Derivative at a point, Standard differentiation Questions in English

(D) Given $f(x) = \log(\sec x + \tan x)$.
To find $f^{\prime}(x)$,we differentiate with respect to $x$ using the chain rule:
$f^{\prime}(x) = \frac{d}{dx} [\log(\sec x + \tan x)]$
$f^{\prime}(x) = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x)$
$f^{\prime}(x) = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)$
Factor out $\sec x$ from the numerator:
$f^{\prime}(x) = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x}$
$f^{\prime}(x) = \sec x$
Now,evaluate at $x = \frac{\pi}{4}$:
$f^{\prime}\left(\frac{\pi}{4}\right) = \sec\left(\frac{\pi}{4}\right) = \sqrt{2}$.

(B) We know that $\Delta = E - 1$,so $\Delta^{2} = (E - 1)^{2} = E^{2} - 2E + 1$.
Given expression is $\left(\frac{\Delta^{2}}{E}\right) x^{3} = \left(\frac{E^{2} - 2E + 1}{E}\right) x^{3}$.
This simplifies to $(E - 2 + E^{-1}) x^{3}$.
Applying the operators: $E(x^{3}) = (x+1)^{3}$,$-2(x^{3}) = -2x^{3}$,and $E^{-1}(x^{3}) = (x-1)^{3}$.
Summing these: $(x+1)^{3} - 2x^{3} + (x-1)^{3}$.
Expanding the terms: $(x^{3} + 3x^{2} + 3x + 1) - 2x^{3} + (x^{3} - 3x^{2} + 3x - 1)$.
Combining like terms: $(x^{3} - 2x^{3} + x^{3}) + (3x^{2} - 3x^{2}) + (3x + 3x) + (1 - 1) = 6x$.

(D) The given series is the Maclaurin series expansion for the exponential function $e^{x}$.
$y = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots = e^{x}$.
Now,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(e^{x})$.
Since the derivative of $e^{x}$ is $e^{x}$,we have:
$\frac{dy}{dx} = e^{x}$.
Substituting the original expression for $y$ back into the result,we get:
$\frac{dy}{dx} = y$.

(D) Given $y = (x^x)^x$.
Using the exponent rule $(a^m)^n = a^{mn}$,we have $y = x^{x \cdot x} = x^{x^2}$.
Taking the natural logarithm on both sides: $\log y = \log(x^{x^2}) = x^2 \log x$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^2 \log x)$.
Using the product rule:
$\frac{1}{y} \frac{dy}{dx} = x^2 \cdot \frac{1}{x} + (\log x) \cdot (2x) = x + 2x \log x$.
$\frac{dy}{dx} = y(x + 2x \log x) = x^{x^2} \cdot x(1 + 2 \log x)$.
Thus,$\frac{dy}{dx} = x \cdot x^{x^2}(1 + 2 \log x)$.

(A) Given,$y=x^{n} \log x+x(\log x)^{n}$.
Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ to both terms:
$\frac{dy}{dx} = \frac{d}{dx}(x^n \log x) + \frac{d}{dx}(x(\log x)^n)$
$= (x^n \cdot \frac{1}{x} + \log x \cdot nx^{n-1}) + (x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} + (\log x)^n \cdot 1)$
$= (x^{n-1} + nx^{n-1} \log x) + (n(\log x)^{n-1} + (\log x)^n)$
$= x^{n-1}(1 + n \log x) + (\log x)^{n-1}(n + \log x)$.

(C) Let $u = e^x$ and $v = \sqrt{x}$.
Differentiating $u$ and $v$ with respect to $x$,we get:
$\frac{du}{dx} = e^x$
$\frac{dv}{dx} = \frac{1}{2\sqrt{x}}$
Using the chain rule,the derivative of $u$ with respect to $v$ is given by:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{e^x}{1/(2\sqrt{x})} = 2\sqrt{x} e^x$.

(A) Given $y = \tan^{-1}\left(\frac{5x-x}{1+5x \cdot x}\right) + \cot^{-1}\left(\frac{\frac{3}{2}-x}{1+\frac{3}{2}x}\right)$.
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we have:
$y = (\tan^{-1}(5x) - \tan^{-1}(x)) + (\cot^{-1}(x) - \cot^{-1}(\frac{3}{2}))$.
Since $\cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x)$,we substitute:
$y = \tan^{-1}(5x) - \tan^{-1}(x) + \frac{\pi}{2} - \tan^{-1}(x) - \cot^{-1}(\frac{3}{2})$.
$y = \tan^{-1}(5x) - 2\tan^{-1}(x) + \frac{\pi}{2} - \cot^{-1}(\frac{3}{2})$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(5x)) - 2\frac{d}{dx}(\tan^{-1}(x)) + 0 - 0$.
$\frac{dy}{dx} = \frac{5}{1+(5x)^2} - \frac{2}{1+x^2} = \frac{5}{1+25x^2} - \frac{2}{1+x^2}$.
Note: Re-evaluating the expression $\cot^{-1}\left(\frac{3-2x}{2+3x}\right) = \tan^{-1}\left(\frac{2+3x}{3-2x}\right) = \tan^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3}x}\right) = \tan^{-1}(\frac{2}{3}) + \tan^{-1}(x)$.
Thus,$y = \tan^{-1}(5x) - \tan^{-1}(x) + \tan^{-1}(\frac{2}{3}) + \tan^{-1}(x) = \tan^{-1}(5x) + \tan^{-1}(\frac{2}{3})$.
$\frac{dy}{dx} = \frac{5}{1+25x^2}$.

