Derivative at a point, Standard differentiation Questions in English

(D) Given the function $f(x) = x - [x] - \cos x$.
In the neighborhood of $x = \frac{\pi}{2}$,the value of $[x]$ is constant because $[x] = [1.57...] = 1$.
Therefore,for $x$ in an interval around $\frac{\pi}{2}$,we can write $f(x) = x - 1 - \cos x$.
Differentiating with respect to $x$,we get $f^{\prime}(x) = \frac{d}{dx}(x - 1 - \cos x) = 1 - 0 - (-\sin x) = 1 + \sin x$.
Now,substituting $x = \frac{\pi}{2}$ into the derivative:
$f^{\prime}\left(\frac{\pi}{2}\right) = 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2$.

(C) Given the function $f(x) = x - \frac{1}{x}$.
To find the derivative $f^{\prime}(x)$,we differentiate with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1})$
$f^{\prime}(x) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^{2}}$.
Now,substitute $x = -1$ into the derivative:
$f^{\prime}(-1) = 1 + \frac{1}{(-1)^{2}}$
$f^{\prime}(-1) = 1 + \frac{1}{1} = 1 + 1 = 2$.

(B) We know that $\pi^{2} \approx 9.86$.
Since $[\cdot]$ is the greatest integer function:
$[\pi^{2}] = [9.86] = 9$.
$[-\pi^{2}] = [-9.86] = -10$.
Substituting these values into the function $f(x)$:
$f(x) = \sin(9x) + \cos(-10x)$.
Since $\cos(-\theta) = \cos(\theta)$,we have:
$f(x) = \sin(9x) + \cos(10x)$.
Now,differentiating $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(\sin(9x)) + \frac{d}{dx}(\cos(10x))$.
Using the chain rule:
$f'(x) = 9 \cos(9x) - 10 \sin(10x)$.

(C) Given,$f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+\tan x}$.
Simplify the expression:
$f(x) = \frac{\sin^{2} x}{1+\frac{\cos x}{\sin x}} + \frac{\cos^{2} x}{1+\frac{\sin x}{\cos x}}$
$f(x) = \frac{\sin^{3} x}{\sin x + \cos x} + \frac{\cos^{3} x}{\cos x + \sin x}$
$f(x) = \frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x}$
Using the identity $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$:
$f(x) = \frac{(\sin x + \cos x)(\sin^{2} x - \sin x \cos x + \cos^{2} x)}{\sin x + \cos x}$
$f(x) = \sin^{2} x - \sin x \cos x + \cos^{2} x$
$f(x) = 1 - \sin x \cos x = 1 - \frac{1}{2} \sin(2x)$.
Now,differentiate with respect to $x$:
$f^{\prime}(x) = -\frac{1}{2} \cdot \cos(2x) \cdot 2 = -\cos(2x)$.
Substitute $x = \frac{\pi}{4}$:
$f^{\prime}\left(\frac{\pi}{4}\right) = -\cos\left(2 \cdot \frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{2}\right) = 0$.

(D) Given,$y = \sin^{n} x \cos nx$.
Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = \sin^{n} x \frac{d}{dx}(\cos nx) + \cos nx \frac{d}{dx}(\sin^{n} x)$
$\frac{dy}{dx} = \sin^{n} x (-n \sin nx) + \cos nx (n \sin^{n-1} x \cos x)$
$\frac{dy}{dx} = n \sin^{n-1} x (\cos x \cos nx - \sin x \sin nx)$
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$\frac{dy}{dx} = n \sin^{n-1} x \cos (nx + x)$
$\frac{dy}{dx} = n \sin^{n-1} x \cos (n+1) x$

(D) Given the function: $f(x) = \frac{g(x) + g(-x)}{2} + \frac{2}{[h(x) + h(-x)]^{-1}}$.
Simplifying the expression,we get: $f(x) = \frac{g(x) + g(-x)}{2} + 2[h(x) + h(-x)]$.
Differentiating both sides with respect to $x$ using the chain rule:
$f^{\prime}(x) = \frac{g^{\prime}(x) - g^{\prime}(-x)}{2} + 2[h^{\prime}(x) - h^{\prime}(-x)]$.
Now,substitute $x = 0$ into the derivative:
$f^{\prime}(0) = \frac{g^{\prime}(0) - g^{\prime}(0)}{2} + 2[h^{\prime}(0) - h^{\prime}(0)]$.
$f^{\prime}(0) = 0 + 2[0] = 0$.

(A) Given,$f(x^{5}) = 5x^{3}$.
Let $x^{5} = y$,which implies $x = y^{1/5}$.
Then $x^{3} = (y^{1/5})^{3} = y^{3/5}$.
Substituting this into the function,we get $f(y) = 5y^{3/5}$.
Replacing $y$ with $x$,we have $f(x) = 5x^{3/5}$.
Now,differentiating $f(x)$ with respect to $x$:
$f'(x) = 5 \cdot \frac{3}{5} x^{(3/5 - 1)} = 3x^{-2/5}$.
This can be rewritten as $f'(x) = \frac{3}{x^{2/5}} = \frac{3}{\sqrt[5]{x^{2}}}$.

