Derivative at a point, Standard differentiation Questions in English

(B) Let $f(x) = |x - 1| + |x - 3|$.
For $1 < x < 3$,we have $x - 1 > 0$ and $x - 3 < 0$.
Therefore,$f(x) = (x - 1) - (x - 3) = x - 1 - x + 3 = 2$.
Since $f(x) = 2$ is a constant function in the interval $(1, 3)$,its derivative $f'(x) = 0$ for all $x \in (1, 3)$.
At the point $x = 2$,which lies within the interval $(1, 3)$,the differential coefficient is $f'(2) = 0$.

(C) Given $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}} = \sqrt{a} \cdot x^{1/2} + a^2 \cdot a^{-1/2} \cdot x^{-1/2} = \sqrt{a} \cdot x^{1/2} + a^{3/2} \cdot x^{-1/2}$.
Differentiating with respect to $x$:
$f'(x) = \sqrt{a} \cdot \frac{1}{2} x^{-1/2} + a^{3/2} \cdot \left( -\frac{1}{2} x^{-3/2} \right)$.
$f'(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^{3/2}}{2x^{3/2}}$.
Now,evaluating at $x = a$:
$f'(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^{3/2}}{2a^{3/2}}$.
$f'(a) = \frac{1}{2} - \frac{1}{2} = 0$.

(C) Let $f(x) = x^6 + 6^x$.
To find the derivative with respect to $x$,we use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the exponential rule $\frac{d}{dx}(a^x) = a^x \ln a$.
Applying these rules:
$\frac{d}{dx}(x^6 + 6^x) = \frac{d}{dx}(x^6) + \frac{d}{dx}(6^x)$
$= 6x^{6-1} + 6^x \ln 6$
$= 6x^5 + 6^x \ln 6$.
Thus,the correct option is $C$.

(B) Given $f(x) = 3|2 + x|$.
For $x < -2$,$|2 + x| = -(2 + x) = -2 - x$.
Thus,for $x < -2$,$f(x) = 3(-2 - x) = -6 - 3x$.
The derivative $f'(x)$ for $x < -2$ is $\frac{d}{dx}(-6 - 3x) = -3$.
Since $x_0 = -3$ lies in the interval $x < -2$,the derivative at $x_0 = -3$ is $f'(-3) = -3$.

(C) Given $y = \cot^{-1}(x^2)$.
Using the chain rule for differentiation,we know that $\frac{d}{dx}(\cot^{-1}(u)) = \frac{-1}{1 + u^2} \cdot \frac{du}{dx}$.
Here,$u = x^2$,so $\frac{du}{dx} = 2x$.
Substituting these into the formula:
$\frac{dy}{dx} = \frac{-1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2)$
$\frac{dy}{dx} = \frac{-1}{1 + x^4} \cdot (2x)$
$\frac{dy}{dx} = \frac{-2x}{1 + x^4}$.

(D) Given the function $y = \log \tan \sqrt{x}$.
Applying the chain rule for differentiation:
$\frac{dy}{dx} = \frac{d}{dx}(\log \tan \sqrt{x})$
$= \frac{1}{\tan \sqrt{x}} \cdot \frac{d}{dx}(\tan \sqrt{x})$
$= \frac{1}{\tan \sqrt{x}} \cdot \sec^2 \sqrt{x} \cdot \frac{d}{dx}(\sqrt{x})$
$= \frac{1}{\tan \sqrt{x}} \cdot \sec^2 \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$
$= \frac{\sec^2 \sqrt{x}}{2\sqrt{x} \tan \sqrt{x}}$.

(D) Given $y = \frac{e^x + e^{-x}}{e^x - e^{-x}}$.
We know that $\cosh x = \frac{e^x + e^{-x}}{2}$ and $\sinh x = \frac{e^x - e^{-x}}{2}$.
Therefore,$y = \frac{\cosh x}{\sinh x} = \coth x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\coth x)$.
Since $\frac{d}{dx}(\coth x) = -\text{cosech}^2 x$,we have $\frac{dy}{dx} = -\text{cosech}^2 x$.

(A) Let the polynomial function be $f(x) = ax^2 + bx + c$.
Then,the derivative is $f'(x) = 2ax + b$.
Given $f(1) = f(-1)$,we have $a(1)^2 + b(1) + c = a(-1)^2 + b(-1) + c$.
This simplifies to $a + b + c = a - b + c$,which implies $2b = 0$,so $b = 0$.
Thus,$f'(x) = 2ax$.
Now,we evaluate the derivatives at $a_1, a_2, a_3$: $f'(a_1) = 2aa_1$,$f'(a_2) = 2aa_2$,and $f'(a_3) = 2aa_3$.
Since $a_1, a_2, a_3$ are in $A.P.$,there exists a common difference $d$ such that $a_2 = a_1 + d$ and $a_3 = a_1 + 2d$.
Then $f'(a_2) - f'(a_1) = 2aa_2 - 2aa_1 = 2a(a_2 - a_1) = 2ad$ and $f'(a_3) - f'(a_2) = 2aa_3 - 2aa_2 = 2a(a_3 - a_2) = 2ad$.
Since the difference between consecutive terms is constant $(2ad)$,the sequence $f'(a_1), f'(a_2), f'(a_3)$ is in $A.P.$