(A) Given $f^{\prime}(x) = \tan ^{-1}(\sec x + \tan x)$.
Simplifying the expression inside the inverse tangent:
$f^{\prime}(x) = \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$
Using half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$:
$f^{\prime}(x) = \tan ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}\right]$
$f^{\prime}(x) = \tan ^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}\right)$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$f^{\prime}(x) = \tan ^{-1}\left(\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}\right) = \tan ^{-1}\left[\tan \left(\frac{\pi}{4} + \frac{x}{2}\right)\right]$
Thus,$f^{\prime}(x) = \frac{\pi}{4} + \frac{x}{2}$.
Integrating with respect to $x$:
$f(x) = \int \left(\frac{\pi}{4} + \frac{x}{2}\right) dx = \frac{\pi x}{4} + \frac{x^2}{4} + C$.
Given $f(0) = 0$,we get $C = 0$.
So,$f(x) = \frac{\pi x + x^2}{4}$.
Therefore,$f(1) = \frac{\pi(1) + (1)^2}{4} = \frac{\pi+1}{4}$.

(C) Given $f^{\prime}(x) = \tan^{-1}(\sec x + \tan x) = \tan^{-1}\left(\frac{1+\sin x}{\cos x}\right)$.
Using trigonometric identities,$\frac{1+\sin x}{\cos x} = \frac{(\cos(x/2) + \sin(x/2))^2}{\cos^2(x/2) - \sin^2(x/2)} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} = \tan(\frac{\pi}{4} + \frac{x}{2})$.
Thus,$f^{\prime}(x) = \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$.
Now,$f(x) = \int f^{\prime}(x) dx = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.
Since $f(0) = 0$,we have $0 = 0 + 0 + C$,so $C = 0$.
Therefore,$f(x) = \frac{x^2 + \pi x}{4}$.
Substituting $x = 1$,we get $f(1) = \frac{1^2 + \pi(1)}{4} = \frac{\pi+1}{4}$.

(B) Let $f(x) = \cot^{-1}(\sqrt{\cos 2x})$.
Using the chain rule,the derivative is $f'(x) = \frac{-1}{1 + (\sqrt{\cos 2x})^2} \times \frac{d}{dx}(\sqrt{\cos 2x})$.
$f'(x) = \frac{-1}{1 + \cos 2x} \times \frac{1}{2\sqrt{\cos 2x}} \times (-\sin 2x \times 2)$.
$f'(x) = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}$.
At $x = \frac{\pi}{6}$,$2x = \frac{\pi}{3}$.
$f'\left(\frac{\pi}{6}\right) = \frac{\sin(\pi/3)}{(1 + \cos(\pi/3))\sqrt{\cos(\pi/3)}}$.
$f'\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}/2}{(1 + 1/2)\sqrt{1/2}} = \frac{\sqrt{3}/2}{(3/2)(1/\sqrt{2})} = \frac{\sqrt{3}}{2} \times \frac{2}{3} \times \sqrt{2} = \frac{\sqrt{6}}{3} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}} = \left(\frac{2}{3}\right)^{1/2}$.

(D) Given $y = \sec(\tan^{-1} x)$.
Using the chain rule,$\frac{dy}{dx} = \sec(\tan^{-1} x) \tan(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x)$.
Since $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$,we have $\frac{dy}{dx} = \sec(\tan^{-1} x) \cdot x \cdot \frac{1}{1+x^2}$.
At $x = 1$,$\tan^{-1}(1) = \frac{\pi}{4}$.
So,$\frac{dy}{dx} = \sec(\frac{\pi}{4}) \cdot 1 \cdot \frac{1}{1+1^2} = \sqrt{2} \cdot \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Alternatively,let $\tan^{-1} x = \theta$,then $\tan \theta = x$. Since $\sec^2 \theta = 1 + \tan^2 \theta$,we have $\sec \theta = \sqrt{1+x^2}$.
Thus,$y = \sqrt{1+x^2}$.
Then $\frac{dy}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$.
At $x = 1$,$\frac{dy}{dx} = \frac{1}{\sqrt{1+1^2}} = \frac{1}{\sqrt{2}}$.

(D) Let $y = \sqrt{\frac{1-\tan x}{1+\tan x}}$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1-\tan x}{1+\tan x}}} \cdot \frac{d}{dx}\left(\frac{1-\tan x}{1+\tan x}\right)$.
Applying the quotient rule to the inner term:
$\frac{d}{dx}\left(\frac{1-\tan x}{1+\tan x}\right) = \frac{(1+\tan x)(-\sec^2 x) - (1-\tan x)(\sec^2 x)}{(1+\tan x)^2} = \frac{-\sec^2 x - \tan x \sec^2 x - \sec^2 x + \tan x \sec^2 x}{(1+\tan x)^2} = \frac{-2\sec^2 x}{(1+\tan x)^2}$.
Now,substitute this back:
$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1+\tan x}{1-\tan x}} \cdot \left(\frac{-2\sec^2 x}{(1+\tan x)^2}\right) = \frac{-\sec^2 x}{\sqrt{1-\tan x} \cdot (1+\tan x)^{3/2}}$.

