Derivative at a point, Standard differentiation Questions in English

(B) Let the given expression be $y = {\left( {{x^{\frac{{\ell + m}}{{m - n}}}}} \right)^{\frac{1}{{n - \ell }}}} \cdot {\left( {{x^{\frac{{m + n}}{{n - \ell }}}}} \right)^{\frac{1}{{\ell - m}}}} \cdot {\left( {{x^{\frac{{n + \ell }}{{\ell - m}}}}} \right)^{\frac{1}{{m - n}}}}$.
Using the property of exponents $(x^a)^b = x^{ab}$,we have:
$y = x^{\frac{\ell + m}{(m - n)(n - \ell)}} \cdot x^{\frac{m + n}{(n - \ell)(\ell - m)}} \cdot x^{\frac{n + \ell}{(\ell - m)(m - n)}}$.
Since the bases are the same,we add the exponents:
$y = x^{\left[ \frac{\ell + m}{(m - n)(n - \ell)} + \frac{m + n}{(n - \ell)(\ell - m)} + \frac{n + \ell}{(\ell - m)(m - n)} \right]}$.
To add these fractions,we find a common denominator,which is $(m - n)(n - \ell)(\ell - m)$.
Multiplying the numerator of each term by the missing factor:
Exponent $= \frac{(\ell + m)(\ell - m) + (m + n)(m - n) + (n + \ell)(n - \ell)}{(m - n)(n - \ell)(\ell - m)}$.
Using the identity $(a+b)(a-b) = a^2 - b^2$:
Exponent $= \frac{(\ell^2 - m^2) + (m^2 - n^2) + (n^2 - \ell^2)}{(m - n)(n - \ell)(\ell - m)} = \frac{0}{(m - n)(n - \ell)(\ell - m)} = 0$.
Thus,$y = x^0 = 1$ for $x \neq 0$.
The derivative of a constant $1$ with respect to $x$ is $\frac{d}{dx}(1) = 0$.

(B) Given $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$.
We can rewrite the numerator as $x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - (\sqrt{3}x)^2$.
Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $y = \frac{(x^2 + 1 - \sqrt{3}x)(x^2 + 1 + \sqrt{3}x)}{x^2 + \sqrt{3}x + 1}$.
Simplifying the expression,we have $y = x^2 - \sqrt{3}x + 1$.
Now,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - \sqrt{3}$.
Comparing this with $\frac{dy}{dx} = ax + b$,we find $a = 2$ and $b = -\sqrt{3}$.
Thus,$a + b = 2 - \sqrt{3}$.
Since $tan(15^\circ) = tan(\frac{\pi}{12}) = 2 - \sqrt{3}$,and $tan(\frac{\pi}{12}) = cot(\frac{\pi}{2} - \frac{\pi}{12}) = cot(\frac{5\pi}{12})$.
Therefore,$a + b = cot\frac{5\pi}{12}$.

(B) Given $h(x) = f(x)g(x)$. Using the product rule,$h'(x) = f'(x)g(x) + g'(x)f(x)$.
At $x = 2$,$h'(2) = f'(2)g(2) + g'(2)f(2) = (-2)(5) + (4)(3) = -10 + 12 = 2$.
Given $F(x) = f(g(x))$. Using the chain rule,$F'(x) = f'(g(x)) \cdot g'(x)$.
At $x = 2$,$F'(2) = f'(g(2)) \cdot g'(2) = f'(5) \cdot 4 = 11 \cdot 4 = 44$.
Now,comparing $F'(2)$ and $h'(2)$,we have $F'(2) = 44$ and $h'(2) = 2$.
Therefore,$F'(2) = 22 \cdot h'(2)$.

(D) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}(e^{\sqrt{x}}) + \frac{d}{dx}(e^{-\sqrt{x}})$
$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x}) + e^{-\sqrt{x}} \cdot \frac{d}{dx}(-\sqrt{x})$
$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot (-\frac{1}{2\sqrt{x}})$
$\frac{dy}{dx} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}$ (This matches option $A$).
Now,consider $y^2 - 4 = (e^{\sqrt{x}} + e^{-\sqrt{x}})^2 - 4$
$y^2 - 4 = e^{2\sqrt{x}} + e^{-2\sqrt{x}} + 2 - 4 = e^{2\sqrt{x}} + e^{-2\sqrt{x}} - 2$
$y^2 - 4 = (e^{\sqrt{x}} - e^{-\sqrt{x}})^2$
Taking the square root,$\sqrt{y^2 - 4} = e^{\sqrt{x}} - e^{-\sqrt{x}}$.
Substituting this into the derivative expression:
$\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \sqrt{y^2 - 4}$ (This matches option $C$).
Since both $A$ and $C$ are correct,the correct option is $D$.

(A) Given that $f$ is differentiable at $x = 0$ and $f'(0) = 1$.
We need to evaluate the limit: $L = \lim_{h \to 0} \frac{f(h) - f(-2h)}{h}$.
By adding and subtracting $f(0)$ in the numerator,we get:
$L = \lim_{h \to 0} \frac{f(h) - f(0) - (f(-2h) - f(0))}{h}$.
$L = \lim_{h \to 0} \left( \frac{f(h) - f(0)}{h} - \frac{f(-2h) - f(0)}{h} \right)$.
$L = \lim_{h \to 0} \frac{f(h) - f(0)}{h} - \lim_{h \to 0} \frac{f(-2h) - f(0)}{h}$.
Since $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = 1$,the first term is $1$.
For the second term,let $k = -2h$. As $h \to 0$,$k \to 0$.
$\lim_{h \to 0} \frac{f(-2h) - f(0)}{h} = \lim_{k \to 0} \frac{f(k) - f(0)}{-k/2} = -2 \lim_{k \to 0} \frac{f(k) - f(0)}{k} = -2 f'(0) = -2(1) = -2$.
Thus,$L = 1 - (-2) = 1 + 2 = 3$.