(D) Given $r = [2\phi + \cos^2(2\phi + \pi/4)]^{1/2}$.
Applying the chain rule,we differentiate with respect to $\phi$:
$\frac{dr}{d\phi} = \frac{1}{2}[2\phi + \cos^2(2\phi + \pi/4)]^{-1/2} \cdot \frac{d}{d\phi}[2\phi + \cos^2(2\phi + \pi/4)]$
$\frac{dr}{d\phi} = \frac{1}{2}[2\phi + \cos^2(2\phi + \pi/4)]^{-1/2} \cdot [2 + 2\cos(2\phi + \pi/4) \cdot (-\sin(2\phi + \pi/4)) \cdot 2]$
$\frac{dr}{d\phi} = \frac{1}{2}[2\phi + \cos^2(2\phi + \pi/4)]^{-1/2} \cdot [2 - 2\sin(4\phi + \pi/2)]$
At $\phi = \pi/4$,the argument $2\phi + \pi/4 = \pi/2 + \pi/4 = 3\pi/4$.
$\cos^2(3\pi/4) = (-1/\sqrt{2})^2 = 1/2$.
$\sin(4(\pi/4) + \pi/2) = \sin(\pi + \pi/2) = \sin(3\pi/2) = -1$.
Substituting these values:
$\frac{dr}{d\phi} = \frac{1}{2}[2(\pi/4) + 1/2]^{-1/2} \cdot [2 - 2(-1)]$
$\frac{dr}{d\phi} = \frac{1}{2}[\pi/2 + 1/2]^{-1/2} \cdot [4]$
$\frac{dr}{d\phi} = 2 \cdot [(\pi + 1)/2]^{-1/2} = 2 \cdot [2/(\pi + 1)]^{1/2}$.

(B) Let $y = f(x) = (1 - x)(2 - x)(3 - x)...(n - x)$.
Using the product rule for differentiation,the derivative is given by:
$\frac{dy}{dx} = \frac{d}{dx}[(1 - x)] \cdot (2 - x)(3 - x)...(n - x) + (1 - x) \cdot \frac{d}{dx}[(2 - x)(3 - x)...(n - x)]$.
Since $\frac{d}{dx}(1 - x) = -1$,we have:
$\frac{dy}{dx} = -1 \cdot (2 - x)(3 - x)...(n - x) + (1 - x) \cdot \frac{d}{dx}[(2 - x)(3 - x)...(n - x)]$.
Now,evaluate at $x = 1$:
$\left. \frac{dy}{dx} \right|_{x=1} = -1 \cdot (2 - 1)(3 - 1)...(n - 1) + (1 - 1) \cdot [\dots]$.
$\left. \frac{dy}{dx} \right|_{x=1} = -1 \cdot (1)(2)(3)...(n - 1) + 0$.
$\left. \frac{dy}{dx} \right|_{x=1} = -1 \cdot (n - 1)!$.
This can be written as $(-1)^1 (n - 1)!$. However,looking at the product $(1-x)(2-x)...(n-x)$,there are $n$ terms. The derivative at $x=1$ is $-(2-1)(3-1)...(n-1) = -(1)(2)...(n-1) = -(n-1)!$. Thus,the correct option is $B$.

(C) Given $y = \frac{1}{4}u^4$ and $u = \frac{2}{3}x^3 + 5$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
First,differentiate $y$ with respect to $u$: $\frac{dy}{du} = \frac{1}{4} \cdot 4u^3 = u^3$.
Next,differentiate $u$ with respect to $x$: $\frac{du}{dx} = \frac{2}{3} \cdot 3x^2 = 2x^2$.
Now,substitute these into the chain rule formula: $\frac{dy}{dx} = u^3 \cdot 2x^2$.
Substitute $u = \frac{2x^3 + 15}{3}$ into the expression: $\frac{dy}{dx} = \left( \frac{2x^3 + 15}{3} \right)^3 \cdot 2x^2$.
$\frac{dy}{dx} = \frac{(2x^3 + 15)^3}{27} \cdot 2x^2 = \frac{2}{27}x^2(2x^3 + 15)^3$.

(A) Given $y = f\left( \frac{5x + 1}{10x^2 - 3} \right)$ and $f'(x) = \cos x$.
Let $u = \frac{5x + 1}{10x^2 - 3}$.
Then $y = f(u)$.
By the chain rule,$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
Since $y = f(u)$,we have $\frac{dy}{du} = f'(u) = \cos(u)$.
Substituting $u$ back,we get $\frac{dy}{du} = \cos \left( \frac{5x + 1}{10x^2 - 3} \right)$.
Therefore,$\frac{dy}{dx} = \cos \left( \frac{5x + 1}{10x^2 - 3} \right) \cdot \frac{d}{dx} \left( \frac{5x + 1}{10x^2 - 3} \right)$.