(B) Given $y = \cos(\sin x^2)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \frac{d}{dx}(\sin x^2)$
$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \cos(x^2) \cdot 2x$
Now,substitute $x = \sqrt{\frac{\pi}{2}}$ into the derivative:
$\frac{dy}{dx} = -\sin(\sin(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) \cdot 2\sqrt{\frac{\pi}{2}}$
Since $\cos(\frac{\pi}{2}) = 0$,the entire expression becomes:
$\frac{dy}{dx} = -\sin(1) \cdot 0 \cdot 2\sqrt{\frac{\pi}{2}} = 0$.

(C) Given $f(x) = 3x^2 + 2x f'(1) + f''(2)$.
Let $f'(1) = a$ and $f''(2) = b$.
Then $f(x) = 3x^2 + 2ax + b$.
Differentiating with respect to $x$,we get $f'(x) = 6x + 2a$.
At $x = 1$,$f'(1) = 6(1) + 2a = 6 + 2a$.
Since $f'(1) = a$,we have $a = 6 + 2a$,which implies $a = -6$.
Now,differentiating $f'(x) = 6x + 2a$ again,we get $f''(x) = 6$.
Thus,$f''(2) = 6$.
Since $f''(2) = b$,we have $b = 6$.
Substituting $a = -6$ and $b = 6$ into the expression for $f(x)$:
$f(x) = 3x^2 + 2(-6)x + 6 = 3x^2 - 12x + 6$.

(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$.
Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$.
Differentiating again,we get $f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$.
Substituting $x=1$ in the expression for $f^{\prime}(x)$:
$f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2) \Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$ (Equation $I$).
Substituting $x=2$ in the expression for $f^{\prime \prime}(x)$:
$f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1) \Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$ (Equation $II$).
Substituting $f^{\prime \prime}(2)$ from Equation $II$ into Equation $I$:
$f^{\prime}(1)+12+2 f^{\prime}(1)=-3 \Rightarrow 3 f^{\prime}(1)=-15 \Rightarrow f^{\prime}(1)=-5$.
Now,find $f^{\prime \prime}(2)$ using Equation $II$:
$f^{\prime \prime}(2)=12+2(-5)=12-10=2$.
Finally,substitute $f^{\prime}(1)$ and $f^{\prime \prime}(2)$ into the original function $f(x)$ at $x=2$:
$f(2)=2^3+2^2(-5)+2(2)+6 = 8-20+4+6 = -2$.

(D) Given $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5 \dots (i)$
Replacing $x$ with $\frac{1}{x}$,we get $8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5 \dots (ii)$
Multiply $(i)$ by $4$ and $(ii)$ by $3$:
$32 f(x)+24 f\left(\frac{1}{x}\right)=4x+20 \dots (iii)$
$18 f(x)+24 f\left(\frac{1}{x}\right)=\frac{3}{x}+15 \dots (iv)$
Subtracting $(iv)$ from $(iii)$:
$14 f(x)=4x-\frac{3}{x}+5 \implies f(x)=\frac{1}{14}\left(4x-\frac{3}{x}+5\right)$
Then $f'(x)=\frac{1}{14}\left(4+\frac{3}{x^2}\right)$
At $x=-1$:
$f(-1)=\frac{1}{14}(-4+3+5)=\frac{4}{14}=\frac{2}{7}$
$f'(-1)=\frac{1}{14}(4+3)=\frac{7}{14}=\frac{1}{2}$
Given $y=x^2 f(x)$,differentiating w.r.t $x$:
$\frac{dy}{dx}=2x f(x)+x^2 f'(x)$
At $x=-1$:
$\left.\frac{dy}{dx}\right|_{x=-1}=2(-1)f(-1)+(-1)^2 f'(-1)$
$= -2\left(\frac{2}{7}\right) + 1\left(\frac{1}{2}\right) = -\frac{4}{7} + \frac{1}{2} = \frac{-8+7}{14} = -\frac{1}{14}$

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

(C) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x) \cdot f^{\prime}(x)$.
At $x=1$,we have:
$\left(\frac{dy}{dx}\right)_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1) \cdot f^{\prime}(1)$.
Given $f(1)=1$ and $f^{\prime}(1)=5$,we substitute these values:
$= f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot 5 + 2(1)(5)$.
Since $f(1)=1$,then $f(f(1)) = f(1) = 1$.
$= f^{\prime}(1) \cdot f^{\prime}(1) \cdot 5 + 10$.
$= 5 \cdot 5 \cdot 5 + 10 = 125 + 10 = 135$.

