}(4) = 3(4)^{2} + 1 = 3(16) + 1 = 48 + 1 = 49$.
Therefore,$g^{\prime}(63) = \frac{1}{49}$.

(A) Given $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$.
Since the limit $\lim _{x \rightarrow 1} \frac{f(x)}{x-1}$ exists,we must have $f(1)=0$.
If $f(1)=0$,then the limit is equal to $f^{\prime}(1)$.
Substituting $x=1$ in the given equation: $f(1)+f^{\prime}(1)+f^{\prime \prime}(1) = 1^{5}+64 = 65$.
Since $f(1)=0$,we have $f^{\prime}(1)+f^{\prime \prime}(1) = 65$.
Differentiating the given equation: $f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x) = 5x^{4}$.
At $x=1$,$f^{\prime}(1)+f^{\prime \prime}(1)+f^{\prime \prime \prime}(1) = 5$.
Since $f^{\prime}(1)+f^{\prime \prime}(1) = 65$,we get $65+f^{\prime \prime \prime}(1) = 5$,so $f^{\prime \prime \prime}(1) = -60$.
Differentiating again: $f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+f^{(4)}(x) = 20x^{3}$.
At $x=1$,$f^{\prime \prime}(1)+f^{\prime \prime \prime}(1)+f^{(4)}(1) = 20$.
Differentiating again: $f^{\prime \prime \prime}(x)+f^{(4)}(x)+f^{(5)}(x) = 60x^{2}$.
At $x=1$,$f^{\prime \prime \prime}(1)+f^{(4)}(1)+f^{(5)}(1) = 60$.
Since $f(x)$ is a polynomial of degree $5$,let $f(x) = ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+g$.
Then $f^{(5)}(x) = 120a$. From $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=x^{5}+64$,the leading coefficient $a=1$.
Thus $f^{(5)}(x) = 120$.
Using $f^{\prime \prime \prime}(1)+f^{(4)}(1)+120 = 60$,we get $f^{\prime \prime \prime}(1)+f^{(4)}(1) = -60$.
From $f^{\prime \prime}(1)+f^{\prime \prime \prime}(1)+f^{(4)}(1) = 20$,we get $f^{\prime \prime}(1) + (-60) = 20$,so $f^{\prime \prime}(1) = 80$.
Finally,$f^{\prime}(1) = 65 - f^{\prime \prime}(1) = 65 - 80 = -15$.

(B) Given that $P(x)$ is a non-zero polynomial satisfying $P(1+x)=P(1-x)$ for all real $x$ and $P(1)=0$.
Differentiating both sides with respect to $x$,we get:
$P'(1+x) = -P'(1-x)$.
Setting $x=0$,we obtain:
$P'(1) = -P'(1) \implies 2P'(1) = 0 \implies P'(1) = 0$.
Since $P(1)=0$ and $P'(1)=0$,by the Factor Theorem,$(x-1)^2$ must be a factor of $P(x)$.
To check if $m$ can be larger,consider $P(x) = (x-1)^2$. This satisfies $P(1+x) = (1+x-1)^2 = x^2$ and $P(1-x) = (1-x-1)^2 = (-x)^2 = x^2$. Thus,$P(1+x)=P(1-x)$ holds.
Since $P(x) = (x-1)^2$ is a valid polynomial,the largest integer $m$ such that $(x-1)^m$ divides $P(x)$ for all such $P(x)$ is $2$.

(D) Given the equation: $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$ (Equation $1$)
Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}\right) + 2 f(x) = x - 10$ (Equation $2$)
Multiply Equation $1$ by $3$ and Equation $2$ by $2$:
$9 f(x) + 6 f\left(\frac{1}{x}\right) = \frac{3}{x} - 30$
$4 f(x) + 6 f\left(\frac{1}{x}\right) = 2x - 20$
Subtracting the second from the first:
$5 f(x) = \frac{3}{x} - 2x - 10$
$f(x) = \frac{3}{5x} - \frac{2x}{5} - 2$
Now,find $f(3)$:
$f(3) = \frac{3}{5(3)} - \frac{2(3)}{5} - 2 = \frac{1}{5} - \frac{6}{5} - 2 = -1 - 2 = -3$
Find $f^{\prime}(x)$:
$f^{\prime}(x) = -\frac{3}{5x^2} - \frac{2}{5}$
Calculate $f^{\prime}\left(\frac{1}{4}\right)$:
$f^{\prime}\left(\frac{1}{4}\right) = -\frac{3}{5(1/16)} - \frac{2}{5} = -\frac{48}{5} - \frac{2}{5} = -\frac{50}{5} = -10$
Finally,calculate $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$:
$|-3 + (-10)| = |-13| = 13$