(A) Given that $f \circ g = I$,where $I$ is the identity function,we have $(f \circ g)(x) = x$ for all $x$ in the domain.
Applying the chain rule to differentiate both sides with respect to $x$,we get $f'(g(x)) \cdot g'(x) = 1$.
Substituting $x = a$ into the equation,we obtain $f'(g(a)) \cdot g'(a) = 1$.
Given $g(a) = b$ and $g'(a) = 2$,we substitute these values into the equation: $f'(b) \cdot 2 = 1$.
Therefore,$f'(b) = 1/2$.

(D) To find the derivative of the composite function $y = F[f\{ \phi (x)\} ]$,we apply the chain rule repeatedly.
Let $u = f\{ \phi (x)\} $ and $v = \phi (x)$. Then $y = F(u)$ and $u = f(v)$.
By the chain rule,$\frac{dy}{dx} = \frac{dF}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.
$1$. The derivative of $F$ with respect to $u$ is $F'(u) = F'[f\{ \phi (x)\} ]$.
$2$. The derivative of $u = f(v)$ with respect to $v$ is $f'(v) = f'\{ \phi (x)\} $.
$3$. The derivative of $v = \phi (x)$ with respect to $x$ is $\phi '(x)$.
Multiplying these together,we get $\frac{dy}{dx} = F'[f\{ \phi (x)\} ] \cdot f'\{ \phi (x)\} \cdot \phi '(x)$.

(B) Given $f(x) = e^x$ and $g(x) = \sin^{-1}x$.
Since $h(x) = f(g(x))$,we have $h(x) = f(\sin^{-1}x) = e^{\sin^{-1}x}$.
Now,differentiate $h(x)$ with respect to $x$ using the chain rule:
$h'(x) = \frac{d}{dx}(e^{\sin^{-1}x}) = e^{\sin^{-1}x} \cdot \frac{d}{dx}(\sin^{-1}x) = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}$.
Therefore,the ratio $h'(x)/h(x)$ is:
$\frac{h'(x)}{h(x)} = \frac{e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1 - x^2}}}{e^{\sin^{-1}x}} = \frac{1}{\sqrt{1 - x^2}}$.

(A) Let $f(x) = \cos^{-1}\left( \sin \sqrt{\frac{1+x}{2}} \right) + x^x$.
Using the identity $\sin \theta = \cos(\frac{\pi}{2} - \theta)$,we have:
$f(x) = \cos^{-1}\left[ \cos \left( \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} \right) \right] + x^x$.
Simplifying the expression:
$f(x) = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$.
Differentiating with respect to $x$:
$f'(x) = 0 - \frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + \frac{d}{dx}(x^x)$.
Since $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$,we get:
$f'(x) = -\frac{1}{2\sqrt{2(1+x)}} + x^x(1 + \ln x)$.
At $x = 1$:
$f'(1) = -\frac{1}{2\sqrt{2(1+1)}} + 1^1(1 + \ln 1) = -\frac{1}{2\sqrt{4}} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

(B) Given the function $f(x) = x + 2$.
First,find the derivative of $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x + 2) = 1$.
Since the derivative $f'(x)$ is a constant value of $1$ for all $x$,the value of $f'(f(x))$ is also $1$ regardless of the input $f(x)$.
Therefore,at $x = 4$,$f'(f(4)) = f'(4 + 2) = f'(6) = 1$.
Thus,the correct option is $B$.

(B) Given equation is $3f(x) - 2f(1/x) = x$ .....$(i)$
Replace $x$ with $1/x$ in equation $(i)$:
$3f(1/x) - 2f(x) = 1/x$
Rearranging the terms,we get:
$-2f(x) + 3f(1/x) = 1/x$ .....$(ii)$
To eliminate $f(1/x)$,multiply equation $(i)$ by $3$ and equation $(ii)$ by $2$,then add them:
$9f(x) - 6f(1/x) = 3x$
$-4f(x) + 6f(1/x) = 2/x$
Adding these two equations:
$5f(x) = 3x + \frac{2}{x}$
$f(x) = \frac{3x}{5} + \frac{2}{5x}$
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = \frac{3}{5} - \frac{2}{5x^2}$
Substitute $x = 2$ to find $f'(2)$:
$f'(2) = \frac{3}{5} - \frac{2}{5(2^2)} = \frac{3}{5} - \frac{2}{20} = \frac{3}{5} - \frac{1}{10}$
$f'(2) = \frac{6 - 1}{10} = \frac{5}{10} = \frac{1}{2}$.

(C) Let $u = \tan^{-1}x$ and $y = \frac{u}{1+u}$.
We need to find $\frac{dy}{du}$.
Using the quotient rule for differentiation,$\frac{dy}{du} = \frac{d}{du} \left( \frac{u}{1+u} \right)$.
$\frac{dy}{du} = \frac{(1+u)(1) - u(1)}{(1+u)^2}$.
$\frac{dy}{du} = \frac{1+u-u}{(1+u)^2} = \frac{1}{(1+u)^2}$.
Substituting $u = \tan^{-1}x$ back,we get $\frac{dy}{du} = \frac{1}{{(1 + \tan^{-1}x)}^2}$.