(B) Given the function $f(x) = e^{x} g(x)$.
Applying the product rule for differentiation,we have $f^{\prime}(x) = \frac{d}{dx}(e^{x}) \cdot g(x) + e^{x} \cdot \frac{d}{dx}(g(x))$.
This simplifies to $f^{\prime}(x) = e^{x} g(x) + e^{x} g^{\prime}(x) = e^{x} (g(x) + g^{\prime}(x))$.
Now,substitute $x = 0$ into the derivative expression:
$f^{\prime}(0) = e^{0} (g(0) + g^{\prime}(0))$.
Given $g(0) = 4$ and $g^{\prime}(0) = 2$,and knowing $e^{0} = 1$:
$f^{\prime}(0) = 1 \cdot (4 + 2) = 6$.

(C) We know that the shift operator $E$ is defined as $E = 1 + \Delta$,where $\Delta$ is the forward difference operator.
Thus,$(1+\Delta)^{n} f(a) = E^{n} f(a)$.
By the definition of the shift operator,$E^{n} f(a) = f(a+nh)$.
Therefore,$(1+\Delta)^{n} f(a) = f(a+nh)$.

(B) Given $f(x) = e^{x} g(x)$.
Applying the product rule for differentiation,we get:
$f^{\prime}(x) = e^{x} g^{\prime}(x) + e^{x} g(x)$.
Substituting $x = 0$:
$f^{\prime}(0) = e^{0} \cdot g^{\prime}(0) + e^{0} \cdot g(0)$.
Since $e^{0} = 1$,$g^{\prime}(0) = 1$,and $g(0) = 2$:
$f^{\prime}(0) = 1 \cdot 1 + 1 \cdot 2$.
$f^{\prime}(0) = 1 + 2 = 3$.

(A) We know that $\Delta \log f(x) = \log f(x+h) - \log f(x) = \log \left[\frac{f(x+h)}{f(x)}\right]$.
Since $f(x+h) = (1 + \Delta) f(x)$,we have $\Delta \log f(x) = \log \left[\frac{(1 + \Delta) f(x)}{f(x)}\right] = \log \left[1 + \frac{\Delta f(x)}{f(x)}\right]$.
Next,for the second term,$\Delta^{2}(3^{x}) = (E - 1)^{2} 3^{x} = (E^{2} - 2E + 1) 3^{x}$.
$= E^{2}(3^{x}) - 2E(3^{x}) + 3^{x} = 3^{x+2} - 2 \cdot 3^{x+1} + 3^{x}$.
$= 3^{x}(9 - 6 + 1) = 4 \cdot 3^{x}$.
Combining both terms,the result is $\log \left[1 + \frac{\Delta f(x)}{f(x)}\right] + 4 \cdot 3^{x}$.

(A) Given $f^{\prime}(x)=k(\cos x-\sin x)$.
Substitute $x=0$ into the derivative: $f^{\prime}(0)=k(\cos 0-\sin 0)=k(1-0)=k$.
Since $f^{\prime}(0)=3$,we have $k=3$.
Now,integrate $f^{\prime}(x)$ to find $f(x)$:
$f(x)=\int 3(\cos x-\sin x) \, dx = 3(\sin x+\cos x)+C$.
Use the condition $f\left(\frac{\pi}{2}\right)=15$ to find $C$:
$f\left(\frac{\pi}{2}\right)=3(\sin \frac{\pi}{2}+\cos \frac{\pi}{2})+C = 3(1+0)+C = 3+C$.
Setting $3+C=15$,we get $C=12$.
Therefore,$f(x)=3(\sin x+\cos x)+12$.

(C) The expression $\lim _{x \rightarrow 1} \frac{G(x)-G(1)}{x-1}$ represents the derivative of $G(x)$ at $x=1$,denoted as $G'(1)$.
Given $G(x) = -\sqrt{25-x^{2}}$.
Using the chain rule,$G'(x) = -\frac{1}{2\sqrt{25-x^{2}}} \cdot (-2x) = \frac{x}{\sqrt{25-x^{2}}}$.
Substituting $x=1$:
$G'(1) = \frac{1}{\sqrt{25-(1)^{2}}} = \frac{1}{\sqrt{24}}$.

(C) Let $f(x) = x^x + x^{x+1} + x^{x+2}$.
We know that $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$.
For $x^{x+1}$,let $y = x^{x+1}$,then $\ln y = (x+1)\ln x$. Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+1}{x} = \ln x + 1 + \frac{1}{x}$. So,$\frac{d}{dx}(x^{x+1}) = x^{x+1}(\ln x + 1 + \frac{1}{x})$.
For $x^{x+2}$,let $y = x^{x+2}$,then $\ln y = (x+2)\ln x$. Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+2}{x} = \ln x + 1 + \frac{2}{x}$. So,$\frac{d}{dx}(x^{x+2}) = x^{x+2}(\ln x + 1 + \frac{2}{x})$.
Now,evaluate at $x=e$:
$\frac{d}{dx}(x^x)|_{x=e} = e^e(1 + \ln e) = e^e(1+1) = 2e^e$.
$\frac{d}{dx}(x^{x+1})|_{x=e} = e^{e+1}(\ln e + 1 + \frac{1}{e}) = e^{e+1}(2 + \frac{1}{e}) = 2e^{e+1} + e^e = e^e(2e+1)$.
$\frac{d}{dx}(x^{x+2})|_{x=e} = e^{e+2}(\ln e + 1 + \frac{2}{e}) = e^{e+2}(2 + \frac{2}{e}) = 2e^{e+2} + 2e^{e+1} = e^e(2e^2+2e)$.
Summing these: $e^e(2 + 2e + 1 + 2e^2 + 2e) = e^e(2e^2 + 4e + 3)$.