(C) By definition,the derivative $f'(1)$ is the limit of the slope of the secant line passing through $P(1, 2)$ and $Q(s, f(s))$ as $s \to 1$.
The slope of the secant line is given by the coefficient of $x$ in the equation $y = \left( \frac{s^2 + 2s - 3}{s - 1} \right) x - 1 - s$.
Thus,$f'(1) = \lim_{s \to 1} \frac{s^2 + 2s - 3}{s - 1}$.
Simplifying the expression:
$f'(1) = \lim_{s \to 1} \frac{(s - 1)(s + 3)}{s - 1}$
$f'(1) = \lim_{s \to 1} (s + 3) = 1 + 3 = 4$.
Therefore,the correct option is $C$.

(D) Given $f(x) = x^{\frac{m}{n}}$,where $m$ and $n$ are odd integers and $0 < m < n$.
First,consider the derivative: $f'(x) = \frac{m}{n} x^{\frac{m}{n} - 1} = \frac{m}{n} x^{\frac{m-n}{n}}$.
Since $m < n$,the exponent $\frac{m-n}{n}$ is negative. Let $k = n - m$,where $k$ is an even positive integer (since $n$ and $m$ are both odd,their difference is even). Then $f'(x) = \frac{m}{n} x^{-\frac{k}{n}} = \frac{m}{n \cdot x^{\frac{k}{n}}}$.
As $x \to 0$,$f'(x) \to \infty$,so $f'(0)$ does not exist. Thus,$f(x)$ is not differentiable at $x = 0$.
Since $k$ is even,$x^{\frac{k}{n}} > 0$ for all $x \neq 0$. Therefore,$f'(x) = \frac{m}{n \cdot x^{\frac{k}{n}}} > 0$ for all $x \neq 0$.
Since $f'(x) > 0$ for all $x \in \mathbb{R} \setminus \{0\}$ and $f(x)$ is continuous at $x = 0$,the function $f(x)$ is strictly increasing on the entire domain $\mathbb{R}$.
Thus,the correct option is $(D)$.

(C) Given $h(x) = \frac{5}{3}(f(x))^3 + \frac{1}{2}(f(x))^2 + 2f(x) + 100$.
To determine the behavior of $h(x)$,we differentiate with respect to $x$:
$h'(x) = \frac{d}{dx} \left[ \frac{5}{3}(f(x))^3 + \frac{1}{2}(f(x))^2 + 2f(x) + 100 \right]$.
Using the chain rule:
$h'(x) = \left( 5(f(x))^2 + f(x) + 2 \right) f'(x)$.
Consider the quadratic expression $Q = 5(f(x))^2 + f(x) + 2$.
The discriminant $D = b^2 - 4ac = (1)^2 - 4(5)(2) = 1 - 40 = -39$.
Since $D < 0$ and the coefficient of $(f(x))^2$ is positive $(5 > 0)$,the quadratic $Q$ is always positive for all real values of $f(x)$.
Therefore,the sign of $h'(x)$ depends solely on the sign of $f'(x)$.
If $f'(x) > 0$,then $h'(x) > 0$,meaning $h(x)$ increases as $f(x)$ increases.
Thus,$h(x)$ increases as $f(x)$ increases.

(C) By the definition of the derivative,$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.
Given the functional equation $f(x + y) = f(x) - 3xy + f(y)$,we substitute $y = h$:
$f(x + h) = f(x) - 3xh + f(h)$.
Substituting this into the derivative definition:
$f'(x) = \lim_{h \to 0} \frac{f(x) - 3xh + f(h) - f(x)}{h}$
$f'(x) = \lim_{h \to 0} \frac{-3xh + f(h)}{h}$
$f'(x) = \lim_{h \to 0} (-3x + \frac{f(h)}{h})$.
Given that $\lim_{h \to 0} \frac{f(h)}{h} = 7$,we have:
$f'(x) = -3x + 7$.

(D) Let $L = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h^2 + 2h}$.
Using the definition of the derivative,$\lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = f'(1)$.
We can rewrite the limit as $\lim_{h \to 0} \left( \frac{f(1+h) - f(1)}{h} \cdot \frac{1}{h+2} \right) = f'(1) \cdot \frac{1}{2} = \frac{f'(1)}{2}$.
Given $f(x) = \frac{\sin(\frac{\pi x}{4})}{x + 1}$,we use the quotient rule: $f'(x) = \frac{(x+1) \cdot \cos(\frac{\pi x}{4}) \cdot \frac{\pi}{4} - \sin(\frac{\pi x}{4}) \cdot 1}{(x+1)^2}$.
At $x = 1$,$f'(1) = \frac{(1+1) \cdot \cos(\frac{\pi}{4}) \cdot \frac{\pi}{4} - \sin(\frac{\pi}{4})}{(1+1)^2} = \frac{2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} - \frac{1}{\sqrt{2}}}{4} = \frac{\frac{\pi}{2\sqrt{2}} - \frac{1}{\sqrt{2}}}{4} = \frac{\pi - 2}{8\sqrt{2}}$.
Thus,the limit is $\frac{f'(1)}{2} = \frac{\pi - 2}{16\sqrt{2}}$.

