Derivative at a point, Standard differentiation Questions in English

(A) Let $y = e^{x\sin x}$.
Applying the chain rule for differentiation,we have:
$\frac{dy}{dx} = \frac{d}{dx}(e^{x\sin x}) = e^{x\sin x} \cdot \frac{d}{dx}(x\sin x)$.
Using the product rule for $\frac{d}{dx}(x\sin x)$:
$\frac{d}{dx}(x\sin x) = x \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(x) = x\cos x + \sin x \cdot 1 = x\cos x + \sin x$.
Therefore,$\frac{dy}{dx} = e^{x\sin x}(x\cos x + \sin x)$.
Thus,the correct option is $A$.

(B) Let $y = \log (\sec x + \tan x)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \frac{d}{dx}(\sec x + \tan x)$
Since $\frac{d}{dx}(\sec x) = \sec x \tan x$ and $\frac{d}{dx}(\tan x) = \sec^2 x$,we have:
$\frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x}$
Factoring out $\sec x$ from the numerator:
$\frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x}$
Canceling the common term $(\sec x + \tan x)$:
$\frac{dy}{dx} = \sec x$.

(D) To find the derivative of $f(x) = x e^{x^2}$,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.
Let $u(x) = x$ and $v(x) = e^{x^2}$.
Then $u'(x) = 1$ and $v'(x) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x = 2x e^{x^2}$.
Applying the product rule:
$\frac{d}{dx}(x e^{x^2}) = (1)(e^{x^2}) + (x)(2x e^{x^2})$
$= e^{x^2} + 2x^2 e^{x^2}$
$= (1 + 2x^2)e^{x^2}$.
Comparing this with the given options,none of the options $A, B, C$ match the result $(1 + 2x^2)e^{x^2}$.
Therefore,the correct choice is $D$.

(C) To find the derivative of $\frac{e^{ax}}{\sin(bx + c)}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Let $u = e^{ax}$ and $v = \sin(bx + c)$.
Then $\frac{du}{dx} = a e^{ax}$ and $\frac{dv}{dx} = b \cos(bx + c)$.
Applying the quotient rule:
$\frac{d}{dx} \left[ \frac{e^{ax}}{\sin(bx + c)} \right] = \frac{\sin(bx + c) \cdot (a e^{ax}) - e^{ax} \cdot (b \cos(bx + c))}{\sin^2(bx + c)}$.
Factoring out $e^{ax}$ in the numerator:
$= \frac{e^{ax}[a \sin(bx + c) - b \cos(bx + c)]}{\sin^2(bx + c)}$.

(C) To find the derivative of the product of two functions,we use the product rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.
Let $u(x) = e^{-ax^2}$ and $v(x) = \log(\sin x)$.
First,find the derivatives of $u(x)$ and $v(x)$:
$u'(x) = e^{-ax^2} \cdot \frac{d}{dx}(-ax^2) = e^{-ax^2} \cdot (-2ax) = -2ax e^{-ax^2}$.
$v'(x) = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Now,apply the product rule:
$\frac{d}{dx} \{ e^{-ax^2} \log(\sin x) \} = (-2ax e^{-ax^2}) \cdot \log(\sin x) + e^{-ax^2} \cdot \cot x$.
Factor out $e^{-ax^2}$:
$= e^{-ax^2} [\cot x - 2ax \log(\sin x)]$.
Thus,the correct option is $C$.

(C) Given $y = \log x \cdot e^{(\tan x + x^2)}$.
Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = \log x \cdot \frac{d}{dx}(e^{(\tan x + x^2)}) + e^{(\tan x + x^2)} \cdot \frac{d}{dx}(\log x)$.
Applying the chain rule for the exponential term:
$\frac{dy}{dx} = \log x \cdot e^{(\tan x + x^2)} \cdot (\sec^2 x + 2x) + e^{(\tan x + x^2)} \cdot \frac{1}{x}$.
Factoring out $e^{(\tan x + x^2)}$:
$\frac{dy}{dx} = e^{(\tan x + x^2)} \left[ \frac{1}{x} + (\sec^2 x + 2x) \log x \right]$.

(A) To find the derivative of the product $e^x \log(1 + x^2)$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.
Let $u = e^x$ and $v = \log(1 + x^2)$.
Then $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) = \frac{2x}{1 + x^2}$.
Applying the product rule:
$\frac{d}{dx} \{ e^x \log(1 + x^2) \} = e^x \cdot \frac{d}{dx}(\log(1 + x^2)) + \log(1 + x^2) \cdot \frac{d}{dx}(e^x)$
$= e^x \left( \frac{2x}{1 + x^2} \right) + \log(1 + x^2) \cdot e^x$
$= e^x \left[ \log(1 + x^2) + \frac{2x}{1 + x^2} \right]$.