(A) For $x \in [1.2, 1.9]$,the greatest integer function $[x]$ is constant because $1 \le x < 2$.
Specifically,for any $x$ in the interval $[1.2, 1.9]$,$[x] = 1$.
Since $f(x) = 1$ is a constant function on the interval $[1.2, 1.9]$,it is continuous and differentiable at every point in the interval.
The derivative of a constant function is zero,so $f'(x) = 0$ for all $x \in [1.2, 1.9]$.
Therefore,option $A$ is correct.

(C) Let $y = \operatorname{cosec}^{-1}(e^x)$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(\operatorname{cosec}^{-1}(e^x))$
We know that $\frac{d}{d u}(\operatorname{cosec}^{-1} u) = \frac{-1}{|u| \sqrt{u^2 - 1}}$.
Here,$u = e^x$,so $\frac{d u}{d x} = e^x$.
Applying the chain rule:
$\frac{d y}{d x} = \frac{-1}{|e^x| \sqrt{(e^x)^2 - 1}} \cdot \frac{d}{d x}(e^x)$
Since $e^x > 0$ for all real $x$,$|e^x| = e^x$.
$\frac{d y}{d x} = \frac{-1}{e^x \sqrt{e^{2 x} - 1}} \cdot e^x$
$\frac{d y}{d x} = \frac{-1}{\sqrt{e^{2 x} - 1}}$
Therefore,the correct option is $C$.

(D) Given $f(x) = 4x^3 + 3x^2 + 3x + 4$.
First,find $f\left(\frac{1}{x}\right)$:
$f\left(\frac{1}{x}\right) = 4\left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2 + 3\left(\frac{1}{x}\right) + 4 = \frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4$.
Now,multiply by $x^3$:
$x^3 \cdot f\left(\frac{1}{x}\right) = x^3 \left(\frac{4}{x^3} + \frac{3}{x^2} + \frac{3}{x} + 4\right) = 4 + 3x + 3x^2 + 4x^3$.
Finally,differentiate with respect to $x$:
$\frac{d}{dx}(4 + 3x + 3x^2 + 4x^3) = 0 + 3 + 6x + 12x^2 = 12x^2 + 6x + 3$.
Thus,the correct option is $D$.

(B) Given the function $f(x) = 1 + x + x^2 + \ldots + x^{1000}$.
This is a geometric series with $1001$ terms,where the first term $a = 1$ and the common ratio $r = x$.
The sum of the series is $f(x) = \frac{x^{1001} - 1}{x - 1}$ for $x \neq 1$.
To find $f^{\prime}(x)$,we use the quotient rule: $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$.
Here $u = x^{1001} - 1$ and $v = x - 1$,so $u^{\prime} = 1001x^{1000}$ and $v^{\prime} = 1$.
$f^{\prime}(x) = \frac{(1001x^{1000})(x - 1) - (x^{1001} - 1)(1)}{(x - 1)^2}$.
Now,substitute $x = -1$:
$f^{\prime}(-1) = \frac{(1001(-1)^{1000})(-1 - 1) - ((-1)^{1001} - 1)}{(-1 - 1)^2}$.
$f^{\prime}(-1) = \frac{(1001 \times 1)(-2) - (-1 - 1)}{(-2)^2}$.
$f^{\prime}(-1) = \frac{-2002 - (-2)}{4} = \frac{-2002 + 2}{4} = \frac{-2000}{4} = -500$.

(C) We know the trigonometric identity $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$.
Given expression is $E = 3 \cos \left(\frac{\pi}{6}+x^{\circ}\right)-4 \cos ^3\left(\frac{\pi}{6}+x^{\circ}\right)$.
This can be rewritten as $E = - (4 \cos ^3\left(\frac{\pi}{6}+x^{\circ}\right) - 3 \cos \left(\frac{\pi}{6}+x^{\circ}\right))$.
Using the identity,$E = - \cos \left(3 \left(\frac{\pi}{6}+x^{\circ}\right)\right) = - \cos \left(\frac{\pi}{2} + 3x^{\circ}\right)$.
Since $\cos(\frac{\pi}{2} + \theta) = -\sin(\theta)$,we have $E = -(-\sin(3x^{\circ})) = \sin(3x^{\circ})$.
Now,we need to differentiate $E$ with respect to $x$.
First,convert $x^{\circ}$ to radians: $x^{\circ} = x \times \frac{\pi}{180}$ radians.
So,$E = \sin\left(3 \times \frac{\pi x}{180}\right) = \sin\left(\frac{\pi x}{60}\right)$.
Now,$\frac{d}{dx} \sin\left(\frac{\pi x}{60}\right) = \cos\left(\frac{\pi x}{60}\right) \times \frac{d}{dx} \left(\frac{\pi x}{60}\right) = \frac{\pi}{60} \cos\left(\frac{\pi x}{60}\right)$.
Converting back to degrees,$\frac{\pi x}{60} = 3x^{\circ}$.
Thus,the derivative is $\frac{\pi}{60} \cos(3x^{\circ})$.
Therefore,the correct option is $C$.