(B) Let $y = {x^6}$ and $z = {x^3}$.
We need to find the derivative of $y$ with respect to $z$,which is $\frac{dy}{dz}$.
Using the chain rule,$\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.
First,differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}({x^6}) = 6{x^5}$.
Next,differentiate $z$ with respect to $x$: $\frac{dz}{dx} = \frac{d}{dx}({x^3}) = 3{x^2}$.
Now,substitute these values into the formula:
$\frac{dy}{dz} = \frac{6{x^5}}{3{x^2}} = 2{x^3}$.
Therefore,the correct option is $B$.

(B) Given that $f(x) = \log x.$
We need to find the derivative of $f(\sin x)$ with respect to $x.$
Let $y = f(\sin x) = \log(\sin x).$
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[\log(\sin x)]$
$\frac{dy}{dx} = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$
$\frac{dy}{dx} = \frac{1}{\sin x} \cdot \cos x$
$\frac{dy}{dx} = \cot x.$
Therefore,the correct option is $B$.

(B) Given $t = \frac{v^2}{2}$.
Differentiating with respect to $t$,we get $\frac{dt}{dt} = \frac{1}{2} \times 2v \frac{dv}{dt}$.
Since $f = \frac{dv}{dt}$,we have $1 = v \cdot f$,which implies $f = \frac{1}{v}$ or $v = f^{-1}$.
Now,differentiate $f = v^{-1}$ with respect to $t$:
$\frac{df}{dt} = -v^{-2} \frac{dv}{dt}$.
Substitute $\frac{dv}{dt} = f$ and $v = \frac{1}{f}$:
$\frac{df}{dt} = -\left( \frac{1}{f} \right)^{-2} \cdot f = -f^2 \cdot f = -f^3$.
Therefore,$-\frac{df}{dt} = -(-f^3) = f^3$.

(D) If $f'(x) = 0$ for all $x$ in the interval $(a, b)$,then the function $f(x)$ does not change its value as $x$ varies within this interval.
Therefore,$f(x) = c$,where $c$ is a constant.
This means the function is a constant function in the interval $(a, b)$.

(A) Let $u = \sqrt{x - 1}$. Since $1 < x < 26$,we have $0 < u < 5$.
Then $x - 1 = u^2$,so $x = u^2 + 1$.
The function becomes $f(x) = u + \sqrt{u^2 + 1 + 24 - 10u} = u + \sqrt{u^2 - 10u + 25}$.
This simplifies to $f(x) = u + \sqrt{(u - 5)^2} = u + |u - 5|$.
Since $0 < u < 5$,$|u - 5| = 5 - u$.
Thus,$f(x) = u + 5 - u = 5$.
Since $f(x) = 5$ is a constant function for $1 < x < 26$,its derivative $f'(x) = 0$ for all $x$ in the interval $(1, 26)$.

(A) Given $y = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log(x + \sqrt{x^2 + a^2})$.
Applying the product rule and chain rule to differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left[ \sqrt{a^2 + x^2} + x \cdot \frac{1}{2\sqrt{a^2 + x^2}} \cdot 2x \right] + \frac{a^2}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left[ 1 + \frac{1}{2\sqrt{x^2 + a^2}} \cdot 2x \right]$
Simplify the first term:
$\frac{1}{2} \left[ \frac{a^2 + x^2 + x^2}{\sqrt{a^2 + x^2}} \right] = \frac{a^2 + 2x^2}{2\sqrt{a^2 + x^2}}$
Simplify the second term:
$\frac{a^2}{2} \cdot \frac{1}{x + \sqrt{x^2 + a^2}} \cdot \left[ \frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}} \right] = \frac{a^2}{2\sqrt{x^2 + a^2}}$
Adding both parts:
$\frac{dy}{dx} = \frac{a^2 + 2x^2 + a^2}{2\sqrt{a^2 + x^2}} = \frac{2(a^2 + x^2)}{2\sqrt{a^2 + x^2}} = \sqrt{a^2 + x^2}$.

(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$
By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.$
Since $f'(x) = \sin(x^2),$ we have $f'(t) = \sin(t^2) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$
Now,calculate $\frac{dt}{dx}$ using the quotient rule:
$\frac{dt}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}.$
Therefore,$\frac{dy}{dx} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$

(C) The function $f(x) = a^x$ is defined for $a > 0$ and $a \neq 1$.
To determine if the function is strictly increasing,we examine its derivative:
$f'(x) = \frac{d}{dx}(a^x) = a^x \ln(a)$.
For the function to be strictly increasing,we require $f'(x) > 0$ for all $x$.
Since $a^x > 0$ for all $x$,the condition $f'(x) > 0$ is satisfied if $\ln(a) > 0$.
This implies $a > e^0$,which means $a > 1$.
Thus,the function $f(x) = a^x$ is strictly increasing when $a > 1$.