(B) Given $f(x) = \sum_{k=1}^{10} kx^k = x + 2x^2 + 3x^3 + \dots + 10x^{10}$.
We know that $f'(x) = \sum_{k=1}^{10} k^2 x^{k-1}$.
Thus,$f(2) = \sum_{k=1}^{10} k(2^k)$ and $f'(2) = \sum_{k=1}^{10} k^2(2^{k-1})$.
Consider the identity $g(x) = \sum_{k=1}^{10} x^k = \frac{x(1-x^{10})}{1-x}$.
Differentiating $g(x)$,we get $g'(x) = \sum_{k=1}^{10} kx^{k-1} = f(x)/x$.
Also,$f(x) = x g'(x)$.
Then $f'(x) = g'(x) + x g''(x)$.
So,$2f(2) + f'(2) = 2(2g'(2)) + (g'(2) + 2g''(2)) = 5g'(2) + 2g''(2)$.
Alternatively,using the property of the sum $\sum k x^k$,we have $f(x) = \frac{x(1-(n+1)x^n + nx^{n+1})}{(1-x)^2}$ for $n=10$.
Evaluating at $x=2$,$f(2) = \frac{2(1-11(2^{10}) + 10(2^{11}))}{(1-2)^2} = 2(1 - 11(1024) + 20480) = 2(1 - 11264 + 20480) = 2(9217) = 18434$.
Calculating $f'(2)$ and substituting,we find $2f(2) + f'(2) = 119(2^{10}) + 1$.
Thus,$n = 10$.

(B) Given $f(x)-f(y) \geq \ln x - \ln y + x - y$.
Rearranging the terms,we get $f(x) - x - \ln x \geq f(y) - y - \ln y$.
Let $g(x) = f(x) - x - \ln x$. Then $g(x) \geq g(y)$ for all $x, y \in (0, \infty)$.
This implies that $g(x)$ is a constant function,say $C$.
So,$f(x) - x - \ln x = C$,which means $f(x) = x + \ln x + C$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = 1 + \frac{1}{x}$.
Now,we need to calculate $\sum_{n=1}^{20} f^{\prime}\left(\frac{1}{n^2}\right)$.
$f^{\prime}\left(\frac{1}{n^2}\right) = 1 + \frac{1}{1/n^2} = 1 + n^2$.
Therefore,$\sum_{n=1}^{20} (1 + n^2) = \sum_{n=1}^{20} 1 + \sum_{n=1}^{20} n^2$.
$= 20 + \frac{20(20+1)(2 \times 20 + 1)}{6} = 20 + \frac{20 \times 21 \times 41}{6}$.
$= 20 + 10 \times 7 \times 41 = 20 + 2870 = 2890$.

(D) Given $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$.
Step $1$: Find the derivatives of $f(x)$.
$f'(x) = 3x^2 + 2x f'(1) + f''(2)$
$f''(x) = 6x + 2f'(1)$
$f'''(x) = 6$
Step $2$: Evaluate the constants.
For $f'''(3)$,since $f'''(x) = 6$,we have $f'''(3) = 6$.
For $f''(2)$,substitute $x=2$ into $f''(x) = 6x + 2f'(1)$:
$f''(2) = 6(2) + 2f'(1) = 12 + 2f'(1)$.
For $f'(1)$,substitute $x=1$ into $f'(x) = 3x^2 + 2x f'(1) + f''(2)$:
$f'(1) = 3(1)^2 + 2(1)f'(1) + f''(2) = 3 + 2f'(1) + f''(2)$.
Step $3$: Solve the system of equations.
From $f''(2) = 12 + 2f'(1)$,substitute into $f'(1) = 3 + 2f'(1) + f''(2)$:
$f'(1) = 3 + 2f'(1) + 12 + 2f'(1)$
$f'(1) = 15 + 4f'(1)$
$-3f'(1) = 15 \implies f'(1) = -5$.
Now find $f''(2)$:
$f''(2) = 12 + 2(-5) = 12 - 10 = 2$.
Step $4$: Calculate $f'(10)$.
$f'(x) = 3x^2 + 2x f'(1) + f''(2) = 3x^2 + 2x(-5) + 2 = 3x^2 - 10x + 2$.
$f'(10) = 3(10)^2 - 10(10) + 2 = 300 - 100 + 2 = 202$.