(B) Let $y = \sqrt{3} \sin \left(2x + \frac{\pi}{3} \right) + \cos \left(2x + \frac{\pi}{3} \right)$.
We can rewrite this expression by multiplying and dividing by $2$:
$y = 2 \left( \frac{\sqrt{3}}{2} \sin \left(2x + \frac{\pi}{3} \right) + \frac{1}{2} \cos \left(2x + \frac{\pi}{3} \right) \right)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,where $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$:
$y = 2 \left( \sin \left(2x + \frac{\pi}{3} \right) \cos \left(\frac{\pi}{6} \right) + \cos \left(2x + \frac{\pi}{3} \right) \sin \left(\frac{\pi}{6} \right) \right)$.
$y = 2 \sin \left(2x + \frac{\pi}{3} + \frac{\pi}{6} \right) = 2 \sin \left(2x + \frac{\pi}{2} \right)$.
Since $\sin(\theta + \frac{\pi}{2}) = \cos \theta$,we have $y = 2 \cos(2x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (2 \cos 2x) = 2 \times (- \sin 2x) \times 2 = -4 \sin 2x$.
Thus,the correct option is $B$.

(A) The given series is the expansion of the exponential function $y = e^x$.
We know that $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$.
Therefore,$y = e^x$.
Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$.
Since $y = e^x$,it follows that $\frac{dy}{dx} = y$.
Thus,the correct option is $A$.

(A) We know that the trigonometric identity for $\sin 2x$ is given by:
$\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$
Substituting this into the expression,we get:
$\frac{d}{dx} (\sin 2x)$
Using the chain rule for differentiation,the derivative of $\sin(ax)$ is $a \cos(ax)$:
$\frac{d}{dx} (\sin 2x) = 2 \cos 2x$
Therefore,the correct option is $A$.

(C) First,convert the angle from degrees to radians: $x^{\circ} = x \times \frac{\pi}{180} \text{ radians}$.
Thus,$\sec(x^{\circ}) = \sec\left(\frac{\pi x}{180}\right)$.
Let $f(x) = \sec\left(\frac{\pi x}{180}\right)$.
Using the chain rule,the derivative is $f'(x) = \sec\left(\frac{\pi x}{180}\right) \tan\left(\frac{\pi x}{180}\right) \times \frac{\pi}{180}$.
At $x = 30$,we have $f'(30) = \sec\left(\frac{30\pi}{180}\right) \tan\left(\frac{30\pi}{180}\right) \times \frac{\pi}{180}$.
This simplifies to $f'(30) = \sec\left(\frac{\pi}{6}\right) \tan\left(\frac{\pi}{6}\right) \times \frac{\pi}{180}$.
Since $\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$ and $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$,we get:
$f'(30) = \left(\frac{2}{\sqrt{3}}\right) \left(\frac{1}{\sqrt{3}}\right) \times \frac{\pi}{180} = \frac{2}{3} \times \frac{\pi}{180} = \frac{\pi}{270}$.
Therefore,the correct option is $C$.

(B) Let $y = \frac{2^x + 3^x}{4^x} = \frac{2^x}{4^x} + \frac{3^x}{4^x} = \left(\frac{2}{4}\right)^x + \left(\frac{3}{4}\right)^x = \left(\frac{1}{2}\right)^x + \left(\frac{3}{4}\right)^x$.
Now,differentiate with respect to $x$ using the formula $\frac{d}{dx}(a^x) = a^x \log a$:
$\frac{dy}{dx} = \frac{d}{dx}\left(\left(\frac{1}{2}\right)^x\right) + \frac{d}{dx}\left(\left(\frac{3}{4}\right)^x\right)$
$\frac{dy}{dx} = \left(\frac{1}{2}\right)^x \log \frac{1}{2} + \left(\frac{3}{4}\right)^x \log \frac{3}{4}$.
Thus,the correct option is $B$.

(A) To find the derivative of $y = 3^{1-2 x}$,we use the chain rule and the formula $\frac{d}{d x}(a^u) = a^u \cdot \log a \cdot \frac{d u}{d x}$.
Here,$a = 3$ and $u = 1 - 2x$.
First,find the derivative of the exponent: $\frac{d}{d x}(1 - 2x) = -2$.
Now,apply the formula:
$\frac{d}{d x}(3^{1-2 x}) = 3^{1-2 x} \cdot \log 3 \cdot \frac{d}{d x}(1 - 2x)$
$= 3^{1-2 x} \cdot \log 3 \cdot (-2)$
$= -2 \cdot 3^{1-2 x} \log 3$.
Thus,the correct option is $A$.