(A) Given the function $f(x) = |\cos x - \sin x|$.
In the neighborhood of $x = \frac{\pi}{6}$,we have $\cos x > \sin x$,so $f(x) = \cos x - \sin x$.
Now,differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x$.
Substitute $x = \frac{\pi}{6}$ into the derivative:
$f^{\prime}\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}\right)$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$,we get:
$f^{\prime}\left(\frac{\pi}{6}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(1 + \sqrt{3})$.

(B) We need to find the derivative of the sum: $\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$.
Using the linearity property of the derivative,we have:
$\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(x^{a}\right)+\frac{d}{d x}\left(a^{x}\right)+\frac{d}{d x}\left(a^{a}\right)$.
$1$. For $\frac{d}{d x}(x^x)$: Let $y = x^x$. Taking $\log$ on both sides,$\log y = x \log x$. Differentiating with respect to $x$,$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{dy}{dx} = x^x(1 + \log x)$.
$2$. For $\frac{d}{d x}(x^a)$: Using the power rule,$\frac{d}{d x}(x^a) = a x^{a-1}$.
$3$. For $\frac{d}{d x}(a^x)$: Using the exponential derivative rule,$\frac{d}{d x}(a^x) = a^x \log a$.
$4$. For $\frac{d}{d x}(a^a)$: Since $a$ is a constant,$a^a$ is a constant,so its derivative is $0$.
Combining these,the result is $x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$.

(B) Let $y = \log _{10} x$ and $z = \log _{x} 10$.
We know that $\log _{a} b = \frac{\ln b}{\ln a}$.
So,$y = \frac{\ln x}{\ln 10}$ and $z = \frac{\ln 10}{\ln x}$.
Thus,$y = \frac{1}{z}$.
We need to find the derivative of $y$ with respect to $z$,which is $\frac{dy}{dz}$.
Since $y = z^{-1}$,differentiating with respect to $z$ gives $\frac{dy}{dz} = -z^{-2} = -\frac{1}{z^{2}}$.
Substituting $z = \frac{1}{y}$ back into the expression,we get $\frac{dy}{dz} = -y^{2}$.
Since $y = \log _{10} x$,the result is $-\left(\log _{10} x\right)^{2}$.

(A) Let $y = e^{ax} \cos bx$.
Using the product rule,the derivative is:
$\frac{dy}{dx} = \frac{d}{dx}(e^{ax}) \cdot \cos bx + e^{ax} \cdot \frac{d}{dx}(\cos bx)$
$\frac{dy}{dx} = ae^{ax} \cos bx - be^{ax} \sin bx$
$\frac{dy}{dx} = e^{ax} (a \cos bx - b \sin bx)$
Let $a = r \cos \alpha$ and $b = r \sin \alpha$.
Then $a^2 + b^2 = r^2(\cos^2 \alpha + \sin^2 \alpha) = r^2$.
Thus,$r = \sqrt{a^2 + b^2}$.
Substituting these into the derivative expression:
$\frac{dy}{dx} = e^{ax} (r \cos \alpha \cos bx - r \sin \alpha \sin bx)$
$\frac{dy}{dx} = re^{ax} (\cos bx \cos \alpha - \sin bx \sin \alpha)$
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:
$\frac{dy}{dx} = re^{ax} \cos(bx + \alpha)$
Therefore,$r = \sqrt{a^2 + b^2}$.

(A) Given,$y = (1 + x^2) \tan^{-1} x - x$.
Applying the product rule for the first term and the power rule for the remaining terms:
$\frac{dy}{dx} = \frac{d}{dx} [(1 + x^2) \tan^{-1} x] - \frac{d}{dx} (x)$
$\frac{dy}{dx} = [(1 + x^2) \cdot \frac{d}{dx}(\tan^{-1} x) + \tan^{-1} x \cdot \frac{d}{dx}(1 + x^2)] - 1$
$\frac{dy}{dx} = [(1 + x^2) \cdot \frac{1}{1 + x^2} + \tan^{-1} x \cdot (2x)] - 1$
$\frac{dy}{dx} = [1 + 2x \tan^{-1} x] - 1$
$\frac{dy}{dx} = 2x \tan^{-1} x$.

(B) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule to differentiate with respect to $x$:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x) \cdot f^{\prime}(x)$.
At $x = 1$:
$\frac{dy}{dx} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1) \cdot f^{\prime}(1)$.
Given $f(1) = 1$ and $f^{\prime}(1) = 3$:
$\frac{dy}{dx} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot 3 + 2(1)(3)$.
Since $f(1) = 1$,then $f(f(1)) = f(1) = 1$.
$\frac{dy}{dx} = f^{\prime}(1) \cdot f^{\prime}(1) \cdot 3 + 6$.
Substituting $f^{\prime}(1) = 3$:
$\frac{dy}{dx} = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

(C) The angular displacement is given by $\theta = \frac{t^{2}}{20} + \frac{t}{5}$.
Angular velocity $\omega$ is the rate of change of angular displacement with respect to time,given by $\omega = \frac{d\theta}{dt}$.
Differentiating $\theta$ with respect to $t$:
$\omega = \frac{d}{dt} \left( \frac{t^{2}}{20} + \frac{t}{5} \right) = \frac{2t}{20} + \frac{1}{5} = \frac{t}{10} + \frac{1}{5}$.
At $t = 4 \ s$,the angular velocity $k$ is:
$k = \left( \frac{4}{10} + \frac{1}{5} \right) = \frac{4}{10} + \frac{2}{10} = \frac{6}{10} = 0.6 \ rad/s$.
We need to find the value of $5k$:
$5k = 5 \times 0.6 = 3$.