(C) Given $f(x) = \frac{(2^x + 2^{-x}) \tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$.
First,calculate $f(0)$:
$f(0) = \frac{(2^0 + 2^0) \tan(0) \sqrt{\tan^{-1}(0-0+1)}}{(0+0+1)^3} = \frac{2 \times 0 \times \sqrt{\pi/4}}{1} = 0$.
Now,use the definition of the derivative at $x=0$:
$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$.
$f'(0) = \lim_{h \to 0} \left[ \frac{2^h + 2^{-h}}{(7h^2 + 3h + 1)^3} \cdot \frac{\tan h}{h} \cdot \sqrt{\tan^{-1}(h^2 - h + 1)} \right]$.
As $h \to 0$:
$\frac{2^h + 2^{-h}}{(7h^2 + 3h + 1)^3} \to \frac{1+1}{1} = 2$.
$\frac{\tan h}{h} \to 1$.
$\sqrt{\tan^{-1}(h^2 - h + 1)} \to \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}}$.
Therefore,$f'(0) = 2 \times 1 \times \sqrt{\frac{\pi}{4}} = 2 \times \frac{\sqrt{\pi}}{2} = \sqrt{\pi}$.

(B) We are given the limit $\alpha = \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) dt}{e^{x^2}-1}$.
We can rewrite the expression as $\alpha = \lim _{x \rightarrow 0} \frac{\int_0^x f(t) dt}{x} \cdot \frac{x^2}{e^{x^2}-1}$.
Since $\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2} = 1$,we have $\lim _{x \rightarrow 0} \frac{x^2}{e^{x^2}-1} = 1$.
Now,applying $L$'$H$ôpital's rule to the first part $\lim _{x \rightarrow 0} \frac{\int_0^x f(t) dt}{x}$ (which is of the form $\frac{0}{0}$):
$\lim _{x \rightarrow 0} \frac{\frac{d}{dx} \int_0^x f(t) dt}{\frac{d}{dx} x} = \lim _{x \rightarrow 0} \frac{f(x)}{1} = f(0)$.
Given $f(0) = \frac{1}{2}$,we get $\alpha = \frac{1}{2} \times 1 = \frac{1}{2}$.
Finally,we calculate $8 \alpha^2 = 8 \times \left(\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2$.

(D) Given $g(x) = h(e^x) \cdot e^{h(x)}$.
Applying the product rule for differentiation,we get:
$g^{\prime}(x) = h(e^x) \cdot \frac{d}{dx}(e^{h(x)}) + e^{h(x)} \cdot \frac{d}{dx}(h(e^x))$
$g^{\prime}(x) = h(e^x) \cdot e^{h(x)} \cdot h^{\prime}(x) + e^{h(x)} \cdot h^{\prime}(e^x) \cdot e^x$
Now,substitute $x = 0$ into the expression:
$g^{\prime}(0) = h(e^0) \cdot e^{h(0)} \cdot h^{\prime}(0) + e^{h(0)} \cdot h^{\prime}(e^0) \cdot e^0$
$g^{\prime}(0) = h(1) \cdot e^{h(0)} \cdot h^{\prime}(0) + e^{h(0)} \cdot h^{\prime}(1) \cdot 1$
Given $h(0)=0$,$h(1)=1$,and $h^{\prime}(0)=h^{\prime}(1)=2$:
$g^{\prime}(0) = (1) \cdot e^0 \cdot (2) + e^0 \cdot (2) \cdot 1$
$g^{\prime}(0) = 1 \cdot 1 \cdot 2 + 1 \cdot 2 \cdot 1$
$g^{\prime}(0) = 2 + 2 = 4$.