(C) Let $u = e^{x^3}$ and $v = \log_e x$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3) = e^{x^3} \cdot 3x^2$.
Next,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx}(\log_e x) = \frac{1}{x}$.
Now,calculate the ratio:
$\frac{du}{dv} = \frac{3x^2 e^{x^3}}{1/x} = 3x^2 e^{x^3} \cdot x = 3x^3 e^{x^3}$.
Thus,the correct option is $C$.

(C) Given $y = |\cos 4x| + |\sin 4x| + |\tan 4x|$.
At $x = \frac{\pi}{6}$,$4x = \frac{4\pi}{6} = \frac{2\pi}{3}$.
In the neighborhood of $x = \frac{\pi}{6}$,$\cos 4x < 0$,$\sin 4x > 0$,and $\tan 4x < 0$.
Therefore,$y = -\cos 4x + \sin 4x - \tan 4x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = -(-\sin 4x \cdot 4) + (\cos 4x \cdot 4) - (\sec^2 4x \cdot 4)$
$\frac{dy}{dx} = 4\sin 4x + 4\cos 4x - 4\sec^2 4x$.
At $x = \frac{\pi}{6}$,$4x = \frac{2\pi}{3}$.
$\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$,$\cos(\frac{2\pi}{3}) = -\frac{1}{2}$,$\sec(\frac{2\pi}{3}) = -2$.
$\frac{dy}{dx} = 4(\frac{\sqrt{3}}{2}) + 4(-\frac{1}{2}) - 4(-2)^2$
$\frac{dy}{dx} = 2\sqrt{3} - 2 - 16 = 2\sqrt{3} - 18$.
Note: The provided options seem to contain a typo. Based on the calculation,the correct value is $2\sqrt{3} - 18$.

(A) Let $u = \ln x$. Then $x = e^u$ and $dx = e^u du$.
We want to find the derivative of $f(x) = e^x (\ln x)^2$ with respect to $u = \ln x$.
Using the chain rule,$\frac{df}{du} = \frac{df}{dx} \cdot \frac{dx}{du}$.
First,calculate $\frac{df}{dx} = \frac{d}{dx}(e^x (\ln x)^2) = e^x (\ln x)^2 + e^x \cdot 2 \ln x \cdot \frac{1}{x} = e^x (\ln x)^2 + \frac{2 e^x \ln x}{x}$.
Since $u = \ln x$,we have $\frac{dx}{du} = \frac{d}{du}(e^u) = e^u = x$.
Thus,$\frac{df}{du} = (e^x (\ln x)^2 + \frac{2 e^x \ln x}{x}) \cdot x = x e^x (\ln x)^2 + 2 e^x \ln x$.
At $x = e$,we have $\ln x = 1$.
Substituting these values: $\frac{df}{du} = e \cdot e^e \cdot (1)^2 + 2 \cdot e^e \cdot 1 = e^{e+1} + 2 e^e = e^e (e + 2)$.

(B) Given $f(x) = x^3 + e^{x/2}$.
Since $g(x) = f^{-1}(x)$,we have $g(f(x)) = x$.
Differentiating both sides with respect to $x$,we get $g'(f(x)) \cdot f'(x) = 1$.
To find $g'(1)$,we set $f(x) = 1$.
$x^3 + e^{x/2} = 1$.
By inspection,at $x = 0$,$0^3 + e^0 = 0 + 1 = 1$. Thus,$f(0) = 1$.
Now,find $f'(x) = 3x^2 + \frac{1}{2}e^{x/2}$.
At $x = 0$,$f'(0) = 3(0)^2 + \frac{1}{2}e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.
Substituting these values into the derivative formula: $g'(f(0)) \cdot f'(0) = 1 \implies g'(1) \cdot \frac{1}{2} = 1$.
Therefore,$g'(1) = 2$.

(B) Given $f(x) = x\sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$.
Dividing by $x$,we get $\frac{f(x)}{x} = \sqrt{x\sqrt{x\sqrt{x\dots\infty}}}$.
Since the expression under the square root is the same as the original function,we have $\frac{f(x)}{x} = \sqrt{f(x)}$.
Squaring both sides,we get $\frac{f(x)^2}{x^2} = f(x)$.
Assuming $f(x) \neq 0$,we divide by $f(x)$ to get $\frac{f(x)}{x^2} = 1$,which implies $f(x) = x^2$.
Now,differentiating with respect to $x$,we get $f'(x) = 2x$.
Substituting $x = 3$,we get $f'(3) = 2(3) = 6$.

(C) Given $h(x) = f(g(f(x)))$.
Using the chain rule for differentiation,we have:
$h'(x) = f'(g(f(x))) \cdot g'(f(x)) \cdot f'(x)$.
Now,we evaluate $h'(x)$ at $x = 2$:
$h'(2) = f'(g(f(2))) \cdot g'(f(2)) \cdot f'(2)$.
Given $f(2) = 1$,substitute this into the expression:
$h'(2) = f'(g(1)) \cdot g'(1) \cdot f'(2)$.
Given $g(1) = 2$,substitute this into the expression:
$h'(2) = f'(2) \cdot g'(1) \cdot f'(2)$.
Given $f'(2) = 4$ and $g'(1) = 4$,substitute these values:
$h'(2) = 4 \cdot 4 \cdot 4 = 64$.