(C) Given $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$.
Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^{2x} + e^{-2x}$ and $v = e^{2x} - e^{-2x}$.
Then $u' = 2e^{2x} - 2e^{-2x} = 2(e^{2x} - e^{-2x})$ and $v' = 2e^{2x} + 2e^{-2x} = 2(e^{2x} + e^{-2x})$.
$\frac{dy}{dx} = \frac{(e^{2x} - e^{-2x}) \cdot 2(e^{2x} - e^{-2x}) - (e^{2x} + e^{-2x}) \cdot 2(e^{2x} + e^{-2x})}{(e^{2x} - e^{-2x})^2}$
$= \frac{2(e^{2x} - e^{-2x})^2 - 2(e^{2x} + e^{-2x})^2}{(e^{2x} - e^{-2x})^2}$
$= \frac{2[(e^{4x} - 2 + e^{-4x}) - (e^{4x} + 2 + e^{-4x})]}{(e^{2x} - e^{-2x})^2}$
$= \frac{2[-4]}{(e^{2x} - e^{-2x})^2} = \frac{-8}{(e^{2x} - e^{-2x})^2}$.

(B) To find the derivative,first convert the angle from degrees to radians: $x^\circ = \frac{\pi x}{180} \text{ radians}$.
Thus,the expression becomes: $\frac{d}{dx}\left[ \frac{2}{\pi }\sin \left( \frac{\pi x}{180} \right) \right]$.
Using the chain rule,$\frac{d}{dx}\sin(ax) = a\cos(ax)$:
$= \frac{2}{\pi } \cdot \frac{\pi }{180} \cos \left( \frac{\pi x}{180} \right)$.
$= \frac{2}{180} \cos (x^\circ) = \frac{1}{90} \cos (x^\circ)$.

(C) Let $y = \cos(\sin(x^2))$.
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:
At $x = \sqrt{\frac{\pi}{2}}$,$x^2 = \frac{\pi}{2}$.
$\frac{dy}{dx} = -\sin(\sin(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) \cdot 2\sqrt{\frac{\pi}{2}}$
Since $\cos(\frac{\pi}{2}) = 0$,the entire expression becomes:
$\frac{dy}{dx} = -\sin(1) \cdot 0 \cdot 2\sqrt{\frac{\pi}{2}} = 0$.

(A) To find the derivative of $e^{ax} \cos(bx + c)$ with respect to $x$,we use the product rule: $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$.
Let $u = e^{ax}$ and $v = \cos(bx + c)$.
Then $\frac{du}{dx} = a e^{ax}$ and $\frac{dv}{dx} = -\sin(bx + c) \cdot b = -b \sin(bx + c)$.
Applying the product rule:
$\frac{d}{dx}[e^{ax} \cos(bx + c)] = e^{ax} \cdot (-b \sin(bx + c)) + \cos(bx + c) \cdot (a e^{ax})$.
Factoring out $e^{ax}$,we get:
$= e^{ax}[a \cos(bx + c) - b \sin(bx + c)]$.
Thus,the correct option is $A$.

(B) Given the expression: $y = (1 + x^{1/4})(1 + x^{1/2})(1 - x^{1/4})$
Rearranging the terms using the commutative property of multiplication:
$y = (1 + x^{1/4})(1 - x^{1/4})(1 + x^{1/2})$
Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$,where $a = 1$ and $b = x^{1/4}$:
$y = (1^2 - (x^{1/4})^2)(1 + x^{1/2})$
$y = (1 - x^{1/2})(1 + x^{1/2})$
Again,using the identity $(a - b)(a + b) = a^2 - b^2$,where $a = 1$ and $b = x^{1/2}$:
$y = 1^2 - (x^{1/2})^2$
$y = 1 - x$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(1 - x)$
$\frac{dy}{dx} = 0 - 1 = -1$

(C) By the definition of the derivative at a point $x_0$,we have:
$f'(x_0) = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$
Given $f(x) = (x - x_0)g(x)$,we find $f(x_0) = (x_0 - x_0)g(x_0) = 0 \cdot g(x_0) = 0$.
Substituting this into the derivative formula:
$f'(x_0) = \lim_{x \to x_0} \frac{(x - x_0)g(x) - 0}{x - x_0}$
$f'(x_0) = \lim_{x \to x_0} g(x)$
Since $g(x)$ is continuous at $x_0$,$\lim_{x \to x_0} g(x) = g(x_0)$.
Therefore,$f'(x_0) = g(x_0)$.