(B) Given that $f(x)$ is an even function,we have $f(-x) = f(x)$.
Differentiating both sides with respect to $x$,we get $f^{\prime}(-x) \cdot (-1) = f^{\prime}(x)$.
This implies $f^{\prime}(-x) = -f^{\prime}(x)$,which means $f^{\prime}(x)$ is an odd function.
Substituting $x = e$,we get $f^{\prime}(-e) = -f^{\prime}(e)$.
Therefore,$f^{\prime}(e) + f^{\prime}(-e) = f^{\prime}(e) - f^{\prime}(e) = 0$.

(C) 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 $y^2 + \left(\frac{d y}{d x}\right)^2$:
$y^2 + \left(\frac{d y}{d x}\right)^2 = (a \sin x + b \cos x)^2 + (a \cos x - b \sin x)^2$
Expanding the squares:
$= (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x)$
Grouping the terms:
$= a^2(\sin^2 x + \cos^2 x) + b^2(\sin^2 x + \cos^2 x)$
Since $\sin^2 x + \cos^2 x = 1$:
$= a^2(1) + b^2(1) = a^2 + b^2$
Since $a$ and $b$ are constants,$a^2 + b^2$ is a constant.
Thus,the expression is a constant.

(C) First,simplify the expression inside the limit: $\frac{1}{y-2} \left( \frac{1}{x} - \frac{1}{x+y-2} \right) = \frac{1}{y-2} \left( \frac{(x+y-2) - x}{x(x+y-2)} \right) = \frac{1}{y-2} \left( \frac{y-2}{x(x+y-2)} \right) = \frac{1}{x(x+y-2)}$.
Now,evaluate the limit as $y \to 2$: $\lim_{y \to 2} \frac{1}{x(x+y-2)} = \frac{1}{x(x+2-2)} = \frac{1}{x^2}$.
Finally,differentiate with respect to $x$: $\frac{d}{dx} \left( \frac{1}{x^2} \right) = \frac{d}{dx} (x^{-2}) = -2x^{-3} = \frac{-2}{x^3}$.

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

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

(C) Given the limit expression: $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$.
We can rewrite the numerator by adding and subtracting $m f(m)$:
$\lim _{x \rightarrow m} \frac{x f(m)-m f(m)+m f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$
$\lim _{x \rightarrow m} \left[ f(m) \frac{x-m}{x-m} - m \frac{f(x)-f(m)}{x-m} \right] + f^{\prime}(m) = f(m)$
$f(m) - m f^{\prime}(m) + f^{\prime}(m) = f(m)$
$-m f^{\prime}(m) + f^{\prime}(m) = 0$
$f^{\prime}(m)(1-m) = 0$
Since it is given that $f^{\prime}(m) \neq 0$,we must have $1-m = 0$,which implies $m = 1$.

(D) Given $f(1)=2$,$f(2)=6$ and $f(x+y)=f(x)+kxy+\frac{4}{3}y^2$.
Substitute $x=1$ and $y=1$ into the functional equation:
$f(1+1) = f(1) + k(1)(1) + \frac{4}{3}(1)^2$
$f(2) = f(1) + k + \frac{4}{3}$
$6 = 2 + k + \frac{4}{3}$
$4 = k + \frac{4}{3}$
$k = 4 - \frac{4}{3} = \frac{8}{3}$.
Now,substitute $k = \frac{8}{3}$ into the original equation:
$f(x+y) = f(x) + \frac{8}{3}xy + \frac{4}{3}y^2$.
To find $f(x)$,we use the definition of the derivative:
$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\frac{8}{3}xh + \frac{4}{3}h^2}{h} = \lim_{h \to 0} (\frac{8}{3}x + \frac{4}{3}h) = \frac{8}{3}x$.
Integrating $f'(x) = \frac{8}{3}x$ with respect to $x$:
$f(x) = \int \frac{8}{3}x \, dx = \frac{8}{3} \cdot \frac{x^2}{2} + C = \frac{4}{3}x^2 + C$.
Using $f(1)=2$:
$2 = \frac{4}{3}(1)^2 + C \Rightarrow C = 2 - \frac{4}{3} = \frac{2}{3}$.
Thus,$f(x) = \frac{4}{3}x^2 + \frac{2}{3}$.

(C) Given $y = \log(\sec(\tan^{-1} x))$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{\sec(\tan^{-1} x)} \cdot \frac{d}{dx}(\sec(\tan^{-1} x))$.
Recall that $\frac{d}{dx}(\sec u) = \sec u \tan u \cdot \frac{du}{dx}$.
Let $u = \tan^{-1} x$,then $\frac{du}{dx} = \frac{1}{1+x^2}$.
So,$\frac{dy}{dx} = \frac{1}{\sec(\tan^{-1} x)} \cdot \sec(\tan^{-1} x) \tan(\tan^{-1} x) \cdot \frac{1}{1+x^2}$.
This simplifies to $\frac{dy}{dx} = \tan(\tan^{-1} x) \cdot \frac{1}{1+x^2} = x \cdot \frac{1}{1+x^2} = \frac{x}{1+x^2}$.
At $x = 1$,$\frac{dy}{dx} = \frac{1}{1+(1)^2} = \frac{1}{2}$.