(A) For $PROPERTY \ 1$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}}$:
$(2) \ f(x)=x^{2/3}, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h^{2/3}-0}{|h|^{1/2}} = \lim_{h \rightarrow 0} \frac{|h|^{2/3}}{|h|^{1/2}} = \lim_{h \rightarrow 0} |h|^{1/6} = 0$. This exists and is finite. So,$(2)$ is correct.
$(4) \ f(x)=|x|, f(0)=0$. $\lim_{h \rightarrow 0} \frac{|h|-0}{|h|^{1/2}} = \lim_{h \rightarrow 0} |h|^{1/2} = 0$. This exists and is finite. So,$(4)$ is correct.
For $PROPERTY \ 2$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$:
$(1) \ f(x)=x|x|, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h|h|}{h^2} = \lim_{h \rightarrow 0} \frac{|h|}{h}$. The $RHL = 1$ and $LHL = -1$. The limit does not exist. So,$(1)$ is incorrect.
$(3) \ f(x)=\sin x, f(0)=0$. $\lim_{h \rightarrow 0} \frac{\sin h}{h^2} = \lim_{h \rightarrow 0} \frac{\sin h}{h} \cdot \frac{1}{h} = 1 \cdot \infty = \infty$. The limit does not exist. So,$(3)$ is incorrect.
Thus,only $(2)$ and $(4)$ are correct.

(A) Given $y = \frac{x^{2/3} - x^{-1/3}}{x^{2/3} + x^{-1/3}}$.
Multiply numerator and denominator by $x^{1/3}$:
$y = \frac{x^{2/3} \cdot x^{1/3} - x^{-1/3} \cdot x^{1/3}}{x^{2/3} \cdot x^{1/3} + x^{-1/3} \cdot x^{1/3}} = \frac{x - 1}{x + 1}$.
Now,differentiate with respect to $x$ using the quotient rule:
$y_1 = \frac{d}{dx} \left( \frac{x-1}{x+1} \right) = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2}$.
$y_1 = \frac{x + 1 - x + 1}{(x+1)^2} = \frac{2}{(x+1)^2}$.
Therefore,$(x+1)^2 y_1 = 2$.

(C) Let $y = f(x) = (1-x)(2-x)(3-x) \dots (n-x)$.
Using the product rule for differentiation,if $y = u_1 u_2 u_3 \dots u_n$,then $\frac{dy}{dx} = u_1' (u_2 u_3 \dots u_n) + u_1 (u_2 u_3 \dots u_n)'$.
At $x=1$,the term $(1-x)$ becomes $0$.
Therefore,in the derivative,all terms containing $(1-x)$ will vanish except for the term where $(1-x)$ is differentiated.
Let $g(x) = (2-x)(3-x) \dots (n-x)$. Then $y = (1-x)g(x)$.
$\frac{dy}{dx} = (-1)g(x) + (1-x)g'(x)$.
At $x=1$,$\frac{dy}{dx} = (-1)g(1) + 0 = -g(1)$.
$g(1) = (2-1)(3-1)(4-1) \dots (n-1) = (1)(2)(3) \dots (n-1) = (n-1)!$.
Thus,$\frac{dy}{dx} \text{ at } x=1 = -(n-1)!$.

(B) Given $y = \log_{e} x^3 + 3 \sin^{-1} x + k x^2$.
Using the property of logarithms,$y = 3 \log_{e} x + 3 \sin^{-1} x + k x^2$.
Differentiating with respect to $x$,we get $y' = \frac{3}{x} + \frac{3}{\sqrt{1 - x^2}} + 2kx$.
Given $y'(\frac{1}{2}) = 2 \sqrt{3}$,substitute $x = \frac{1}{2}$ into the derivative:
$y'(\frac{1}{2}) = \frac{3}{1/2} + \frac{3}{\sqrt{1 - (1/2)^2}} + 2k(\frac{1}{2}) = 2 \sqrt{3}$.
$6 + \frac{3}{\sqrt{3/4}} + k = 2 \sqrt{3}$.
$6 + \frac{3}{\sqrt{3}/2} + k = 2 \sqrt{3}$.
$6 + \frac{6}{\sqrt{3}} + k = 2 \sqrt{3}$.
$6 + 2 \sqrt{3} + k = 2 \sqrt{3}$.
$k = -6$.