(B) To find the derivative of $e^{x^3}$ with respect to $x$,we use the chain rule.
Let $y = e^{x^3}$.
By the chain rule,$\frac{dy}{dx} = \frac{d}{dx}(e^{x^3}) = e^{x^3} \cdot \frac{d}{dx}(x^3)$.
Since $\frac{d}{dx}(x^3) = 3x^2$,we have:
$\frac{dy}{dx} = e^{x^3} \cdot 3x^2 = 3x^2e^{x^3}$.
Thus,the correct option is $B$.

(A) The derivative of the inverse trigonometric function $\sin^{-1}x$ with respect to $x$ is a standard result in calculus.
By definition,$\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$ for $|x| < 1$.
Therefore,the correct option is $A$.

(B) Given $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$.
Applying the quotient rule for differentiation,$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{d}{dx}(\sin^{-1} x) - \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1 - x^2})}{1 - x^2}$.
Since $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}(\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$,we have:
$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{1}{\sqrt{1 - x^2}} - \sin^{-1} x \cdot \left( \frac{-x}{\sqrt{1 - x^2}} \right)}{1 - x^2}$.
$\frac{dy}{dx} = \frac{1 + \frac{x \sin^{-1} x}{\sqrt{1 - x^2}}}{1 - x^2}$.
Multiplying both sides by $(1 - x^2)$,we get:
$(1 - x^2)\frac{dy}{dx} = 1 + x \left( \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \right)$.
Since $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$,we substitute $y$ into the expression:
$(1 - x^2)\frac{dy}{dx} = 1 + xy$.

(C) The derivative of the inverse trigonometric function $\sec^{-1} x$ with respect to $x$ is a standard formula in calculus.
By definition,if $y = \sec^{-1} x$,then $x = \sec y$.
Differentiating both sides with respect to $y$,we get $\frac{dx}{dy} = \sec y \tan y$.
Since $\tan y = \sqrt{\sec^2 y - 1} = \sqrt{x^2 - 1}$,we have $\frac{dx}{dy} = x\sqrt{x^2 - 1}$.
Therefore,$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{x\sqrt{x^2 - 1}}$.
Thus,the correct option is $C$.

(B) Given $f(2) = 4$ and $f'(2) = 1$.
We need to evaluate $\mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2}$.
Adding and subtracting $2f(2)$ in the numerator:
$= \mathop {\lim }\limits_{x \to 2} \frac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2}$
$= \mathop {\lim }\limits_{x \to 2} \left[ \frac{f(2)(x - 2)}{x - 2} - \frac{2(f(x) - f(2))}{x - 2} \right]$
$= f(2) - 2 \mathop {\lim }\limits_{x \to 2} \frac{f(x) - f(2)}{x - 2}$
$= f(2) - 2f'(2)$
Substituting the values: $4 - 2(1) = 4 - 2 = 2$.
Alternatively,using $L$-Hospital rule:
$= \mathop {\lim }\limits_{x \to 2} \frac{\frac{d}{dx}(xf(2) - 2f(x))}{\frac{d}{dx}(x - 2)}$
$= \mathop {\lim }\limits_{x \to 2} \frac{f(2) - 2f'(x)}{1} = f(2) - 2f'(2) = 4 - 2(1) = 2$.

(C) To find the derivative of $\log \left( x + \frac{1}{x} \right)$,we use the chain rule:
$\frac{d}{dx} \left[ \log \left( x + \frac{1}{x} \right) \right] = \frac{1}{x + \frac{1}{x}} \times \frac{d}{dx} \left( x + \frac{1}{x} \right)$
Since $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2}$,we have:
$= \frac{1}{x + \frac{1}{x}} \times \left( 1 - \frac{1}{x^2} \right)$
$= \frac{1 - \frac{1}{x^2}}{x + \frac{1}{x}}$
Thus,the correct option is $C$.

(B) We know that $\sin^{-1}(\sqrt{1 - x}) = \cos^{-1}(\sqrt{x})$.
Substituting this into the given equation:
$y = \cos^{-1}(\sqrt{x}) + \cos^{-1}(\sqrt{x}) = 2\cos^{-1}(\sqrt{x})$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \left( -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \right) \cdot \frac{d}{dx}(\sqrt{x})$
$\frac{dy}{dx} = -\frac{2}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx} = -\frac{1}{\sqrt{x(1 - x)}}$.