(C) We are given the expression $f(x) = \frac{1+x^2+x^4}{1+x+x^2}$.
First,we simplify the numerator: $1+x^2+x^4 = (1+x^2)^2 - x^2 = (1+x^2-x)(1+x^2+x) = (x^2-x+1)(x^2+x+1)$.
Thus,$f(x) = \frac{(x^2-x+1)(x^2+x+1)}{x^2+x+1} = x^2-x+1$.
Now,we differentiate $f(x)$ with respect to $x$:
$\frac{d}{dx}(x^2-x+1) = 2x-1$.
Comparing this with $ax+b$,we get $a=2$ and $b=-1$.
Therefore,$(a, b) = (2, -1)$.

(D) Given $f(0)=0$ and $f^{\prime}(0)=3$.
Let $y = f(f(f(f(f(x)))))$.
Using the chain rule,the derivative is:
$\frac{dy}{dx} = f^{\prime}(f(f(f(f(x))))) \cdot f^{\prime}(f(f(f(x)))) \cdot f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x)$.
At $x=0$:
Since $f(0)=0$,we have $f(f(0)) = f(0) = 0$,and so on.
Thus,$\left. \frac{dy}{dx} \right|_{x=0} = f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) \cdot f^{\prime}(0) = [f^{\prime}(0)]^5$.
Substituting $f^{\prime}(0)=3$:
$[3]^5 = 243$.

(A) Let $y = \tan^{-1}(\cos \sqrt{x}) + \sec^{-1}(e^x)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(\cos \sqrt{x})) + \frac{d}{dx}(\sec^{-1}(e^x))$
$= \frac{1}{1 + (\cos \sqrt{x})^2} \cdot (-\sin \sqrt{x}) \cdot \frac{1}{2\sqrt{x}} + \frac{1}{|e^x| \sqrt{(e^x)^2 - 1}} \cdot e^x$
$= \frac{-\sin \sqrt{x}}{2\sqrt{x}(1 + \cos^2 \sqrt{x})} + \frac{1}{\sqrt{e^{2x} - 1}}$.
Now,substitute $x = \frac{\pi^2}{4}$,so $\sqrt{x} = \frac{\pi}{2}$:
$\frac{dy}{dx} = \frac{-\sin(\frac{\pi}{2})}{2(\frac{\pi}{2})(1 + \cos^2(\frac{\pi}{2}))} + \frac{1}{\sqrt{e^{2(\frac{\pi^2}{4})} - 1}}$
$= \frac{-1}{\pi(1 + 0)} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}}$
$= \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi}$.

(D) Given the equation: $8 f(x) + 6 f(\frac{1}{x}) = x + 5$ $(1)$
Replacing $x$ with $\frac{1}{x}$ in $(1)$,we get: $8 f(\frac{1}{x}) + 6 f(x) = \frac{1}{x} + 5$ $(2)$
Multiply $(1)$ by $4$ and $(2)$ by $3$:
$32 f(x) + 24 f(\frac{1}{x}) = 4x + 20$
$18 f(x) + 24 f(\frac{1}{x}) = \frac{3}{x} + 15$
Subtracting the two equations: $(32 - 18) f(x) = 4x - \frac{3}{x} + 5$
$14 f(x) = 4x - \frac{3}{x} + 5 \Rightarrow f(x) = \frac{4x^2 + 5x - 3}{14x}$
Now,$x^2 f(x) = x^2 \left( \frac{4x^2 + 5x - 3}{14x} \right) = \frac{4x^3 + 5x^2 - 3x}{14}$
Differentiating with respect to $x$: $\frac{d}{dx} (x^2 f(x)) = \frac{1}{14} (12x^2 + 10x - 3)$
At $x = 1$: $\frac{d}{dx} (x^2 f(x)) = \frac{1}{14} (12(1)^2 + 10(1) - 3) = \frac{12 + 10 - 3}{14} = \frac{19}{14}$

(C) Given $f(x) = |x^2 - 3x + 2|$.
We can factorize the quadratic expression as $f(x) = |(x - 1)(x - 2)|$.
The expression inside the absolute value is positive for $x < 1$ and $x > 2$,and negative for $1 < x < 2$.
Thus,$f(x) = \begin{cases} x^2 - 3x + 2 & \text{if } x \leq 1 \text{ or } x \geq 2 \\ -(x^2 - 3x + 2) & \text{if } 1 < x < 2 \end{cases}$.
Differentiating with respect to $x$:
$f'(x) = \begin{cases} 2x - 3 & \text{if } x < 1 \text{ or } x > 2 \\ -2x + 3 & \text{if } 1 < x < 2 \end{cases}$.
Comparing this with the given options,for $x > 2$,$f'(x) = 2x - 3$,which matches option $C$.