(A) Given $y = 5 \sin x + 6 \cos x$.
First,find the derivative $y_1 = \frac{dy}{dx} = 5 \cos x - 6 \sin x$.
Now,calculate $y^2 + (y_1)^2$:
$y^2 = (5 \sin x + 6 \cos x)^2 = 25 \sin^2 x + 36 \cos^2 x + 60 \sin x \cos x$.
$(y_1)^2 = (5 \cos x - 6 \sin x)^2 = 25 \cos^2 x + 36 \sin^2 x - 60 \sin x \cos x$.
Adding these two expressions:
$y^2 + (y_1)^2 = (25 \sin^2 x + 36 \cos^2 x + 60 \sin x \cos x) + (25 \cos^2 x + 36 \sin^2 x - 60 \sin x \cos x)$.
$y^2 + (y_1)^2 = 25(\sin^2 x + \cos^2 x) + 36(\cos^2 x + \sin^2 x)$.
Since $\sin^2 x + \cos^2 x = 1$,we get:
$y^2 + (y_1)^2 = 25(1) + 36(1) = 61$.

(A) Let $y = e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$.
We know that $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.
So,the first term becomes $e^{\pi/2}$,which is a constant. The derivative of a constant is $0$.
Now,differentiate the remaining terms: $\frac{d}{dx} \left[ \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin ^{-1} \frac{x}{a} \right]$.
Using the product rule for the first part: $\frac{d}{dx} \left( \frac{x}{2} \sqrt{a^2-x^2} \right) = \frac{1}{2} \sqrt{a^2-x^2} + \frac{x}{2} \cdot \frac{1}{2\sqrt{a^2-x^2}} \cdot (-2x) = \frac{1}{2} \sqrt{a^2-x^2} - \frac{x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-x^2-x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-2x^2}{2\sqrt{a^2-x^2}}$.
For the second part: $\frac{d}{dx} \left( \frac{a^2}{2} \sin ^{-1} \frac{x}{a} \right) = \frac{a^2}{2} \cdot \frac{1}{\sqrt{1-(x/a)^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{1}{\sqrt{(a^2-x^2)/a^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{a}{\sqrt{a^2-x^2}} \cdot \frac{1}{a} = \frac{a^2}{2\sqrt{a^2-x^2}}$.
Adding these results: $\frac{a^2-2x^2}{2\sqrt{a^2-x^2}} + \frac{a^2}{2\sqrt{a^2-x^2}} = \frac{2a^2-2x^2}{2\sqrt{a^2-x^2}} = \frac{2(a^2-x^2)}{2\sqrt{a^2-x^2}} = \sqrt{a^2-x^2}$.

(A) To find the derivative of $f(x) = \sin(x^2) + \cos(x^2)$,we use the chain rule.
Let $u = x^2$,then $\frac{du}{dx} = 2x$.
The derivative is given by $\frac{d}{dx}(\sin(u) + \cos(u)) = \frac{d}{du}(\sin(u) + \cos(u)) \cdot \frac{du}{dx}$.
$= (\cos(u) - \sin(u)) \cdot 2x$.
Substituting $u = x^2$ back into the expression,we get:
$= 2x(\cos(x^2) - \sin(x^2))$.
Thus,the correct option is $A$.

(A) Let $y = \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2$.
Expanding the expression using $(a+b)^2 = a^2 + 2ab + b^2$,we get:
$y = (\sqrt{x})^2 + 2(\sqrt{x})\left(\frac{1}{\sqrt{x}}\right) + \left(\frac{1}{\sqrt{x}}\right)^2$
$y = x + 2 + \frac{1}{x}$
Now,differentiate with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(x) + \frac{d}{d x}(2) + \frac{d}{d x}(x^{-1})$
$\frac{d y}{d x} = 1 + 0 - x^{-2}$
$\frac{d y}{d x} = 1 - \frac{1}{x^2}$
Thus,the correct option is $A$.

(D) Let $\theta = \cot ^{-1} \sqrt{\frac{2+x}{2-x}}$. Then $\cot \theta = \sqrt{\frac{2+x}{2-x}}$.
From the right-angled triangle with base $\sqrt{2+x}$ and perpendicular $\sqrt{2-x}$,the hypotenuse is $\sqrt{(\sqrt{2+x})^2 + (\sqrt{2-x})^2} = \sqrt{2+x+2-x} = \sqrt{4} = 2$.
Thus,$\cos \theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{\sqrt{2+x}}{2}$.
Now,the expression becomes $\frac{d}{d x}\left[\cos ^2 \theta\right] = \frac{d}{d x}\left[\left(\frac{\sqrt{2+x}}{2}\right)^2\right]$.
$= \frac{d}{d x}\left(\frac{2+x}{4}\right) = \frac{d}{d x}\left(\frac{1}{2} + \frac{x}{4}\right) = 0 + \frac{1}{4} = \frac{1}{4}$.

(D) Given the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$.
To find $f'(x)$,we differentiate $f(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \dots + \frac{2x}{2} + 1$.
Simplifying this,we get $f'(x) = x^{99} + x^{98} + \dots + x + 1$.
Now,substituting $x = 0$ into the derivative:
$f'(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1$.

(A) Given $f(x+y)=f(x)f(y)$.
Putting $x=0, y=5$,we get $f(5)=f(0)f(5)$.
Since $f(5)=3 \neq 0$,we have $f(0)=1$.
Now,$f^{\prime}(5) = \lim_{h \to 0} \frac{f(5+h)-f(5)}{h}$.
Using the given functional equation,$f(5+h)=f(5)f(h)$.
So,$f^{\prime}(5) = \lim_{h \to 0} \frac{f(5)f(h)-f(5)}{h} = f(5) \lim_{h \to 0} \frac{f(h)-1}{h}$.
Since $f(0)=1$,this is $f(5) \lim_{h \to 0} \frac{f(h)-f(0)}{h} = f(5)f^{\prime}(0)$.
Substituting the values,$f^{\prime}(5) = 3 \times 2 = 6$.