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

(D) Let $f(x) = \tan x - x$.
To find the derivative with respect to $x$,we differentiate both terms:
$\frac{d}{dx}(\tan x - x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(x)$.
We know that $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(x) = 1$.
Therefore,$f'(x) = \sec^2 x - 1$.
Using the trigonometric identity $1 + \tan^2 x = \sec^2 x$,we have $\sec^2 x - 1 = \tan^2 x$.
Thus,the derivative is $\tan^2 x$.

(B) Given $f(x) = (\log _{\cot x}\tan x)(\log _{\tan x}\cot x)^{-1}$.
We know that $\log _{a}b = \frac{1}{\log _{b}a}$.
Let $u = \log _{\cot x}\tan x$. Then $\log _{\tan x}\cot x = \frac{1}{u}$.
Substituting this into the expression for $f(x)$:
$f(x) = u \cdot (u^{-1})^{-1} = u \cdot u = u^2$.
Since $\tan x = (\cot x)^{-1}$,we have $u = \log _{\cot x}(\cot x)^{-1} = -1 \cdot \log _{\cot x}(\cot x) = -1$.
Thus,$f(x) = (-1)^2 = 1$.
Since $f(x) = 1$ is a constant function,its derivative $f'(x) = 0$ for all $x$ in the domain.
Therefore,$f'(2) = 0$.

(A) Let $y = \sqrt{\sec \sqrt{x}} = (\sec \sqrt{x})^{1/2}$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2}(\sec \sqrt{x})^{-1/2} \cdot \frac{d}{dx}(\sec \sqrt{x})$
$= \frac{1}{2\sqrt{\sec \sqrt{x}}} \cdot (\sec \sqrt{x} \tan \sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x})$
$= \frac{1}{2\sqrt{\sec \sqrt{x}}} \cdot \sec \sqrt{x} \tan \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$
$= \frac{1}{4\sqrt{x}} \cdot \frac{\sec \sqrt{x}}{\sqrt{\sec \sqrt{x}}} \cdot \tan \sqrt{x}$
$= \frac{1}{4\sqrt{x}} \cdot (\sec \sqrt{x})^{1/2} \cdot \frac{\sin \sqrt{x}}{\cos \sqrt{x}}$
$= \frac{1}{4\sqrt{x}} \cdot (\sec \sqrt{x})^{1/2} \cdot \sec \sqrt{x} \cdot \sin \sqrt{x}$
$= \frac{1}{4\sqrt{x}} (\sec \sqrt{x})^{3/2} \sin \sqrt{x}$.

(C) Given the function $f(x) = x^2 - 6x + 8$.
To find the value of $x$ for which $f'(x)$ vanishes,we first find the derivative of $f(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x^2 - 6x + 8) = 2x - 6$.
Now,set the derivative equal to zero:
$2x - 6 = 0$.
Solving for $x$:
$2x = 6 \Rightarrow x = 3$.
Since $x = 3$ lies within the interval $[2, 4]$,the value of $x$ for which $f'(x)$ vanishes is $3$.

(B) Given the function $f(x) = e^x g(x)$.
Applying the product rule for differentiation,we get:
$f'(x) = \frac{d}{dx}(e^x) \cdot g(x) + e^x \cdot \frac{d}{dx}(g(x))$
$f'(x) = e^x g(x) + e^x g'(x)$
Now,substitute $x = 0$ into the derivative expression:
$f'(0) = e^0 g(0) + e^0 g'(0)$
Since $e^0 = 1$,$g(0) = 2$,and $g'(0) = 1$,we have:
$f'(0) = 1 \cdot 2 + 1 \cdot 1$
$f'(0) = 2 + 1 = 3$.

(C) Given the function $y = e^x \log x$.
To find $\frac{dy}{dx}$,we use the product rule of differentiation: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.
Here,$u = e^x$ and $v = \log x$.
Then,$\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = \frac{1}{x}$.
Applying the product rule:
$\frac{dy}{dx} = e^x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(e^x)$
$\frac{dy}{dx} = e^x \cdot \frac{1}{x} + \log x \cdot e^x$
Factoring out $e^x$,we get:
$\frac{dy}{dx} = e^x \left( \frac{1}{x} + \log x \right)$.
Thus,the correct option is $C$.

(C) Given that $y = \sec(x^{\circ})$.
First,convert the angle from degrees to radians: $x^{\circ} = \frac{\pi x}{180} \text{ radians}$.
So,$y = \sec\left(\frac{\pi x}{180}\right)$.
Now,differentiate with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{d}{dx} \left[ \sec\left(\frac{\pi x}{180}\right) \right]$
$= \sec\left(\frac{\pi x}{180}\right) \tan\left(\frac{\pi x}{180}\right) \cdot \frac{d}{dx} \left( \frac{\pi x}{180} \right)$
$= \sec(x^{\circ}) \tan(x^{\circ}) \cdot \frac{\pi}{180}$
$= \frac{\pi}{180} \sec(x^{\circ}) \tan(x^{\circ})$.