(B) Given functions are $f(x) = \sqrt{x^2 + 1}$,$g(x) = \frac{x + 1}{x^2 + 1}$,and $h(x) = 2x - 3$.
First,we find the derivatives of each function:
$f'(x) = \frac{d}{dx}(\sqrt{x^2 + 1}) = \frac{1}{2\sqrt{x^2 + 1}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 1}}$.
$g'(x) = \frac{d}{dx}(\frac{x + 1}{x^2 + 1}) = \frac{(x^2 + 1)(1) - (x + 1)(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2 - 2x}{(x^2 + 1)^2} = \frac{-x^2 - 2x + 1}{(x^2 + 1)^2}$.
$h'(x) = \frac{d}{dx}(2x - 3) = 2$.
Now,we evaluate the composite expression $f' [h'(g'(x))]$.
Since $h'(x) = 2$ is a constant function,$h'(g'(x)) = 2$ for any value of $x$.
Therefore,$f' [h'(g'(x))] = f'(2)$.
Substituting $x = 2$ into the expression for $f'(x)$:
$f'(2) = \frac{2}{\sqrt{2^2 + 1}} = \frac{2}{\sqrt{4 + 1}} = \frac{2}{\sqrt{5}}$.
Thus,the correct option is $B$.

(B) Given $f(x) = 2x^2 + 3x - 5$.
First,find the derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(2x^2 + 3x - 5) = 4x + 3$.
Now,calculate $f'(0)$:
$f'(0) = 4(0) + 3 = 3$.
Next,calculate $f'(-1)$:
$f'(-1) = 4(-1) + 3 = -4 + 3 = -1$.
Finally,evaluate $f'(0) + 3f'(-1)$:
$f'(0) + 3f'(-1) = 3 + 3(-1) = 3 - 3 = 0$.

(A) Let $y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}}$.
Using the identity $1+\cos 2A = 2 \cos^2 A$,we simplify the expression step by step:
$y = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 4 x)}}}} = \frac{2}{\sqrt{2+\sqrt{2+\sqrt{4 \cos^2 2 x}}}} = \frac{2}{\sqrt{2+\sqrt{2+2 \cos 2 x}}}$.
Continuing the simplification:
$y = \frac{2}{\sqrt{2+\sqrt{2(1+\cos 2 x)}}} = \frac{2}{\sqrt{2+\sqrt{4 \cos^2 x}}} = \frac{2}{\sqrt{2+2 \cos x}} = \frac{2}{\sqrt{2(1+\cos x)}} = \frac{2}{\sqrt{4 \cos^2 \frac{x}{2}}}$.
Thus,$y = \frac{2}{2 \cos \frac{x}{2}} = \sec \frac{x}{2}$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \sec \frac{x}{2} \tan \frac{x}{2} \cdot \frac{1}{2}$.
Comparing this with $k \sec \frac{x}{2} \tan \frac{x}{2}$,we get $k = \frac{1}{2}$.

(A) Let $y = e^{3x} \sin 4x$.
Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = e^{3x} \frac{d}{dx}(\sin 4x) + \sin 4x \frac{d}{dx}(e^{3x})$
$\frac{dy}{dx} = e^{3x} (4 \cos 4x) + \sin 4x (3 e^{3x})$
$\frac{dy}{dx} = e^{3x} (3 \sin 4x + 4 \cos 4x)$
To simplify $3 \sin 4x + 4 \cos 4x$,we multiply and divide by $\sqrt{3^2 + 4^2} = 5$:
$\frac{dy}{dx} = 5 e^{3x} \left( \frac{3}{5} \sin 4x + \frac{4}{5} \cos 4x \right)$
Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$,so $\tan \alpha = \frac{4}{3}$,which means $\alpha = \tan^{-1} \frac{4}{3}$.
Then $\frac{dy}{dx} = 5 e^{3x} (\cos \alpha \sin 4x + \sin \alpha \cos 4x)$
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$\frac{dy}{dx} = 5 e^{3x} \sin(4x + \alpha) = 5 e^{3x} \sin \left(4x + \tan^{-1} \frac{4}{3}\right)$.

(D) Given,$f(x) = \frac{x}{1+x}$.
We need to find $g(x) = f(f(x))$.
$g(x) = f\left(\frac{x}{1+x}\right) = \frac{\frac{x}{1+x}}{1 + \frac{x}{1+x}}$.
Multiplying the numerator and denominator by $(1+x)$,we get:
$g(x) = \frac{x}{1+x+x} = \frac{x}{2x+1}$.
Now,differentiate $g(x)$ with respect to $x$ using the quotient rule $\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$g^{\prime}(x) = \frac{(1)(2x+1) - (x)(2)}{(2x+1)^2}$.
$g^{\prime}(x) = \frac{2x+1 - 2x}{(2x+1)^2} = \frac{1}{(2x+1)^2}$.