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

(D) The expression $\mathop{\lim}\limits_{x \to 1} \frac{G(x) - G(1)}{x - 1}$ represents the derivative of $G(x)$ at $x = 1$,denoted as $G'(1)$.
Given $G(x) = -(25 - x^2)^{1/2}$.
Differentiating with respect to $x$ using the chain rule:
$G'(x) = -\frac{1}{2}(25 - x^2)^{-1/2} \cdot (-2x) = \frac{x}{\sqrt{25 - x^2}}$.
Evaluating at $x = 1$:
$G'(1) = \frac{1}{\sqrt{25 - 1^2}} = \frac{1}{\sqrt{24}}$.
Thus,the limit is $\frac{1}{\sqrt{24}}$.

(C) Let $L = \lim_{x \to 0} \left\{ \frac{f(1 + x)}{f(1)} \right\}^{\frac{1}{x}}$.
Taking natural logarithm on both sides,$\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{f(1 + x)}{f(1)} \right)$.
Using the definition of the derivative or $L$'Hopital's rule,$\ln L = \lim_{x \to 0} \frac{\ln f(1 + x) - \ln f(1)}{x}$.
Applying $L$'Hopital's rule,$\ln L = \lim_{x \to 0} \frac{f'(1 + x)}{f(1 + x)} = \frac{f'(1)}{f(1)}$.
Given $f(1) = 3$ and $f'(1) = 6$,we have $\ln L = \frac{6}{3} = 2$.
Therefore,$L = e^2$.

(A) Let $L = \lim_{x \to 1} \int_4^{f(x)} \frac{2t}{x - 1} dt$.
Since the integrand $\frac{2t}{x-1}$ does not depend on $t$ in the denominator,we can write the integral as $\frac{1}{x-1} \int_4^{f(x)} 2t dt$.
The integral of $2t$ is $t^2$,so $\int_4^{f(x)} 2t dt = [t^2]_4^{f(x)} = (f(x))^2 - 16$.
Thus,$L = \lim_{x \to 1} \frac{(f(x))^2 - 16}{x - 1}$.
Using the difference of squares,$(f(x))^2 - 16 = (f(x) - 4)(f(x) + 4)$.
Since $f(1) = 4$,as $x \to 1$,$f(x) \to 4$,so $f(x) + 4 \to 8$.
Therefore,$L = \lim_{x \to 1} \frac{(f(x) - 4)(f(x) + 4)}{x - 1} = \left( \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} \right) \times \left( \lim_{x \to 1} (f(x) + 4) \right)$.
This simplifies to $f'(1) \times (4 + 4) = 8f'(1)$.

(B) To find the derivative of $f(x) = e^{\sqrt{1 - x^2}} \cdot \tan x$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.
Let $u = e^{\sqrt{1 - x^2}}$ and $v = \tan x$.
Then $\frac{dv}{dx} = \sec^2 x$.
Using the chain rule for $u$,$\frac{du}{dx} = e^{\sqrt{1 - x^2}} \cdot \frac{d}{dx}(\sqrt{1 - x^2}) = e^{\sqrt{1 - x^2}} \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x e^{\sqrt{1 - x^2}}}{\sqrt{1 - x^2}}$.
Applying the product rule:
$\frac{d}{dx} (u v) = e^{\sqrt{1 - x^2}} \cdot \sec^2 x + \tan x \cdot \left( -\frac{x e^{\sqrt{1 - x^2}}}{\sqrt{1 - x^2}} \right)$.
Factoring out $e^{\sqrt{1 - x^2}}$,we get:
$= e^{\sqrt{1 - x^2}} \left[ \sec^2 x - \frac{x \tan x}{\sqrt{1 - x^2}} \right]$.

(A) Given $y = 5x(1 - x)^{-2/3} + \cos^2(2x + 1)$.
Applying the product rule and chain rule:
$\frac{dy}{dx} = 5 \left[ (1 - x)^{-2/3} + x \cdot \left( -\frac{2}{3} \right)(1 - x)^{-5/3}(-1) \right] + 2\cos(2x + 1)(-\sin(2x + 1))(2)$.
$\frac{dy}{dx} = 5(1 - x)^{-2/3} + \frac{10x}{3(1 - x)^{5/3}} - 4\cos(2x + 1)\sin(2x + 1)$.
Using $2\sin\theta\cos\theta = \sin(2\theta)$,we have $4\cos(2x + 1)\sin(2x + 1) = 2\sin(4x + 2)$.
$\frac{dy}{dx} = \frac{5(1 - x) + \frac{10x}{3}}{(1 - x)^{5/3}} - 2\sin(4x + 2) = \frac{5 - 5x + \frac{10x}{3}}{(1 - x)^{5/3}} - 2\sin(4x + 2)$.
$\frac{dy}{dx} = \frac{5(3 - 3x + 2x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2) = \frac{5(3 - x)}{3(1 - x)^{5/3}} - 2\sin(4x + 2)$.