(C) Given $y = \sqrt{\sin \sqrt{x}}$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{\sin \sqrt{x}})$
$= \frac{1}{2\sqrt{\sin \sqrt{x}}} \times \frac{d}{dx}(\sin \sqrt{x})$
$= \frac{1}{2\sqrt{\sin \sqrt{x}}} \times \cos \sqrt{x} \times \frac{d}{dx}(\sqrt{x})$
$= \frac{1}{2\sqrt{\sin \sqrt{x}}} \times \cos \sqrt{x} \times \frac{1}{2\sqrt{x}}$
$= \frac{\cos \sqrt{x}}{4\sqrt{x} \sqrt{\sin \sqrt{x}}}$
Thus,the correct option is $C$.

(B) Let $f(x) = \sin 2x \cos 2x \cos 3x + \log_2 2^{x+3}$.
Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$ and the property $\log_a a^k = k$,we get:
$f(x) = \frac{1}{2} \sin 4x \cos 3x + (x + 3)$.
Using the product-to-sum formula $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$,we have:
$f(x) = \frac{1}{2} \cdot \frac{1}{2} [\sin(4x+3x) + \sin(4x-3x)] + x + 3$
$f(x) = \frac{1}{4} [\sin 7x + \sin x] + x + 3$.
Differentiating with respect to $x$:
$f'(x) = \frac{1}{4} [7 \cos 7x + \cos x] + 1$.
Evaluating at $x = \pi$:
$f'(\pi) = \frac{1}{4} [7 \cos(7\pi) + \cos(\pi)] + 1$
Since $\cos(7\pi) = -1$ and $\cos(\pi) = -1$:
$f'(\pi) = \frac{1}{4} [7(-1) + (-1)] + 1$
$f'(\pi) = \frac{1}{4} [-8] + 1 = -2 + 1 = -1$.

(A) Given the function $f(x) = (\sqrt{x} + \frac{1}{\sqrt{x}})^2$.
Expanding the expression,we get $f(x) = x + \frac{1}{x} + 2$.
Now,find the first derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x + x^{-1} + 2) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.
We are given that $f'(x) = \frac{3}{4}$.
So,$1 - \frac{1}{x^2} = \frac{3}{4}$.
Rearranging the terms,we get $\frac{1}{x^2} = 1 - \frac{3}{4} = \frac{1}{4}$.
This implies $x^2 = 4$,which gives $x = \pm 2$.

(D) Given $y = \frac{(1 - x)^2}{x^2}$.
Expanding the numerator,we get $y = \frac{1 - 2x + x^2}{x^2}$.
Dividing each term by $x^2$,we get $y = \frac{1}{x^2} - \frac{2x}{x^2} + \frac{x^2}{x^2} = x^{-2} - 2x^{-1} + 1$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(x^{-2}) - 2\frac{d}{dx}(x^{-1}) + \frac{d}{dx}(1)$.
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$,we get $\frac{dy}{dx} = -2x^{-3} - 2(-1)x^{-2} + 0$.
$\frac{dy}{dx} = -\frac{2}{x^3} + \frac{2}{x^2}$.
Thus,the correct option is $D$.

(A) To find the derivative of $\sin(2x^2)$ with respect to $x$,we use the chain rule.
Let $y = \sin(2x^2)$.
Applying the chain rule: $\frac{dy}{dx} = \frac{d}{dx}(\sin(2x^2)) = \cos(2x^2) \cdot \frac{d}{dx}(2x^2)$.
Now,differentiate $2x^2$ with respect to $x$: $\frac{d}{dx}(2x^2) = 2 \cdot 2x = 4x$.
Therefore,$\frac{dy}{dx} = \cos(2x^2) \cdot 4x = 4x \cos(2x^2)$.
Thus,the correct option is $A$.

(A) We are given the function $y = \cosh^{-1}(\sec x)$.
Using the chain rule,the derivative is $\frac{dy}{dx} = \frac{d}{dx} \cosh^{-1}(\sec x)$.
The derivative of $\cosh^{-1}(u)$ is $\frac{1}{\sqrt{u^2 - 1}} \cdot \frac{du}{dx}$.
Here,$u = \sec x$,so $\frac{du}{dx} = \sec x \tan x$.
Substituting these into the formula:
$\frac{dy}{dx} = \frac{1}{\sqrt{\sec^2 x - 1}} \cdot (\sec x \tan x)$.
Since $\sec^2 x - 1 = \tan^2 x$,we have $\sqrt{\sec^2 x - 1} = \tan x$ (assuming $\tan x > 0$ for the domain).
Therefore,$\frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec x \tan x = \sec x$.