(B) Given the equation: $5 f(x) + 3 f\left(\frac{1}{x}\right) = x + 2$ $(1)$
Replace $x$ with $\frac{1}{x}$ in equation $(1)$: $5 f\left(\frac{1}{x}\right) + 3 f(x) = \frac{1}{x} + 2$ $(2)$
Multiply $(1)$ by $5$ and $(2)$ by $3$:
$25 f(x) + 15 f\left(\frac{1}{x}\right) = 5x + 10$
$9 f(x) + 15 f\left(\frac{1}{x}\right) = \frac{3}{x} + 6$
Subtract the second from the first: $(25 - 9) f(x) = 5x - \frac{3}{x} + 4$
$16 f(x) = 5x - \frac{3}{x} + 4 \implies f(x) = \frac{5x^2 + 4x - 3}{16x}$
Given $y = x f(x) = \frac{5x^2 + 4x - 3}{16}$
Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{16} (10x + 4)$
At $x = 1$: $\frac{dy}{dx} = \frac{1}{16} (10(1) + 4) = \frac{14}{16} = \frac{7}{8}$

(A) The greatest integer function $[x]$ is a step function that takes integer values for all $x \in \mathbb{R}$.
For any integer $n$,$[x] = n$ in the interval $[n, n+1)$.
At $x = -1$,we consider the left-hand and right-hand derivatives.
For $x$ in the neighborhood of $-1$,specifically for $x \in [-1, 0)$,$[x] = -1$.
Thus,for $x \in [-1, 0)$,the function $f(x) = \sin(\pi[x]) = \sin(\pi(-1)) = \sin(-\pi) = 0$.
Since the function is constant $(0)$ in the interval $[-1, 0)$,its derivative $\frac{d}{dx} \sin(\pi[x])$ is $0$ at $x = -1$ (considering the right-hand derivative).
Since the function is constant in the interval $[-2, -1)$,$[x] = -2$,so $f(x) = \sin(-2\pi) = 0$.
Therefore,the derivative is $0$.

(C) Given $y = \sin^{98} x \cdot \cos^{39} x$.
Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = \sin^{98} x \cdot \frac{d}{dx}(\cos^{39} x) + \cos^{39} x \cdot \frac{d}{dx}(\sin^{98} x)$
$\frac{dy}{dx} = \sin^{98} x \cdot (39 \cos^{38} x \cdot (-\sin x)) + \cos^{39} x \cdot (98 \sin^{97} x \cdot \cos x)$
$\frac{dy}{dx} = -39 \sin^{99} x \cdot \cos^{38} x + 98 \sin^{97} x \cdot \cos^{40} x$
Rearranging the terms:
$\frac{dy}{dx} = 98 \sin^{97} x \cdot \cos^{40} x - 39 \sin^{99} x \cdot \cos^{38} x$.
Note: The provided options seem to have typographical errors in the exponents. Based on the derivation,the correct expression is $98 \sin^{97} x \cos^{40} x - 39 \sin^{99} x \cos^{38} x$.

(A) Given,$y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$.
Squaring both sides,we get $y^2=2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)$.
Differentiating with respect to $x$:
$2 y \frac{d y}{d x} = 2 + 2 \cos \left(2 x+\frac{\pi}{4}\right) \cdot \left(-\sin \left(2 x+\frac{\pi}{4}\right)\right) \cdot 2$.
Using the identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we have:
$2 y \frac{d y}{d x} = 2 - 2 \sin \left(4 x+\frac{\pi}{2}\right) = 2 - 2 \cos(4x)$.
So,$\frac{d y}{d x} = \frac{1 - \cos(4x)}{y}$.
At $x = \frac{\pi}{4}$,$y = \sqrt{2(\frac{\pi}{4}) + \cos^2(\frac{\pi}{2} + \frac{\pi}{4})} = \sqrt{\frac{\pi}{2} + \sin^2(\frac{\pi}{4})} = \sqrt{\frac{\pi}{2} + \frac{1}{2}} = \sqrt{\frac{\pi+1}{2}}$.
Substituting these values:
$\left. \frac{d y}{d x} \right|_{x=\frac{\pi}{4}} = \frac{1 - \cos(\pi)}{\sqrt{\frac{\pi+1}{2}}} = \frac{1 - (-1)}{\frac{\sqrt{\pi+1}}{\sqrt{2}}} = \frac{2 \sqrt{2}}{\sqrt{\pi+1}}$.

(D) Given the function $f(x) = x^4 - x^3 + 7x^2 + 14$.
To find the derivative $f^{\prime}(x)$,we apply the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f^{\prime}(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(x^3) + 7 \cdot \frac{d}{dx}(x^2) + \frac{d}{dx}(14)$
$f^{\prime}(x) = 4x^3 - 3x^2 + 7(2x) + 0$
$f^{\prime}(x) = 4x^3 - 3x^2 + 14x$
Now,substitute $x = 5$ into the derivative expression:
$f^{\prime}(5) = 4(5)^3 - 3(5)^2 + 14(5)$
$f^{\prime}(5) = 4(125) - 3(25) + 70$
$f^{\prime}(5) = 500 - 75 + 70$
$f^{\prime}(5) = 495$.