(D) Let $g(x) = f(f(f(x))) + (f(x))^2$.
We need to find $g^{\prime}(1)$.
Using the chain rule,the derivative of $f(f(f(x)))$ is $f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)$.
At $x=1$,this is $f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1)$.
Given $f(1)=1$ and $f^{\prime}(1)=3$,we substitute these values:
$f^{\prime}(f(f(1))) = f^{\prime}(f(1)) = f^{\prime}(1) = 3$.
$f^{\prime}(f(1)) = f^{\prime}(1) = 3$.
$f^{\prime}(1) = 3$.
So,the derivative of $f(f(f(x)))$ at $x=1$ is $3 \cdot 3 \cdot 3 = 27$.
The derivative of $(f(x))^2$ is $2f(x) \cdot f^{\prime}(x)$.
At $x=1$,this is $2f(1) \cdot f^{\prime}(1) = 2(1)(3) = 6$.
Therefore,$g^{\prime}(1) = 27 + 6 = 33$.

(C) Given $y = \tan^{-1} \left[ \frac{4 \sin 2x}{\cos 2x - 6 \sin^2 x} \right]$.
Using double angle formulas,$\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$.
Substituting these into the expression:
$y = \tan^{-1} \left[ \frac{4(2 \sin x \cos x)}{(\cos^2 x - \sin^2 x) - 6 \sin^2 x} \right] = \tan^{-1} \left[ \frac{8 \sin x \cos x}{\cos^2 x - 7 \sin^2 x} \right]$.
Divide numerator and denominator by $\cos^2 x$:
$y = \tan^{-1} \left[ \frac{8 \tan x}{1 - 7 \tan^2 x} \right]$.
Let $f(x) = \frac{8 \tan x}{1 - 7 \tan^2 x}$. Then $y = \tan^{-1}(f(x))$.
$\frac{dy}{dx} = \frac{1}{1 + (f(x))^2} \cdot f'(x)$.
At $x = 0$,$f(0) = 0$,so $\frac{dy}{dx} = f'(0)$.
Using the quotient rule for $f(x) = \frac{u}{v}$,$f'(x) = \frac{u'v - uv'}{v^2}$.
$u = 8 \tan x \implies u' = 8 \sec^2 x$.
$v = 1 - 7 \tan^2 x \implies v' = -14 \tan x \sec^2 x$.
At $x = 0$,$u(0) = 0, u'(0) = 8, v(0) = 1, v'(0) = 0$.
$f'(0) = \frac{8(1) - 0(0)}{1^2} = 8$.
Thus,$\frac{dy}{dx}$ at $x = 0$ is $8$.

(C) Given $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$.
Simplify the terms:
$f(x) = \frac{\sin^2 x}{1+\frac{\cos x}{\sin x}} + \frac{\cos^2 x}{1+\frac{\sin x}{\cos x}}$
$f(x) = \frac{\sin^3 x}{\sin x + \cos x} + \frac{\cos^3 x}{\cos x + \sin x}$
$f(x) = \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x}$
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$:
$f(x) = \frac{(\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)}{\sin x + \cos x}$
$f(x) = \sin^2 x + \cos^2 x - \sin x \cos x$
$f(x) = 1 - \frac{1}{2}(2 \sin x \cos x) = 1 - \frac{1}{2} \sin(2x)$.
Now,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = 0 - \frac{1}{2} \cos(2x) \cdot 2 = -\cos(2x)$.
Calculate $f^{\prime}\left(\frac{\pi}{6}\right)$:
$f^{\prime}\left(\frac{\pi}{6}\right) = -\cos\left(2 \cdot \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$.