(C) First,simplify the expression inside the derivative using polynomial division and partial fractions.
Let $f(x) = \frac{2x^3 + 3x^2 + x - 3}{x^2 + x - 2}$.
Since $x^2 + x - 2 = (x - 1)(x + 2)$,we perform division:
$2x^3 + 3x^2 + x - 3 = (2x + 1)(x^2 + x - 2) + 0x - 1$,which is not quite right. Let's re-evaluate:
$\frac{2x^3 + 3x^2 + x - 3}{x^2 + x - 2} = 2x + 1 + \frac{-2x - 1}{x^2 + x - 2}$.
Using partial fractions for $\frac{-2x - 1}{(x - 1)(x + 2)} = \frac{P}{x - 1} + \frac{Q}{x + 2}$.
$-2x - 1 = P(x + 2) + Q(x - 1)$.
For $x = 1$: $-3 = 3P \implies P = -1$.
For $x = -2$: $3 = -3Q \implies Q = -1$.
So,$f(x) = 2x + 1 - \frac{1}{x - 1} - \frac{1}{x + 2}$.
Now,differentiate with respect to $x$:
$\frac{d}{dx}f(x) = 2 + \frac{1}{(x - 1)^2} + \frac{1}{(x + 2)^2}$.
Comparing this with $A + \frac{B}{(x - 1)^2} + \frac{C}{(x + 2)^2}$,we get $A = 2$,$B = 1$,$C = 1$.
Therefore,$A - B + C = 2 - 1 + 1 = 2$.

(B) Let $f(x) = |x - 1| + |x - 3|$.
For $x$ in the interval $(1, 3)$,we have $x - 1 > 0$ and $x - 3 < 0$.
Thus,$|x - 1| = x - 1$ and $|x - 3| = -(x - 3) = -x + 3$.
Substituting these into the function,we get $f(x) = (x - 1) + (-x + 3) = 2$.
Since $f(x) = 2$ for all $x \in (1, 3)$,the function is constant in this neighborhood of $x = 2$.
The derivative of a constant function is $0$.
Therefore,$f'(2) = 0$.

(B) First,find the point of intersection by setting the curves equal: $3^{x-1} \log x = x^x - 1$. At $x = 1$,$3^0 \log 1 = 1^1 - 1$,which gives $1 \times 0 = 0$. Thus,the curves intersect at $(1, 0)$.
For the first curve $y = 3^{x-1} \log x$,the derivative is $\frac{dy}{dx} = 3^{x-1} \log 3 \log x + 3^{x-1} \cdot \frac{1}{x}$.
At $x = 1$,$m_1 = 3^0 \log 3 \log 1 + 3^0 \cdot \frac{1}{1} = 0 + 1 = 1$.
For the second curve $y = x^x - 1$,let $u = x^x$,then $\log u = x \log x$. Differentiating gives $\frac{1}{u} \frac{du}{dx} = \log x + 1$,so $\frac{du}{dx} = x^x (1 + \log x)$.
Thus,$\frac{dy}{dx} = x^x (1 + \log x)$.
At $x = 1$,$m_2 = 1^1 (1 + \log 1) = 1(1 + 0) = 1$.
Since $m_1 = m_2 = 1$,the tangents are parallel,meaning the angle of intersection $\theta = 0$.
Therefore,$\cos \theta = \cos 0 = 1$.

(C) Given: $2f(\sin x) + f(\cos x) = x$ --- $(1)$
Replace $x$ with $\frac{\pi}{2} - x$:
$2f(\sin(\frac{\pi}{2} - x)) + f(\cos(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$
$2f(\cos x) + f(\sin x) = \frac{\pi}{2} - x$ --- $(2)$
Multiply equation $(2)$ by $2$:
$4f(\cos x) + 2f(\sin x) = \pi - 2x$ --- $(3)$
Subtract equation $(1)$ from equation $(3)$:
$(4f(\cos x) + 2f(\sin x)) - (2f(\sin x) + f(\cos x)) = (\pi - 2x) - x$
$3f(\cos x) = \pi - 3x$
$f(\cos x) = \frac{\pi}{3} - x$
Let $t = \cos x$,then $x = \cos^{-1} t$. Substituting this:
$f(t) = \frac{\pi}{3} - \cos^{-1} t$
Thus,$f(x) = \frac{\pi}{3} - \cos^{-1} x$
Differentiating with respect to $x$:
$f'(x) = \frac{d}{dx}(\frac{\pi}{3} - \cos^{-1} x) = 0 - (-\frac{1}{\sqrt{1 - x^2}}) = \frac{1}{\sqrt{1 - x^2}}$

(B) Since $f(x)$ is a polynomial of degree $3$,we can express it using Taylor expansion about $x=3$ as:
$f(x) = f(3) + f'(3)(x-3) + \frac{f''(3)}{2!}(x-3)^2 + \frac{f'''(3)}{3!}(x-3)^3$
Given $f(3)=1$,$f'(3)=-1$,$f''(3)=0$,and $f'''(3)=12$,we substitute these values:
$f(x) = 1 + (-1)(x-3) + \frac{0}{2}(x-3)^2 + \frac{12}{6}(x-3)^3$
$f(x) = 1 - (x-3) + 2(x-3)^3$
Now,differentiate $f(x)$ with respect to $x$:
$f'(x) = -1 + 2 \times 3(x-3)^2 = -1 + 6(x-3)^2$
To find $f'(1)$,substitute $x=1$:
$f'(1) = -1 + 6(1-3)^2 = -1 + 6(-2)^2 = -1 + 6(4) = -1 + 24 = 23$