(A) Given,$f(x) = \sqrt{x + 2 \sqrt{2x - 4}} + \sqrt{x - 2 \sqrt{2x - 4}}$.
We can rewrite the expression inside the square roots by setting $u = \sqrt{x-2}$,so $x-2 = u^2$ and $x = u^2 + 2$.
Then $2x-4 = 2(x-2) = 2u^2$,so $\sqrt{2x-4} = \sqrt{2}u$.
Substituting these into $f(x)$:
$f(x) = \sqrt{u^2 + 2 + 2\sqrt{2}u} + \sqrt{u^2 + 2 - 2\sqrt{2}u}$
$f(x) = \sqrt{(u + \sqrt{2})^2} + \sqrt{(u - \sqrt{2})^2}$
$f(x) = |u + \sqrt{2}| + |u - \sqrt{2}| = |\sqrt{x-2} + \sqrt{2}| + |\sqrt{x-2} - \sqrt{2}|$.
For $x \geq 4$,$\sqrt{x-2} \geq \sqrt{2}$,so $|\sqrt{x-2} - \sqrt{2}| = \sqrt{x-2} - \sqrt{2}$.
Thus,$f(x) = \sqrt{x-2} + \sqrt{2} + \sqrt{x-2} - \sqrt{2} = 2\sqrt{x-2}$.
Differentiating with respect to $x$:
$f^{\prime}(x) = 2 \times \frac{1}{2\sqrt{x-2}} = \frac{1}{\sqrt{x-2}}$.
Therefore,$10 \times f^{\prime}(102) = 10 \times \frac{1}{\sqrt{102-2}} = 10 \times \frac{1}{\sqrt{100}} = 10 \times \frac{1}{10} = 1$.

(C) We know that $e^{\log _e f(x)} = f(x)$.
Given expression is $\frac{d}{d x}\left(e^{\log _e \sqrt{1+\tan ^2 x}}\right)$.
Using the property,this simplifies to $\frac{d}{d x}\left(\sqrt{1+\tan ^2 x}\right)$.
Since $1+\tan ^2 x = \sec ^2 x$,we have $\sqrt{1+\tan ^2 x} = \sqrt{\sec ^2 x} = |\sec x|$.
Assuming $\sec x > 0$,we differentiate $\sec x$ with respect to $x$.
$\frac{d}{d x}(\sec x) = \sec x \tan x$.

(A) Given the expression $\frac{d}{dx} \left( \frac{x^4 + x^2 + 1}{x^2 + x + 1} \right)$.
First,simplify the fraction inside the derivative:
$\frac{x^4 + x^2 + 1}{x^2 + x + 1} = \frac{(x^2 + x + 1)(x^2 - x + 1)}{x^2 + x + 1} = x^2 - x + 1$.
Now,differentiate the simplified expression with respect to $x$:
$\frac{d}{dx} (x^2 - x + 1) = 2x - 1$.
Comparing $2x - 1$ with $ax + b$,we get $a = 2$ and $b = -1$.
Therefore,$a - b = 2 - (-1) = 2 + 1 = 3$.

(B) Given $f(x) = \frac{x}{1+|x|}$.
To find $f^{\prime}(0)$,we use the definition of the derivative:
$f^{\prime}(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$.
Since $f(0) = \frac{0}{1+|0|} = 0$,we have:
$f^{\prime}(0) = \lim_{h \to 0} \frac{\frac{h}{1+|h|} - 0}{h} = \lim_{h \to 0} \frac{1}{1+|h|}$.
Now,we evaluate the left-hand and right-hand limits:
Left-hand limit $(h \to 0^-)$: $\lim_{h \to 0^-} \frac{1}{1-h} = 1$.
Right-hand limit $(h \to 0^+)$: $\lim_{h \to 0^+} \frac{1}{1+h} = 1$.
Since both limits are equal,$f^{\prime}(0) = 1$.

(B) Given $y = x^x + x^7 + 7^x + 7^7$.
To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$ separately.
$1$. For $x^x$: Let $u = x^x$. Taking $\log$ on both sides,$\log u = x \log x$. Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{du}{dx} = x^x(1 + \log_e x)$.
$2$. For $x^7$: Using the power rule,$\frac{d}{dx}(x^7) = 7x^6$.
$3$. For $7^x$: Using the exponential rule $\frac{d}{dx}(a^x) = a^x \log_e a$,we get $\frac{d}{dx}(7^x) = 7^x \log_e 7$.
$4$. For $7^7$: Since $7^7$ is a constant,its derivative is $0$.
Combining these,$\frac{dy}{dx} = x^x(1 + \log_e x) + 7x^6 + 7^x \log_e 7 + 0$.
Therefore,$\frac{dy}{dx} = x^x(1 + \log_e x) + 7x^6 + 7^x \log_e 7$.

(D) Given $y = \frac{1}{4} \log \left(\frac{1+x}{1-x}\right) - \frac{1}{2} \tan^{-1}(x)$.
Using logarithmic properties,$y = \frac{1}{4} \log(1+x) - \frac{1}{4} \log(1-x) - \frac{1}{2} \tan^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{4} \left(\frac{1}{1+x}\right) - \frac{1}{4} \left(\frac{-1}{1-x}\right) - \frac{1}{2} \left(\frac{1}{1+x^2}\right)$
$\frac{dy}{dx} = \frac{1}{4(1+x)} + \frac{1}{4(1-x)} - \frac{1}{2(1+x^2)}$
$\frac{dy}{dx} = \frac{1}{4} \left(\frac{1-x+1+x}{1-x^2}\right) - \frac{1}{2(1+x^2)} = \frac{1}{4} \left(\frac{2}{1-x^2}\right) - \frac{1}{2(1+x^2)} = \frac{1}{2(1-x^2)} - \frac{1}{2(1+x^2)}$.
At $x = \frac{1}{\sqrt{2}}$,$x^2 = \frac{1}{2}$.
$\left(\frac{dy}{dx}\right)_{x=1/\sqrt{2}} = \frac{1}{2(1-1/2)} - \frac{1}{2(1+1/2)} = \frac{1}{2(1/2)} - \frac{1}{2(3/2)} = 1 - \frac{1}{3} = \frac{2}{3}$.