(B) Given $f(x) = \sqrt{1 + \cos^2(x^2)}$.
Using the chain rule,let $u = 1 + \cos^2(x^2)$,so $f(x) = \sqrt{u}$.
$f^{\prime}(x) = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{1 + \cos^2(x^2)}} \cdot \frac{d}{dx}(1 + \cos^2(x^2))$.
Now,$\frac{d}{dx}(\cos^2(x^2)) = 2\cos(x^2) \cdot (-\sin(x^2)) \cdot 2x = -2x \sin(2x^2)$.
Thus,$f^{\prime}(x) = \frac{-2x \sin(2x^2)}{2\sqrt{1 + \cos^2(x^2)}} = \frac{-x \sin(2x^2)}{\sqrt{1 + \cos^2(x^2)}}$.
At $x = \frac{\sqrt{\pi}}{2}$,$x^2 = \frac{\pi}{4}$.
$f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right) = \frac{-\frac{\sqrt{\pi}}{2} \sin(2 \cdot \frac{\pi}{4})}{\sqrt{1 + \cos^2(\frac{\pi}{4})}} = \frac{-\frac{\sqrt{\pi}}{2} \sin(\frac{\pi}{2})}{\sqrt{1 + (\frac{1}{\sqrt{2}})^2}} = \frac{-\frac{\sqrt{\pi}}{2} \cdot 1}{\sqrt{1 + \frac{1}{2}}} = \frac{-\frac{\sqrt{\pi}}{2}}{\sqrt{\frac{3}{2}}} = -\frac{\sqrt{\pi}}{2} \cdot \sqrt{\frac{2}{3}} = -\frac{\sqrt{\pi}}{\sqrt{2} \cdot \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{\pi}}{\sqrt{6}} = -\sqrt{\frac{\pi}{6}}$.

(D) Given $y=a \sin x+b \cos x$.
Differentiating with respect to $x$,we get $\frac{d y}{d x}=a \cos x-b \sin x$.
Now,consider the expression $y^2+\left(\frac{d y}{d x}\right)^2$:
$y^2+\left(\frac{d y}{d x}\right)^2 = (a \sin x+b \cos x)^2+(a \cos x-b \sin x)^2$
$= (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x)$
$= a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x)$
$= a^2(1) + b^2(1) = a^2+b^2$.
Since $a$ and $b$ are constants,$a^2+b^2$ is also a constant.
Therefore,$y^2+\left(\frac{d y}{d x}\right)^2$ is a constant.

(A) Given $f(x) = \cos^{-1} x$ and $g(x) = e^x$.
We define $h(x) = g(f(x)) = e^{\cos^{-1} x}$.
Now,we differentiate $h(x)$ with respect to $x$ using the chain rule:
$h'(x) = \frac{d}{dx}(e^{\cos^{-1} x}) = e^{\cos^{-1} x} \cdot \frac{d}{dx}(\cos^{-1} x)$.
Since $\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}}$,we have:
$h'(x) = e^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)$.
Now,we calculate the ratio $\frac{h'(x)}{h(x)}$:
$\frac{h'(x)}{h(x)} = \frac{e^{\cos^{-1} x} \cdot \left( \frac{-1}{\sqrt{1-x^2}} \right)}{e^{\cos^{-1} x}} = \frac{-1}{\sqrt{1-x^2}}$.
Thus,the correct option is $A$.

(A) Given $g(x) = [f(2f(x) + 2)]^2$.
Applying the chain rule to differentiate $g(x)$ with respect to $x$:
$g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot \frac{d}{dx}[2f(x) + 2]$
$g'(x) = 2[f(2f(x) + 2)] \cdot f'(2f(x) + 2) \cdot 2f'(x)$
$g'(x) = 4 \cdot f(2f(x) + 2) \cdot f'(2f(x) + 2) \cdot f'(x)$
Now,substitute $x = 0$:
$g'(0) = 4 \cdot f(2f(0) + 2) \cdot f'(2f(0) + 2) \cdot f'(0)$
Given $f(0) = -1$ and $f'(0) = 1$:
$g'(0) = 4 \cdot f(2(-1) + 2) \cdot f'(2(-1) + 2) \cdot (1)$
$g'(0) = 4 \cdot f(0) \cdot f'(0) \cdot 1$
$g'(0) = 4 \cdot (-1) \cdot 1 \cdot 1 = -4$.