(D) Given $f'(ln x) = \begin{cases} 1 & 0 < x \le 1 \\ x & x > 1 \end{cases}$.
Let $ln x = t$,then $x = e^t$.
For $x > 1$,$t > 0$,so $f'(t) = e^t$.
Integrating,$f(t) = e^t + C_1$.
Since $f$ is continuous at $t=0$,we use the limit from the left: $f(0) = 0$.
$f(0) = e^0 + C_1 = 0 \implies 1 + C_1 = 0 \implies C_1 = -1$.
Thus,$f(x) = e^x - 1$ for $x > 0$.
For $0 < x \le 1$,$t \le 0$,so $f'(t) = 1$.
Integrating,$f(t) = t + C_2$.
Given $f(0) = 0$,$0 + C_2 = 0 \implies C_2 = 0$.
Thus,$f(x) = x$ for $x \le 0$.
Combining these,$f(x) = \begin{cases} x & x \le 0 \\ e^x - 1 & x > 0 \end{cases}$.

(B) Let $f(x) = \int\limits_a^x \ln^2 t \, dt$.
Then the given expression is $\mathop {\text{Limit}}\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
By the definition of the derivative,this is equal to $f'(x)$.
Using the Fundamental Theorem of Calculus (Leibniz's Rule),we have $f'(x) = \frac{d}{dx} \int\limits_a^x \ln^2 t \, dt = \ln^2 x$.
Therefore,the limit is $\ln^2 x$.

(A) Given $f'(\sin x) = \cos^2 x$.
Let $t = \sin x$,then $\cos^2 x = 1 - \sin^2 x = 1 - t^2$.
So,$f'(t) = 1 - t^2$.
Integrating both sides with respect to $t$,we get:
$f(t) = \int (1 - t^2) dt = t - \frac{t^3}{3} + C$.
Given $f(1) = 1$,substitute $t = 1$:
$1 = 1 - \frac{1^3}{3} + C \Rightarrow 1 = 1 - \frac{1}{3} + C \Rightarrow C = \frac{1}{3}$.
Thus,$f(t) = t - \frac{t^3}{3} + \frac{1}{3}$.
Replacing $t$ with $x$,we get $f(x) = x - \frac{x^3}{3} + \frac{1}{3}$.

(D) Given $f''(x) = x^{1/3}$.
Integrating with respect to $x$:
$f'(x) = \int x^{1/3} dx = \frac{x^{4/3}}{4/3} + C_1 = \frac{3}{4}x^{4/3} + C_1$.
Comparing this with the given statements:
Statement $I$ $(f'(x) = \frac{3}{4}x^{4/3} + 9)$ is possible if $C_1 = 9$.
Statement $IV$ $(f'(x) = \frac{3}{4}x^{4/3} - 4)$ is possible if $C_1 = -4$.
Now,integrating $f'(x)$ to find $f(x)$:
$f(x) = \int (\frac{3}{4}x^{4/3} + C_1) dx = \frac{3}{4} \cdot \frac{x^{7/3}}{7/3} + C_1x + C_2 = \frac{9}{28}x^{7/3} + C_1x + C_2$.
Statements $II$ and $III$ involve $f(x) = \frac{9}{28}x^{7/3} + K$. This form is only possible if $C_1 = 0$. If $C_1 = 0$,then $f'(x) = \frac{3}{4}x^{4/3}$,which does not match $I$ or $IV$. Thus,$II$ and $III$ cannot be true simultaneously with the provided $f'(x)$ options. Therefore,only $I$ and $IV$ are valid forms for $f'(x)$.

(B) Given $y = f \left( \frac{3x + 4}{5x + 6} \right)$ and $f'(x) = \tan(x^2)$.
Applying the chain rule to differentiate with respect to $x$:
$\frac{dy}{dx} = f' \left( \frac{3x + 4}{5x + 6} \right) \cdot \frac{d}{dx} \left( \frac{3x + 4}{5x + 6} \right)$
Using the quotient rule for the derivative of the inner function:
$\frac{d}{dx} \left( \frac{3x + 4}{5x + 6} \right) = \frac{3(5x + 6) - 5(3x + 4)}{(5x + 6)^2} = \frac{15x + 18 - 15x - 20}{(5x + 6)^2} = \frac{-2}{(5x + 6)^2}$
Substituting $f'(x) = \tan(x^2)$ into the expression:
$\frac{dy}{dx} = \tan \left( \frac{3x + 4}{5x + 6} \right)^2 \cdot \frac{-2}{(5x + 6)^2}$
Thus,the correct option is $B$.

(D) Let $x = \sin^2 \theta$ and $\sqrt{x} = \sin \alpha$. Then $x = \sin^2 \alpha$ and $\sqrt{x} = \sin \alpha$.
Substituting these into the expression:
$y = \sin^{-1} (\sin^2 \alpha \sqrt{1 - \sin^2 \alpha} + \sin \alpha \sqrt{1 - \sin^4 \alpha})$
This substitution is complex. Let's use $x = \sin^2 \theta$ and $\sqrt{x} = \sin \phi$.
Actually,let $x = \sin^2 \alpha$ and $\sqrt{x} = \sin \beta$.
Then $y = \sin^{-1} (\sin^2 \alpha \cos \beta + \sin \beta \cos \alpha)$ is not standard.
Let $x = \sin^2 \theta$ and $\sqrt{x} = \sin \phi$.
Let $x = \sin^2 A$ and $\sqrt{x} = \sin B$.
Consider $y = \sin^{-1} (x \sqrt{1 - (\sqrt{x})^2} + \sqrt{x} \sqrt{1 - x^2})$.
Let $x = \sin A$ and $\sqrt{x} = \sin B$.
Then $y = \sin^{-1} (\sin A \cos B + \sin B \cos A) = \sin^{-1} (\sin(A + B)) = A + B$.
Since $x = \sin A$,$A = \sin^{-1} x$.
Since $\sqrt{x} = \sin B$,$B = \sin^{-1} \sqrt{x}$.
So,$y = \sin^{-1} x + \sin^{-1} \sqrt{x}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$
$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2\sqrt{x(1 - x)}}$.
Comparing this with $\frac{dy}{dx} = \frac{1}{2\sqrt{x(1 - x)}} + p$,we get $p = \frac{1}{\sqrt{1 - x^2}}$.