(B) The left-hand derivative $(LHD)$ at $x = k$ is defined as $\lim_{h \to 0^+} \frac{f(k-h) - f(k)}{-h}$.
Given $f(x) = \{x\} \sin(\pi x)$,we have $f(k) = \{k\} \sin(\pi k) = 0 \times 0 = 0$.
For a small $h > 0$,$k-h$ lies in the interval $(k-1, k)$,so $\{k-h\} = k-h - (k-1) = 1-h$.
Thus,$f(k-h) = (1-h) \sin(\pi(k-h)) = (1-h) \sin(k\pi - \pi h) = (1-h) (\sin(k\pi)\cos(\pi h) - \cos(k\pi)\sin(\pi h))$.
Since $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$,we get $f(k-h) = (1-h) (0 - (-1)^k \sin(\pi h)) = -(1-h) (-1)^k \sin(\pi h) = (1-h) (-1)^{k+1} \sin(\pi h)$.
Now,$LHD = \lim_{h \to 0^+} \frac{(1-h) (-1)^{k+1} \sin(\pi h) - 0}{-h} = \lim_{h \to 0^+} \frac{(1-h) (-1)^{k+1} \sin(\pi h)}{-h}$.
Using $\lim_{h \to 0} \frac{\sin(\pi h)}{\pi h} = 1$,we get $LHD = (-1)^{k+1} \times (-1) \times \pi = (-1)^{k+2} \pi = (-1)^k \pi$.

(B) Given that $f'(x) > 0$,$f(x)$ is a strictly increasing function.
Given that $g'(x) < 0$,$g(x)$ is a strictly decreasing function.
Since $g(x)$ is strictly decreasing,for any $x_1 < x_2$,we have $g(x_1) > g(x_2)$.
Specifically,for $x < x+1$,we have $g(x) > g(x+1)$.
Since $f(x)$ is strictly increasing,applying $f$ to both sides of the inequality $g(x) > g(x+1)$ preserves the inequality sign.
Therefore,$f(g(x)) > f(g(x+1))$.
Thus,option $(b)$ is correct.

(C) Given $f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3)$.
Let $f'(1) = a$,$f''(2) = b$,and $f'''(3) = c$.
Then $f(x) = x^3 + ax^2 + bx + c$.
Now,find the derivatives:
$f'(x) = 3x^2 + 2ax + b$
$f''(x) = 6x + 2a$
$f'''(x) = 6$
Using the definitions of $a, b, c$:
$c = f'''(3) = 6$.
$b = f''(2) = 6(2) + 2a = 12 + 2a \Rightarrow 2a - b = -12$.
$a = f'(1) = 3(1)^2 + 2a(1) + b = 3 + 2a + b \Rightarrow a + b = -3$.
Adding the two equations: $(2a - b) + (a + b) = -12 - 3 \Rightarrow 3a = -15 \Rightarrow a = -5$.
Substituting $a = -5$ into $a + b = -3$: $-5 + b = -3 \Rightarrow b = 2$.
Thus,$f(x) = x^3 - 5x^2 + 2x + 6$.
Calculating $f(2)$: $f(2) = (2)^3 - 5(2)^2 + 2(2) + 6 = 8 - 20 + 4 + 6 = -2$.

(B) Given $\frac{f'(x)}{f(x)} = 1$ for all $x \in R$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = Ae^x$.
Using the condition $f(1) = 2$,we have $Ae^1 = 2$,so $A = 2e^{-1}$.
Thus,$f(x) = 2e^{-1} \cdot e^x = 2e^{x-1}$.
Consequently,$f'(x) = 2e^{x-1}$.
Given $h(x) = f(f(x))$,by the chain rule,$h'(x) = f'(f(x)) \cdot f'(x)$.
At $x = 1$,$h'(1) = f'(f(1)) \cdot f'(1)$.
Since $f(1) = 2$,we have $h'(1) = f'(2) \cdot f'(1)$.
Substituting the values,$f'(2) = 2e^{2-1} = 2e$ and $f'(1) = 2e^{1-1} = 2$.
Therefore,$h'(1) = (2e) \cdot (2) = 4e$.

(A) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = f'(f(f(x))) \cdot f'(f(x)) \cdot f'(x) + 2f(x) \cdot f'(x)$.
At $x = 1$,we have $f(1) = 1$ and $f'(1) = 3$.
Substituting these values:
$\frac{dy}{dx} = f'(f(f(1))) \cdot f'(f(1)) \cdot f'(1) + 2f(1) \cdot f'(1)$.
Since $f(1) = 1$,this becomes:
$\frac{dy}{dx} = f'(f(1)) \cdot f'(1) \cdot f'(1) + 2(1)(3)$.
$\frac{dy}{dx} = f'(1) \cdot 3 \cdot 3 + 6$.
$\frac{dy}{dx} = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

(A) Given the function $y = x^2 + \cos(2x) + e^{ax}$.
To find the derivative $\frac{dy}{dx}$,we differentiate each term with respect to $x$ using the sum rule:
$\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(\cos(2x)) + \frac{d}{dx}(e^{ax})$.
Using the power rule,$\frac{d}{dx}(x^2) = 2x$.
Using the chain rule for the trigonometric function,$\frac{d}{dx}(\cos(2x)) = -\sin(2x) \cdot \frac{d}{dx}(2x) = -\sin(2x) \cdot 2 = -2\sin(2x)$.
Using the chain rule for the exponential function,$\frac{d}{dx}(e^{ax}) = e^{ax} \cdot \frac{d}{dx}(ax) = e^{ax} \cdot a = ae^{ax}$.
Combining these results,we get $\frac{dy}{dx} = 2x - 2\sin(2x) + ae^{ax}$.
Thus,the correct option is $A$.