(D) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x)f^{\prime}(x)$.
At $x = 1$,we have $f(1) = 1$ and $f^{\prime}(1) = 3$.
Substituting these values:
$f(f(1)) = f(1) = 1$.
$f(f(f(1))) = f(1) = 1$.
Therefore,$\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1)f^{\prime}(1)$.
$= f^{\prime}(1) \cdot f^{\prime}(1) \cdot f^{\prime}(1) + 2(1)(3)$.
$= 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

(B) Given $h(x) = \sqrt{4f(x) + 3g(x)}$.
At $x = 1$,$h(1) = \sqrt{4f(1) + 3g(1)} = \sqrt{4(4) + 3(3)} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Now,differentiate $h(x)$ with respect to $x$ using the chain rule:
$h'(x) = \frac{1}{2\sqrt{4f(x) + 3g(x)}} \cdot \frac{d}{dx}(4f(x) + 3g(x))$
$h'(x) = \frac{4f'(x) + 3g'(x)}{2\sqrt{4f(x) + 3g(x)}}$
Substitute $x = 1$ into the derivative:
$h'(1) = \frac{4f'(1) + 3g'(1)}{2\sqrt{4f(1) + 3g(1)}}$
Given $f'(1) = 4$ and $g'(1) = 3$:
$h'(1) = \frac{4(4) + 3(3)}{2(5)} = \frac{16 + 9}{10} = \frac{25}{10} = \frac{5}{2}$.

(D) $y = \log \tan \left(\frac{x}{2}\right) + \sin^{-1}(\cos x)$
Since $\cos x = \sin \left(\frac{\pi}{2} - x\right)$,we have:
$y = \log \tan \left(\frac{x}{2}\right) + \sin^{-1} \left[ \sin \left(\frac{\pi}{2} - x\right) \right] = \log \tan \left(\frac{x}{2}\right) + \frac{\pi}{2} - x$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \sec^2 \left(\frac{x}{2}\right) \cdot \frac{1}{2} + 0 - 1$
$\frac{dy}{dx} = \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} \cdot \frac{1}{\cos^2 \left(\frac{x}{2}\right)} \cdot \frac{1}{2} - 1$
$\frac{dy}{dx} = \frac{1}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} - 1$
Using the identity $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$:
$\frac{dy}{dx} = \frac{1}{\sin x} - 1 = \operatorname{cosec} x - 1$

(A) Given $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$.
Using the property $\operatorname{cosec}^{-1}(u) = \sin^{-1}(1/u)$,we get:
$f(x)=\sin ^{-1}\left[\frac{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}{10}\right]$.
We can write this as $f(x)=\sin ^{-1}\left[\frac{6}{10} \sin \left(2^x\right)-\frac{8}{10} \cos \left(2^x\right)\right]$.
Let $\cos \alpha = \frac{6}{10}$ and $\sin \alpha = \frac{8}{10}$.
Then $f(x)=\sin ^{-1}\left[\sin \left(2^x\right) \cos \alpha - \cos \left(2^x\right) \sin \alpha\right]$.
Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have:
$f(x)=\sin ^{-1}\left[\sin \left(2^x-\alpha\right)\right] = 2^x - \alpha$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(2^x - \alpha) = 2^x \log 2 - 0 = 2^x \log 2$.

(A) Given $y = \log \sqrt{\tan x} = \frac{1}{2} \log(\tan x)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \times \frac{1}{\tan x} \times \sec^2 x$.
Using the identities $\tan x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we get:
$\frac{dy}{dx} = \frac{1}{2} \times \frac{\cos x}{\sin x} \times \frac{1}{\cos^2 x} = \frac{1}{2 \sin x \cos x} = \frac{1}{\sin(2x)}$.
Now,substitute $x = \frac{\pi}{4}$:
$\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = \frac{1}{\sin(2 \times \frac{\pi}{4})} = \frac{1}{\sin(\frac{\pi}{2})} = \frac{1}{1} = 1$.

(A) Given that $g(a)=b$,$g^{\prime}(a)=2$,and $f(g(x))=x$ (since $f \circ g = I$).
By differentiating both sides of $f(g(x))=x$ with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
Substituting $x=a$ into the equation,we have:
$f^{\prime}(g(a)) \cdot g^{\prime}(a) = 1$.
Since $g(a)=b$ and $g^{\prime}(a)=2$,the equation becomes:
$f^{\prime}(b) \cdot 2 = 1$.
Therefore,$f^{\prime}(b) = \frac{1}{2}$.