(B) Given $y = f \left( \frac{2x - 1}{x^2 + 1} \right)$ and $f'(x) = \sin x$.
Applying the chain rule,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = f' \left( \frac{2x - 1}{x^2 + 1} \right) \cdot \frac{d}{dx} \left( \frac{2x - 1}{x^2 + 1} \right)$.
Since $f'(x) = \sin x$,we have $f' \left( \frac{2x - 1}{x^2 + 1} \right) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)$.
Now,differentiate $u = \frac{2x - 1}{x^2 + 1}$ using the quotient rule:
$\frac{du}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{2 + 2x - 2x^2}{(x^2 + 1)^2} = \frac{2(1 + x - x^2)}{(x^2 + 1)^2}$.
Substituting these back into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{2(1 + x - x^2)}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)$.
Thus,the correct option is $B$.

(C) Given the definition $D^*f(x) = \lim_{h \to 0} \frac{f^2(x + h) - f^2(x)}{h}$.
This is the definition of the derivative of the function $g(x) = f^2(x) = [f(x)]^2$.
Using the chain rule,$D^*f(x) = \frac{d}{dx} [f(x)]^2 = 2f(x) \cdot f'(x)$.
Given $f(x) = x \ln x$,we find $f'(x)$ using the product rule:
$f'(x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.
Now,substitute $f(x)$ and $f'(x)$ into the expression for $D^*f(x)$:
$D^*f(x) = 2(x \ln x)(\ln x + 1)$.
Evaluate at $x = e$:
$D^*f(e) = 2(e \ln e)(\ln e + 1) = 2(e \cdot 1)(1 + 1) = 2e(2) = 4e$.

(D) Given $f(x) = \frac{1}{2} \cdot 5^{2x+1} = \frac{5}{2} \cdot 5^{2x}$.
Calculating the derivative: $f'(x) = \frac{5}{2} \cdot 5^{2x} \cdot \ln 5 \cdot 2 = 5 \cdot 5^{2x} \ln 5$.
Given $g(x) = 5^x + 4x \ln 5$.
Calculating the derivative: $g'(x) = 5^x \ln 5 + 4 \ln 5 = \ln 5 (5^x + 4)$.
We need to solve $f'(x) > g'(x)$:
$5 \cdot 5^{2x} \ln 5 > \ln 5 (5^x + 4)$.
Since $\ln 5 > 0$,we can divide by $\ln 5$:
$5 \cdot (5^x)^2 > 5^x + 4$.
Let $t = 5^x$. Since $x \in \mathbb{R}$,$t > 0$.
The inequality becomes $5t^2 - t - 4 > 0$.
Factoring the quadratic: $(5t + 4)(t - 1) > 0$.
Since $t > 0$,$5t + 4$ is always positive.
Thus,we must have $t - 1 > 0$,which implies $t > 1$.
Substituting back $t = 5^x$: $5^x > 1$.
Since $5^x$ is an increasing function,$x > 0$.

(D) Given that $f \circ g$ is an identity function,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$f'(g(x)) \cdot g'(x) = 1$.
Substitute $x = a$ into the equation:
$f'(g(a)) \cdot g'(a) = 1$.
Given $g(a) = b$ and $g'(a) = 2$,we substitute these values:
$f'(b) \cdot 2 = 1$.
Therefore,$f'(b) = 1/2$.

(B) Given the equation $f(x^2) = x^3$ for $x > 0$.
To find $f'(4)$,we differentiate both sides of the equation with respect to $x$ using the chain rule:
$\frac{d}{dx}[f(x^2)] = \frac{d}{dx}[x^3]$
$f'(x^2) \cdot (2x) = 3x^2$
For $x > 0$,we can divide by $2x$:
$f'(x^2) = \frac{3x^2}{2x} = \frac{3}{2}x$
We want to find $f'(4)$. Since $x^2 = 4$ and $x > 0$,we have $x = 2$.
Substituting $x = 2$ into the expression for $f'(x^2)$:
$f'(4) = \frac{3}{2}(2) = 3$
Thus,the value of $f'(4)$ is $3$.