(A) Given $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{\sec^2 \alpha}\right)+\csc^2 \alpha}$.
Since $1+\tan^2 \alpha = \sec^2 \alpha$,we have $y(\alpha)=\sqrt{2\left(\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}}{\frac{1}{\cos^2 \alpha}}\right)+\csc^2 \alpha}$.
Simplifying the expression inside the square root: $2\left(\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}\right) \cdot \cos^2 \alpha + \csc^2 \alpha = 2\frac{\cos \alpha}{\sin \alpha} + \csc^2 \alpha = 2\cot \alpha + \csc^2 \alpha$.
Note that $2\cot \alpha + \csc^2 \alpha = 2\cot \alpha + 1 + \cot^2 \alpha = (1+\cot \alpha)^2$.
Thus,$y(\alpha) = \sqrt{(1+\cot \alpha)^2} = |1+\cot \alpha|$.
For $\alpha \in \left(\frac{3\pi}{4}, \pi\right)$,$\cot \alpha < -1$,so $1+\cot \alpha < 0$.
Therefore,$y(\alpha) = -(1+\cot \alpha) = -1 - \cot \alpha$.
Differentiating with respect to $\alpha$: $\frac{dy}{d\alpha} = -(-\csc^2 \alpha) = \csc^2 \alpha$.
At $\alpha = \frac{5\pi}{6}$,$\csc \alpha = \csc \frac{5\pi}{6} = 2$.
So,$\frac{dy}{d\alpha} = (2)^2 = 4$.

(C) Given that $f \circ g$ is the identity function,we have $f(g(x)) = x$ for all $x \in R$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
We are given that $g(a) = b$ and $g^{\prime}(a) = 5$.
Substituting $x = a$ into the differentiated equation:
$f^{\prime}(g(a)) \cdot g^{\prime}(a) = 1$.
Substituting the known values $g(a) = b$ and $g^{\prime}(a) = 5$:
$f^{\prime}(b) \cdot 5 = 1$.
Therefore,$f^{\prime}(b) = 1/5$.

(C) Given $f^{\prime}(x)=\tan^{-1}(\sec x+\tan x)$.
We simplify the expression inside the inverse tangent:
$f^{\prime}(x)=\tan^{-1}\left(\frac{1+\sin x}{\cos x}\right) = \tan^{-1}\left(\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}\right) = \tan^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}\right)$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$f^{\prime}(x)=\tan^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) = \tan^{-1}\left(\tan(\frac{\pi}{4}+\frac{x}{2})\right)$
Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$,we have $0 < \frac{\pi}{4}+\frac{x}{2} < \frac{\pi}{2}$,so $f^{\prime}(x) = \frac{\pi}{4} + \frac{x}{2}$.
Integrating with respect to $x$:
$f(x) = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.
Given $f(0)=0$,we find $C=0$.
Thus,$f(x) = \frac{\pi x + x^2}{4}$.
For $x=1$,$f(1) = \frac{\pi(1) + (1)^2}{4} = \frac{\pi+1}{4}$.

(A) The given function is a composite function $f(x) = v(u(x))$,where $u(x) = x^{2}$ and $v(t) = \sin(t)$.
By the chain rule,the derivative is given by $\frac{df}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx}$.
Here,$\frac{dv}{dt} = \frac{d}{dt}(\sin t) = \cos t$ and $\frac{dt}{dx} = \frac{d}{dx}(x^{2}) = 2x$.
Substituting these values,we get $\frac{df}{dx} = \cos(t) \cdot 2x$.
Replacing $t$ with $x^{2}$,we obtain $\frac{df}{dx} = 2x \cos(x^{2})$.

(A) Let $f(x) = \tan(2 x+3)$.
To find the derivative,we use the chain rule.
Let $u = 2 x+3$. Then $f(x) = \tan(u)$.
The derivative of $\tan(u)$ with respect to $u$ is $\sec^{2}(u)$.
The derivative of $u = 2 x+3$ with respect to $x$ is $\frac{du}{dx} = 2$.
By the chain rule,$\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.
Substituting the values,we get $\frac{df}{dx} = \sec^{2}(u) \cdot 2$.
Replacing $u$ with $2 x+3$,we get $\frac{df}{dx} = 2 \sec^{2}(2 x+3)$.