(A) Given $f(x) = \frac{1}{\sqrt{x^2 + a^2} + \sqrt{x^2 + b^2}}$.
Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2})$:
$f(x) = \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{(\sqrt{x^2 + a^2} + \sqrt{x^2 + b^2})(\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2})}$
$f(x) = \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{(x^2 + a^2) - (x^2 + b^2)}$
$f(x) = \frac{1}{a^2 - b^2} [\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}]$
Now,differentiating with respect to $x$:
$f'(x) = \frac{1}{a^2 - b^2} \left[ \frac{d}{dx}(\sqrt{x^2 + a^2}) - \frac{d}{dx}(\sqrt{x^2 + b^2}) \right]$
$f'(x) = \frac{1}{a^2 - b^2} \left[ \frac{1}{2\sqrt{x^2 + a^2}} \cdot 2x - \frac{1}{2\sqrt{x^2 + b^2}} \cdot 2x \right]$
$f'(x) = \frac{x}{a^2 - b^2} \left[ \frac{1}{\sqrt{x^2 + a^2}} - \frac{1}{\sqrt{x^2 + b^2}} \right]$.

(D) Given the function $f(x) = x|x|$.
We can define the absolute value function as:
$f(x) = \begin{cases} x(-x) = -x^2, & x < 0 \\ x(x) = x^2, & x \ge 0 \end{cases}$
Now,we find the derivative $f'(x)$ for both cases:
For $x < 0$,$f'(x) = \frac{d}{dx}(-x^2) = -2x$.
For $x > 0$,$f'(x) = \frac{d}{dx}(x^2) = 2x$.
At $x = 0$,we check the left-hand derivative and right-hand derivative:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} (-h) = 0$.
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} (h) = 0$.
Since $LHD = RHD = 0$,the derivative at $x = 0$ is $0$.
Combining these results,$f'(x) = \begin{cases} -2x, & x < 0 \\ 2x, & x \ge 0 \end{cases} = 2|x|$.

(D) Let $y = \sqrt{\sqrt{x} + 1}$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{\sqrt{x} + 1}) = \frac{1}{2\sqrt{\sqrt{x} + 1}} \cdot \frac{d}{dx}(\sqrt{x} + 1)$.
Since $\frac{d}{dx}(\sqrt{x} + 1) = \frac{1}{2\sqrt{x}}$,we have:
$\frac{dy}{dx} = \frac{1}{2\sqrt{\sqrt{x} + 1}} \cdot \frac{1}{2\sqrt{x}}$.
$\frac{dy}{dx} = \frac{1}{4\sqrt{x}\sqrt{\sqrt{x} + 1}}$.
Combining the terms under the square root:
$\frac{dy}{dx} = \frac{1}{4\sqrt{x(\sqrt{x} + 1)}}$.
Thus,the correct option is $D$.

(A) Given the function $y = e^{\sqrt{x}}$.
To find the derivative $\frac{dy}{dx}$,we apply the chain rule.
Let $u = \sqrt{x} = x^{1/2}$. Then $y = e^u$.
Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.
$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u = e^{\sqrt{x}}$.
$\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
Therefore,$\frac{dy}{dx} = e^{\sqrt{x}} \times \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$.

(B) Given $f(x) = \cos^{-1}\left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right]$.
Let $u = \log x$. Then the expression becomes $\cos^{-1}\left[ \frac{1 - u^2}{1 + u^2} \right]$.
Using the trigonometric substitution $u = \tan \theta$,we know that $\cos^{-1}\left[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right] = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1} u$.
Thus,$f(x) = 2\tan^{-1}(\log x)$.
Differentiating with respect to $x$ using the chain rule:
$f'(x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{d}{dx}(\log x) = 2 \cdot \frac{1}{1 + (\log x)^2} \cdot \frac{1}{x}$.
Now,evaluating at $x = e$:
$f'(e) = 2 \cdot \frac{1}{1 + (\log e)^2} \cdot \frac{1}{e}$.
Since $\log e = 1$,we have $f'(e) = 2 \cdot \frac{1}{1 + 1^2} \cdot \frac{1}{e} = 2 \cdot \frac{1}{2} \cdot \frac{1}{e} = \frac{1}{e}$.