(D) Given the function $y = f(x^2 + 2)$.
Applying the chain rule to differentiate with respect to $x$,we get:
$\frac{dy}{dx} = f'(x^2 + 2) \cdot \frac{d}{dx}(x^2 + 2)$
$\frac{dy}{dx} = f'(x^2 + 2) \cdot (2x)$
Now,evaluate the derivative at $x = 1$:
$\left. \frac{dy}{dx} \right|_{x=1} = f'(1^2 + 2) \cdot (2 \cdot 1)$
$\left. \frac{dy}{dx} \right|_{x=1} = f'(3) \cdot 2$
Given that $f'(3) = 5$,substitute this value into the expression:
$\left. \frac{dy}{dx} \right|_{x=1} = 5 \cdot 2 = 10$

(A) By definition,a function $f(x)$ is even if $f(-x) = f(x)$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$
$f^{\prime}(-x) \cdot (-1) = f^{\prime}(x)$
$-f^{\prime}(-x) = f^{\prime}(x)$
$f^{\prime}(-x) = -f^{\prime}(x)$
Since $f^{\prime}(-x) = -f^{\prime}(x)$,the derivative $f^{\prime}(x)$ is an odd function.

(B) Given the function: $g(x) = \frac{x^{200}}{200} + \frac{x^{199}}{199} + \dots + \frac{x^2}{2} + x + 5$.
To find $g'(x)$,we differentiate $g(x)$ with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$g'(x) = \frac{200x^{199}}{200} + \frac{199x^{198}}{199} + \dots + \frac{2x}{2} + 1 + 0$.
Simplifying the expression,we get:
$g'(x) = x^{199} + x^{198} + \dots + x + 1$.
Now,substitute $x = 0$ into the derivative:
$g'(0) = 0^{199} + 0^{198} + \dots + 0 + 1$.
$g'(0) = 1$.

(C) Given the function $f(x) = 3 + |x|$.
We know that the derivative of $|x|$ is given by:
$f'(x) = \frac{d}{dx}(3 + |x|) = \frac{d}{dx}(|x|) = \text{sgn}(x)$
Where $\text{sgn}(x)$ is the signum function defined as:
$f'(x) = \begin{cases} 1, & x > 0 \\ -1, & x < 0 \end{cases}$
The function $f'(x)$ is undefined at $x = 0$.
Looking at the values of $f'(x)$,it takes only two values: $1$ for $x > 0$ and $-1$ for $x < 0$.
The minimum value of this function $f'(x)$ is $-1$.

(C) Given $y = \frac{\cos x}{1+\sin x}$.
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(-\sin x)(1+\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2}$
$\frac{dy}{dx} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2} = \frac{-(\sin x + \sin^2 x + \cos^2 x)}{(1+\sin x)^2}$
Since $\sin^2 x + \cos^2 x = 1$,we have:
$\frac{dy}{dx} = \frac{-(1+\sin x)}{(1+\sin x)^2} = \frac{-1}{1+\sin x}$. This matches option $(a)$.
Now,simplify $\frac{-1}{1+\sin x}$ using trigonometric identities:
$1+\sin x = 1+\cos\left(\frac{\pi}{2}-x\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$.
Therefore,$\frac{dy}{dx} = \frac{-1}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} = -\frac{1}{2}\sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$. This matches option $(c)$.
Thus,both $(a)$ and $(c)$ are correct.

(B) Given the function is $f(x) = \frac{a}{x^{4}} - \frac{b}{x^{2}} + \cos x$.
To find the derivative with respect to $x$,we use the power rule $\frac{d}{dx}(x^{n}) = nx^{n-1}$ and the derivative of $\cos x$ is $-\sin x$.
$\frac{d}{dx} f(x) = \frac{d}{dx}(ax^{-4} - bx^{-2} + \cos x)$
$= a(-4x^{-5}) - b(-2x^{-3}) - \sin x$
$= -\frac{4a}{x^{5}} + \frac{2b}{x^{3}} - \sin x$
Comparing this with the expression $ma + nb - p$,we get:
$m = -\frac{4}{x^{5}}$,$n = \frac{2}{x^{3}}$,and $p = \sin x$.

(C) $y = (\cos x^{2})^{2}$
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2(\cos x^{2}) \cdot \frac{d}{dx}(\cos x^{2})$
$\frac{dy}{dx} = 2(\cos x^{2}) \cdot (-\sin x^{2}) \cdot \frac{d}{dx}(x^{2})$
$\frac{dy}{dx} = 2(\cos x^{2}) \cdot (-\sin x^{2}) \cdot (2x)$
$\frac{dy}{dx} = -4x \sin x^{2} \cos x^{2}$
Using the trigonometric identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have:
$\frac{dy}{dx} = -2x(2 \sin x^{2} \cos x^{2}) = -2x \sin 2x^{2}$