(D) The function $f(x)$ is defined only if the argument of the square root is non-negative,i.e.,$\ln(2a - a^2) \ge 0$.
This implies $2a - a^2 \ge 1$,which simplifies to $a^2 - 2a + 1 \le 0$,or $(a - 1)^2 \le 0$.
Since the square of a real number cannot be negative,we must have $(a - 1)^2 = 0$,which means $a = 1$.
Substituting $a = 1$ into the expression for $f(x)$,we get $f(x) = 4x^3 - 6x^2 \cos 2 + 3x \sin 2 \sin 6$.
Now,find the derivative $f'(x) = 12x^2 - 12x \cos 2 + 3 \sin 2 \sin 6$.
Evaluating at $x = 1/2$,we get $f'(1/2) = 12(1/4) - 12(1/2) \cos 2 + 3 \sin 2 \sin 6 = 3 - 6 \cos 2 + 3 \sin 2 \sin 6$.
Using the identity $\sin 2 \sin 6 = \frac{1}{2}(\cos(2-6) - \cos(2+6)) = \frac{1}{2}(\cos 4 - \cos 8)$,we have $f'(1/2) = 3 - 6 \cos 2 + \frac{3}{2}(\cos 4 - \cos 8)$.
Since $\cos 2 \approx -0.416$,$-6 \cos 2 \approx 2.496$. Thus $f'(1/2) = 3 + 2.496 + \dots > 0$.

(C) Given $f(x) = (kx + e^x) h(x)$.
Using the product rule for differentiation,$f'(x) = \frac{d}{dx}(kx + e^x) \cdot h(x) + (kx + e^x) \cdot h'(x)$.
$f'(x) = (k + e^x) h(x) + (kx + e^x) h'(x)$.
Now,substitute $x = 0$ into the derivative expression:
$f'(0) = (k + e^0) h(0) + (k(0) + e^0) h'(0)$.
$f'(0) = (k + 1) h(0) + (0 + 1) h'(0)$.
$f'(0) = (k + 1) h(0) + h'(0)$.
Given $h(0) = 5$,$h'(0) = -2$,and $f'(0) = 18$,we substitute these values:
$18 = (k + 1)(5) + (-2)$.
$18 = 5k + 5 - 2$.
$18 = 5k + 3$.
$15 = 5k$.
$k = 3$.

(D) Given $f(x) = (x^x)^x = x^{x^2}$.
Taking the natural logarithm on both sides,$\ln(f(x)) = x^2 \ln(x)$.
Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = 2x \ln(x) + x^2 \cdot \frac{1}{x} = 2x \ln(x) + x$.
So,$f'(x) = f(x) (2x \ln(x) + x) = x^{x^2} (2x \ln(x) + x)$.
At $x = 1$,$f'(1) = 1^1 (2(1) \ln(1) + 1) = 1(0 + 1) = 1$.
Now,$g(x) = x^{(x^x)}$.
Taking the natural logarithm,$\ln(g(x)) = x^x \ln(x)$.
Differentiating with respect to $x$,$\frac{g'(x)}{g(x)} = \frac{d}{dx}(x^x) \ln(x) + x^x \cdot \frac{1}{x}$.
We know $\frac{d}{dx}(x^x) = x^x(1 + \ln(x))$.
So,$\frac{g'(x)}{g(x)} = x^x(1 + \ln(x)) \ln(x) + x^{x-1}$.
At $x = 1$,$g(1) = 1^{(1^1)} = 1$.
$g'(1) = g(1) [1^1(1 + \ln(1)) \ln(1) + 1^{1-1}] = 1 [1(1 + 0)(0) + 1] = 1(0 + 1) = 1$.
Thus,$f'(1) = 1$ and $g'(1) = 1$.

(B) Let $u = \sqrt{a - x}$ and $v = \sqrt{x - b}$. Then $u^2 = a - x$ and $v^2 = x - b$.
Note that $u^2 + v^2 = a - b$. The numerator is $u^3 - (b - x)v$. Since $b - x = -v^2$,the numerator is $u^3 + v^3$.
Thus,$y = \frac{u^3 + v^3}{u + v} = u^2 - uv + v^2$.
Substituting back,$y = (a - x) - \sqrt{(a - x)(x - b)} + (x - b) = a - b - \sqrt{-x^2 + (a + b)x - ab}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{2\sqrt{-x^2 + (a + b)x - ab}} \cdot (-2x + a + b)$.
$\frac{dy}{dx} = \frac{2x - (a + b)}{2\sqrt{(a - x)(x - b)}}$.

(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$.
We can rewrite the coefficients using binomial expansion properties:
$N = \cos 6x + \cos 4x + 5\cos 4x + 5\cos 2x + 10\cos 2x + 10$.
Using the identity $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$ and $1 + \cos 2x = 2\cos^2 x$:
$N = 2\cos 5x \cos x + 5(2\cos 3x \cos x) + 10(2\cos^2 x)$.
$N = 2\cos x (\cos 5x + 5\cos 3x + 10\cos x)$.
Let the denominator be $D = \cos 5x + 5\cos 3x + 10\cos x$.
Then $y = \frac{N}{D} = \frac{2\cos x (\cos 5x + 5\cos 3x + 10\cos x)}{\cos 5x + 5\cos 3x + 10\cos x} = 2\cos x$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(2\cos x) = -2\sin x$.