Let $y = \sin \left(\cos \left(x^{2}\right)\right).$
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left[ \sin \left(\cos \left(x^{2}\right)\right) \right]$
$= \cos \left(\cos \left(x^{2}\right)\right) \cdot \frac{d}{dx} \left( \cos \left(x^{2}\right) \right)$
$= \cos \left(\cos \left(x^{2}\right)\right) \cdot \left( -\sin \left(x^{2}\right) \right) \cdot \frac{d}{dx} \left( x^{2} \right)$
$= \cos \left(\cos \left(x^{2}\right)\right) \cdot (-\sin \left(x^{2}\right)) \cdot (2x)$
$= -2x \sin \left(x^{2}\right) \cos \left(\cos \left(x^{2}\right)\right).$

Let $f(x) = \sin(x^{2}+5)$.
Using the chain rule,we differentiate the outer function $\sin(u)$ and multiply it by the derivative of the inner function $u = x^{2}+5$.
$\frac{d}{dx}[\sin(x^{2}+5)] = \cos(x^{2}+5) \cdot \frac{d}{dx}(x^{2}+5)$.
Since $\frac{d}{dx}(x^{2}) = 2x$ and $\frac{d}{dx}(5) = 0$,we have $\frac{d}{dx}(x^{2}+5) = 2x$.
Therefore,$\frac{d}{dx}[\sin(x^{2}+5)] = \cos(x^{2}+5) \cdot 2x = 2x \cos(x^{2}+5)$.

Let $f(x) = \cos (\sin x)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d}{dx}[\cos (\sin x)] = -\sin (\sin x) \cdot \frac{d}{dx}(\sin x)$
Since $\frac{d}{dx}(\sin x) = \cos x$,we substitute this back:
$= -\sin (\sin x) \cdot \cos x$
$= -\cos x \sin (\sin x)$

Let $f(x) = \sin (ax+b)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d}{dx}[\sin (ax+b)] = \cos (ax+b) \cdot \frac{d}{dx}(ax+b)$
$= \cos (ax+b) \cdot [\frac{d}{dx}(ax) + \frac{d}{dx}(b)]$
$= \cos (ax+b) \cdot (a + 0)$
$= a \cos (ax+b)$

Let $f(x) = \frac{\sin (a x+b)}{\cos (c x+d)}$.
Using the quotient rule $\frac{d}{dx} \left[ \frac{g(x)}{h(x)} \right] = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$,where $g(x) = \sin(ax+b)$ and $h(x) = \cos(cx+d)$.
First,find the derivatives using the chain rule:
$g'(x) = \frac{d}{dx} [\sin(ax+b)] = \cos(ax+b) \cdot \frac{d}{dx}(ax+b) = a \cos(ax+b)$.
$h'(x) = \frac{d}{dx} [\cos(cx+d)] = -\sin(cx+d) \cdot \frac{d}{dx}(cx+d) = -c \sin(cx+d)$.
Now,substitute these into the quotient rule formula:
$f'(x) = \frac{[a \cos(ax+b)] \cdot [\cos(cx+d)] - [\sin(ax+b)] \cdot [-c \sin(cx+d)]}{[\cos(cx+d)]^2}$.
$f'(x) = \frac{a \cos(ax+b) \cos(cx+d) + c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)}$.
This can be simplified as:
$f'(x) = \frac{a \cos(ax+b) \cos(cx+d)}{\cos^2(cx+d)} + \frac{c \sin(ax+b) \sin(cx+d)}{\cos^2(cx+d)}$.
$f'(x) = a \cos(ax+b) \sec(cx+d) + c \sin(ax+b) \tan(cx+d) \sec(cx+d)$.

(N/A) Let $y = \cos(x^{3}) \cdot \sin^{2}(x^{5})$.
Using the product rule $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = \cos(x^{3}) \cdot \frac{d}{dx}[\sin^{2}(x^{5})] + \sin^{2}(x^{5}) \cdot \frac{d}{dx}[\cos(x^{3})]$
Applying the chain rule:
$\frac{dy}{dx} = \cos(x^{3}) \cdot [2 \sin(x^{5}) \cdot \cos(x^{5}) \cdot 5x^{4}] + \sin^{2}(x^{5}) \cdot [-\sin(x^{3}) \cdot 3x^{2}]$
Simplifying the expression:
$\frac{dy}{dx} = 10x^{4} \sin(x^{5}) \cos(x^{5}) \cos(x^{3}) - 3x^{2} \sin(x^{3}) \sin^{2}(x^{5})$

Let $y = 2 \sqrt{\cot \left(x^{2}\right)}$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{2\sqrt{\cot(x^2)}} \cdot \frac{d}{dx}[\cot(x^2)]$
$= \frac{1}{\sqrt{\cot(x^2)}} \cdot [-\csc^2(x^2) \cdot \frac{d}{dx}(x^2)]$
$= \frac{1}{\sqrt{\frac{\cos(x^2)}{\sin(x^2)}}} \cdot [-\frac{1}{\sin^2(x^2)} \cdot 2x]$
$= -\sqrt{\frac{\sin(x^2)}{\cos(x^2)}} \cdot \frac{2x}{\sin^2(x^2)}$
$= -\frac{2x}{\sqrt{\cos(x^2) \sin(x^2) \sin(x^2)}}$
$= -\frac{2x}{\sin(x^2) \sqrt{\sin(x^2)\cos(x^2)}}$
Multiplying numerator and denominator by $\sqrt{2}$:
$= -\frac{2\sqrt{2}x}{\sin(x^2) \sqrt{2\sin(x^2)\cos(x^2)}}$
$= -\frac{2\sqrt{2}x}{\sin(x^2) \sqrt{\sin(2x^2)}}$