(D) Given,$y = \frac{e^x \log x}{x^2}$.
Applying the quotient rule for differentiation,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = e^x \log x$ and $v = x^2$.
$\frac{du}{dx} = e^x \log x + e^x \cdot \frac{1}{x} = e^x \left( \log x + \frac{1}{x} \right)$.
$\frac{dv}{dx} = 2x$.
Therefore,$\frac{dy}{dx} = \frac{x^2 \left[ e^x \left( \log x + \frac{1}{x} \right) \right] - (e^x \log x)(2x)}{(x^2)^2}$.
$\frac{dy}{dx} = \frac{x^2 e^x \log x + x e^x - 2x e^x \log x}{x^4}$.
$\frac{dy}{dx} = \frac{x e^x \log x + e^x - 2 e^x \log x}{x^3}$.
$\frac{dy}{dx} = \frac{e^x [x \log x + 1 - 2 \log x]}{x^3}$.
$\frac{dy}{dx} = \frac{e^x [1 + (x - 2) \log x]}{x^3}$.
Thus,option $D$ is correct.

(C) Given $f(x) = \log_{x^2} (\ln x)$.
Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$,we have:
$f(x) = \frac{\ln(\ln x)}{\ln x^2} = \frac{\ln(\ln x)}{2 \ln x}$.
Now,differentiate with respect to $x$ using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:
$f'(x) = \frac{1}{2} \left[ \frac{(\frac{1}{x \ln x})(\ln x) - (\ln(\ln x))(\frac{1}{x})}{(\ln x)^2} \right]$.
$f'(x) = \frac{1}{2} \left[ \frac{\frac{1}{x} - \frac{\ln(\ln x)}{x}}{(\ln x)^2} \right] = \frac{1 - \ln(\ln x)}{2x(\ln x)^2}$.
At $x = e$,$\ln x = 1$ and $\ln(\ln x) = \ln(1) = 0$.
$f'(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}$.

(C) Given that $\frac{f(x)}{g(x)} = c$,where $c$ is a non-zero constant.
Taking the derivative of both sides with respect to $x$,we get $\left(\frac{f(x)}{g(x)}\right)^{\prime} = 0$.
Since $\beta(x) = \left(\frac{f(x)}{g(x)}\right)^{\prime}$,it follows that $\beta(x) = 0$.
Also,from $\frac{f(x)}{g(x)} = c$,we have $f(x) = c \cdot g(x)$.
Differentiating both sides,$f^{\prime}(x) = c \cdot g^{\prime}(x)$,which implies $\frac{f^{\prime}(x)}{g^{\prime}(x)} = c$.
Since $\alpha(x) = \frac{f^{\prime}(x)}{g^{\prime}(x)}$,we have $\alpha(x) = c$.
Now,substituting these values into the expression $\frac{\alpha(x) - \beta(x)}{\alpha(x) + \beta(x)}$,we get $\frac{c - 0}{c + 0} = \frac{c}{c} = 1$.

(C) Given $y = \cos(x^{\circ})$ and $z = \cos x$.
Since $x^{\circ} = \frac{\pi x}{180}$ radians,we have $y = \cos\left(\frac{\pi x}{180}\right)$.
Differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = -\sin\left(\frac{\pi x}{180}\right) \cdot \frac{\pi}{180} = -\frac{\pi}{180} \sin(x^{\circ})$.
Differentiating $z$ with respect to $x$:
$\frac{dz}{dx} = -\sin x$.
Now,using the chain rule:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-\frac{\pi}{180} \sin(x^{\circ})}{-\sin x} = \frac{\pi}{180} \sin(x^{\circ}) \operatorname{cosec} x$.

(D) Let us evaluate each option:
$A$: $\frac{d}{dx}[\sec^{-1}(\cosh x)] = \frac{1}{|\cosh x|\sqrt{\cosh^2 x - 1}} \cdot \sinh x = \frac{\sinh x}{\cosh x \sinh x} = \text{sech } x$. (True)
$B$: $\frac{d}{dx}[\cos^{-1}(\text{sech } x)] = -\frac{1}{\sqrt{1-\text{sech}^2 x}} \cdot (-\text{sech } x \tanh x) = \frac{\text{sech } x \tanh x}{\tanh x} = \text{sech } x$. (True)
$C$: $\frac{d}{dx}[\tan^{-1}(\sinh x)] = \frac{1}{1+\sinh^2 x} \cdot \cosh x = \frac{\cosh x}{\cosh^2 x} = \text{sech } x$. (True)
$D$: $\frac{d}{dx}[\tan^{-1}(\tan \frac{x}{2})] = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$. Since $\frac{1}{2} \neq \sec x$,this statement is false.