(C) Given the function $f(x) = |x^2 - x|$.
For $x$ near $2$,$x^2 - x$ is positive because $2^2 - 2 = 2 > 0$.
Since $x^2 - x > 0$ in an open interval around $x = 2$,we can write $f(x) = x^2 - x$.
Now,find the derivative $f'(x) = \frac{d}{dx}(x^2 - x) = 2x - 1$.
Evaluating at $x = 2$,we get $f'(2) = 2(2) - 1 = 4 - 1 = 3$.

(D) Let $f(x) = \log_{\sqrt{x}} \left(\frac{1}{x}\right)$.
Using the base change formula $\log_a b = \frac{\log b}{\log a}$,we have:
$f(x) = \frac{\log(1/x)}{\log(\sqrt{x})}$
Since $\log(1/x) = -\log x$ and $\log(\sqrt{x}) = \log(x^{1/2}) = \frac{1}{2} \log x$,we substitute these into the expression:
$f(x) = \frac{-\log x}{\frac{1}{2} \log x}$
For $x > 0$ and $x \neq 1$,we can cancel $\log x$:
$f(x) = \frac{-1}{1/2} = -2$
Since $f(x) = -2$ is a constant function,its derivative with respect to $x$ is:
$f'(x) = \frac{d}{dx}(-2) = 0$.

(B) Let $f(x) = |x - 1| + |x - 5|$.
We define the function based on the intervals of $x$:
$f(x) = \begin{cases} -(x - 1) - (x - 5), & x < 1 \\ (x - 1) - (x - 5), & 1 \le x \le 5 \\ (x - 1) + (x - 5), & x > 5 \end{cases}$
Simplifying the expressions:
$f(x) = \begin{cases} 6 - 2x, & x < 1 \\ 4, & 1 \le x \le 5 \\ 2x - 6, & x > 5 \end{cases}$
Since $x = 3$ lies in the interval $(1, 5)$,we have $f(x) = 4$ for all $x$ in this interval.
Therefore,the derivative $f'(x) = \frac{d}{dx}(4) = 0$ for $x \in (1, 5)$.
Thus,at $x = 3$,the value is $0$.

(C) Let $y = \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x}$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we have:
$y = \frac{(\tan 2x - \tan x)(\tan 2x + \tan x)}{(1 + \tan 2x \tan x)(1 - \tan 2x \tan x)}$.
Using the tangent addition and subtraction formulas $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ and $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$y = \tan(2x - x) \tan(2x + x) = \tan x \tan 3x$.
Now,we need to find $\frac{d}{dx} [y \cdot \cot 3x]$.
Substituting $y$,we get $\frac{d}{dx} [\tan x \tan 3x \cdot \cot 3x]$.
Since $\tan 3x \cdot \cot 3x = 1$,the expression simplifies to $\frac{d}{dx} [\tan x]$.
Therefore,$\frac{d}{dx} [\tan x] = \sec^2 x$.

(D) Given $y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x \cdot \sqrt{x}} \right)$.
Using the formula $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left( \frac{a - b}{1 + ab} \right)$,we let $a = \sqrt{x}$ and $b = x$.
Thus,$y = \tan^{-1}(\sqrt{x}) - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$y' = \frac{d}{dx}(\tan^{-1}(\sqrt{x})) - \frac{d}{dx}(\tan^{-1}(x))$
$y' = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x}) - \frac{1}{1 + x^2}$
$y' = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} - \frac{1}{1 + x^2}$.
Now,evaluating at $x = 1$:
$y'(1) = \frac{1}{1 + 1} \cdot \frac{1}{2\sqrt{1}} - \frac{1}{1 + 1^2}$
$y'(1) = \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.

(B) Let $y = 10^{x \tan x}$.
Using the chain rule for differentiation,$\frac{d}{dx}(a^u) = a^u \ln a \cdot \frac{du}{dx}$.
Here,$a = 10$ and $u = x \tan x$.
First,find the derivative of $u = x \tan x$ using the product rule:
$\frac{d}{dx}(x \tan x) = \frac{d}{dx}(x) \cdot \tan x + x \cdot \frac{d}{dx}(\tan x) = 1 \cdot \tan x + x \cdot \sec^2 x = \tan x + x \sec^2 x$.
Now,differentiate $10^{x \tan x}$:
$\frac{d}{dx}(10^{x \tan x}) = 10^{x \tan x} \cdot \ln 10 \cdot (\tan x + x \sec^2 x)$.
Substitute this back into the original expression:
$10^{-x \tan x} \left[ 10^{x \tan x} \cdot \ln 10 \cdot (\tan x + x \sec^2 x) \right]$
$= 10^{-x \tan x + x \tan x} \cdot \ln 10 \cdot (\tan x + x \sec^2 x)$
$= 10^0 \cdot \ln 10 \cdot (\tan x + x \sec^2 x)$
$= \ln 10 (\tan x + x \sec^2 x)$.