Let $f(x) = \cos (\sqrt{x})$.
Using the chain rule,we differentiate the outer function $\cos(u)$ where $u = \sqrt{x}$,and then multiply by the derivative of the inner function $u = \sqrt{x}$.
$\frac{d}{dx} [\cos (\sqrt{x})] = -\sin (\sqrt{x}) \cdot \frac{d}{dx} (\sqrt{x})$
Since $\frac{d}{dx} (\sqrt{x}) = \frac{d}{dx} (x^{1/2}) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}$,we substitute this back into the expression:
$= -\sin (\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$
$= -\frac{\sin (\sqrt{x})}{2\sqrt{x}}$

(A) Given the equation $x-y=\pi$.
Method $1$: Express $y$ in terms of $x$:
$y = x - \pi$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(\pi)$.
Since the derivative of $x$ is $1$ and the derivative of a constant $\pi$ is $0$,we get:
$\frac{dy}{dx} = 1 - 0 = 1$.
Method $2$: Differentiating the given equation $x-y=\pi$ directly with respect to $x$:
$\frac{d}{dx}(x) - \frac{d}{dx}(y) = \frac{d}{dx}(\pi)$.
$1 - \frac{dy}{dx} = 0$.
$\frac{dy}{dx} = 1$.

(A) Let $y = \sin^{-1} x$. Then,$x = \sin y$.
Differentiating both sides with respect to $x$,we get:
$1 = \cos y \frac{dy}{dx}$
This implies that $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\cos(\sin^{-1} x)}$.
Observe that this is defined only for $\cos y \neq 0$,i.e.,$y \neq \pm \frac{\pi}{2}$,which means $x \neq \pm 1$,or $x \in (-1, 1)$.
Since $\cos^2 y = 1 - \sin^2 y = 1 - x^2$ and $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,$\cos y$ must be positive.
Therefore,$\cos y = \sqrt{1 - x^2}$.
Thus,for $x \in (-1, 1)$,$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.

(A) Let $y = \tan^{-1} x$. Then,$x = \tan y$.
Differentiating both sides with respect to $x$,we get:
$1 = \sec^2 y \cdot \frac{dy}{dx}$
This implies that:
$\frac{dy}{dx} = \frac{1}{\sec^2 y}$
Using the trigonometric identity $\sec^2 y = 1 + \tan^2 y$,we have:
$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y}$
Since $x = \tan y$,we substitute $x$ back into the equation:
$\frac{dy}{dx} = \frac{1}{1 + x^2}$

(A) Let $y = e^{-x}$.
Using the chain rule for differentiation,we have:
$\frac{dy}{dx} = \frac{d}{dx}(e^{-x})$
Since the derivative of $e^u$ with respect to $u$ is $e^u$,we apply the chain rule:
$\frac{dy}{dx} = e^{-x} \cdot \frac{d}{dx}(-x)$
$\frac{dy}{dx} = e^{-x} \cdot (-1)$
$\frac{dy}{dx} = -e^{-x}$

Let $y = \sin (\log x).$
Using the chain rule,we have:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{d}{dx}(\log x)$
Since $\frac{d}{dx}(\log x) = \frac{1}{x},$ we get:
$\frac{dy}{dx} = \cos (\log x) \cdot \frac{1}{x} = \frac{\cos (\log x)}{x}$

(N/A) Let $y = \cos^{-1}(e^x).$
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos^{-1}(e^x))$
Since $\frac{d}{dx}(\cos^{-1}(u)) = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$,where $u = e^x$:
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-(e^x)^2}} \cdot \frac{d}{dx}(e^x)$
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-e^{2x}}} \cdot e^x$
$\frac{dy}{dx} = \frac{-e^x}{\sqrt{1-e^{2x}}}$

(B) Let $y = e^{\cos x}$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(e^{\cos x})$
Since $\frac{d}{du}(e^u) = e^u$ and $\frac{d}{dx}(\cos x) = -\sin x$,we apply the chain rule:
$\frac{dy}{dx} = e^{\cos x} \cdot \frac{d}{dx}(\cos x)$
$\frac{dy}{dx} = e^{\cos x} \cdot (-\sin x)$
$\frac{dy}{dx} = -(\sin x) e^{\cos x}$

(A) Let $y = \frac{e^{x}}{\sin x}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}}$,where $u = e^{x}$ and $v = \sin x$:
$\frac{dy}{dx} = \frac{\sin x \cdot \frac{d}{dx}(e^{x}) - e^{x} \cdot \frac{d}{dx}(\sin x)}{(\sin x)^{2}}$
Since $\frac{d}{dx}(e^{x}) = e^{x}$ and $\frac{d}{dx}(\sin x) = \cos x$,we get:
$\frac{dy}{dx} = \frac{\sin x \cdot e^{x} - e^{x} \cdot \cos x}{\sin^{2} x}$
Factoring out $e^{x}$ from the numerator:
$\frac{dy}{dx} = \frac{e^{x}(\sin x - \cos x)}{\sin^{2} x}$,where $x \neq n\pi, n \in Z$.

(A) Let $y = e^{\sin ^{-1} x}$.
Differentiating with respect to $x$,we use the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}(e^{\sin ^{-1} x})$
$\frac{dy}{dx} = e^{\sin ^{-1} x} \cdot \frac{d}{dx}(\sin ^{-1} x)$
Since the derivative of $\sin ^{-1} x$ is $\frac{1}{\sqrt{1-x^2}}$,we have:
$\frac{dy}{dx} = e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$
$\frac{dy}{dx} = \frac{e^{\sin ^{-1} x}}{\sqrt{1-x^2}}$,where $x \in (-1, 1)$.