(A) Given the limit: $\mathop {\lim }\limits_{x \to a} \frac{xf(a) - af(x)}{x - a}$.
To evaluate this,we add and subtract $af(a)$ in the numerator:
$= \mathop {\lim }\limits_{x \to a} \frac{xf(a) - af(a) + af(a) - af(x)}{x - a}$
$= \mathop {\lim }\limits_{x \to a} \left[ \frac{f(a)(x - a)}{x - a} - \frac{a(f(x) - f(a))}{x - a} \right]$
$= \mathop {\lim }\limits_{x \to a} f(a) - a \mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$
Since $f(x)$ is differentiable at $x = a$,$\mathop {\lim }\limits_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$.
Therefore,the limit is $f(a) - af'(a)$.

(B) Given the relationship $f'(a + b) = f'(a) + f'(b)$.
Let us test the options:
For option $(A)$,$f(x) = x$,then $f'(x) = 1$. Thus,$f'(a+b) = 1$ and $f'(a) + f'(b) = 1 + 1 = 2$. Since $1 \neq 2$,this is incorrect.
For option $(B)$,$f(x) = x^2$,then $f'(x) = 2x$. Thus,$f'(a+b) = 2(a+b) = 2a + 2b$ and $f'(a) + f'(b) = 2a + 2b$. Since $2a + 2b = 2a + 2b$,this is correct.
For option $(C)$,$f(x) = x^3$,then $f'(x) = 3x^2$. Thus,$f'(a+b) = 3(a+b)^2 = 3(a^2 + 2ab + b^2)$ and $f'(a) + f'(b) = 3a^2 + 3b^2$. These are not equal.
Therefore,the correct function is $f(x) = x^2$.

(B) Given the function $y = (1 + x^2)\tan^{-1}x - x.$
To find $\frac{dy}{dx},$ we apply the product rule to the first term $(1 + x^2)\tan^{-1}x$ and the power rule to the second term $-x.$
Using the product rule $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx},$ where $u = (1 + x^2)$ and $v = \tan^{-1}x$:
$\frac{dy}{dx} = \frac{d}{dx}[(1 + x^2)\tan^{-1}x] - \frac{d}{dx}(x)$
$\frac{dy}{dx} = (1 + x^2) \cdot \frac{d}{dx}(\tan^{-1}x) + \tan^{-1}x \cdot \frac{d}{dx}(1 + x^2) - 1$
Since $\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1 + x^2}$ and $\frac{d}{dx}(1 + x^2) = 2x$:
$\frac{dy}{dx} = (1 + x^2) \cdot \frac{1}{1 + x^2} + \tan^{-1}x \cdot (2x) - 1$
$\frac{dy}{dx} = 1 + 2x\tan^{-1}x - 1$
$\frac{dy}{dx} = 2x\tan^{-1}x.$
Thus,the correct option is $B$.

(A) Given $y = \frac{a + b{x^{3/2}}}{{x^{5/4}}} = a{x^{-5/4}} + b{x^{1/4}}$.
Differentiating with respect to $x$:
$y' = a(-\frac{5}{4}){x^{-9/4}} + b(\frac{1}{4}){x^{-3/4}}$.
Setting $y' = 0$ at $x = 5$:
$-\frac{5a}{4}{x^{-9/4}} + \frac{b}{4}{x^{-3/4}} = 0$.
Multiply by $4{x^{9/4}}$:
$-5a + b{x^{6/4}} = 0$.
$-5a + b{x^{3/2}} = 0$.
At $x = 5$,we have $b(5)^{3/2} = 5a$.
$\frac{a}{b} = \frac{5^{3/2}}{5} = 5^{3/2 - 1} = 5^{1/2} = \sqrt{5}$.
Thus,the ratio $a:b = \sqrt{5} : 1$.

(A) Given $y = \sec(\tan^{-1}x)$.
Let $\tan^{-1}x = \theta$,then $x = \tan\theta$.
This implies $\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + x^2}$.
So,$y = \sqrt{1 + x^2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2})$
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^2}} \cdot \frac{d}{dx}(1 + x^2)$
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x)$
$\frac{dy}{dx} = \frac{x}{\sqrt{1 + x^2}}$.