Derivative at a point, Standard differentiation Questions in English

(B) Given $f(x) = 2x^6 + 3x^4 + 4x^2$.
First,we check if $f(x)$ is an even or odd function.
$f(-x) = 2(-x)^6 + 3(-x)^4 + 4(-x)^2 = 2x^6 + 3x^4 + 4x^2 = f(x)$.
Since $f(-x) = f(x)$,$f(x)$ is an even function.
Now,we find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(2x^6 + 3x^4 + 4x^2) = 12x^5 + 12x^3 + 8x$.
To check if $f'(x)$ is even or odd,we evaluate $f'(-x)$:
$f'(-x) = 12(-x)^5 + 12(-x)^3 + 8(-x) = -12x^5 - 12x^3 - 8x = -(12x^5 + 12x^3 + 8x) = -f'(x)$.
Since $f'(-x) = -f'(x)$,$f'(x)$ is an odd function.

(B) The function is defined as $f(x) = \frac{x}{1 + |x|}$.
We examine the left-hand derivative and right-hand derivative at $x = 0$.
For $x < 0$,$|x| = -x$,so $f(x) = \frac{x}{1 - x}$.
The derivative is $f'(x) = \frac{(1-x)(1) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2}$.
Thus,the left-hand derivative $f'(0^-) = \lim_{x \to 0^-} \frac{1}{(1-x)^2} = 1$.
For $x > 0$,$|x| = x$,so $f(x) = \frac{x}{1 + x}$.
The derivative is $f'(x) = \frac{(1+x)(1) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2}$.
Thus,the right-hand derivative $f'(0^+) = \lim_{x \to 0^+} \frac{1}{(1+x)^2} = 1$.
Since $f'(0^-) = f'(0^+) = 1$,the derivative $f'(0) = 1$.

(A) Given that $f(x)$ is differentiable at $x = 1$,the derivative is defined as $f'(1) = \mathop {\lim }\limits_{h \to 0} \frac{f(1 + h) - f(1)}{h}$.
We are given $\mathop {\lim }\limits_{h \to 0} \frac{f(1 + h)}{h} = 5$.
For this limit to exist and be finite,the numerator $f(1 + h)$ must approach $0$ as $h \to 0$,which implies $f(1) = 0$.
Substituting $f(1) = 0$ into the definition of the derivative,we get:
$f'(1) = \mathop {\lim }\limits_{h \to 0} \frac{f(1 + h) - 0}{h} = \mathop {\lim }\limits_{h \to 0} \frac{f(1 + h)}{h} = 5$.

(D) Given $f(x+y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$.
First,note that $f(0) = f(0+0) = f(0)f(0)$,so $f(0) = 1$ (assuming $f(x) \neq 0$).
Using the definition of the derivative:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h}$
$= f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$
Since $f(h) = 1 + \sin(3h)g(h)$,we have $f(h) - 1 = \sin(3h)g(h)$.
Substituting this into the limit:
$f'(x) = f(x) \lim_{h \to 0} \frac{\sin(3h)g(h)}{h}$
$= f(x) \lim_{h \to 0} \left( \frac{\sin(3h)}{3h} \times 3 \times g(h) \right)$
$= f(x) \times 1 \times 3 \times g(0) = 3f(x)g(0)$.
Thus,the correct option is $D$.

(D) We know that the derivative is defined as $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.
Since $f(x + y) = f(x) + f(y)$,we have $f(x + h) = f(x) + f(h)$.
Substituting this into the derivative formula,we get $f'(x) = \lim_{h \to 0} \frac{f(x) + f(h) - f(x)}{h} = \lim_{h \to 0} \frac{f(h)}{h}$.
Given $f(x) = x^2 g(x)$,we have $f(h) = h^2 g(h)$.
Thus,$f'(x) = \lim_{h \to 0} \frac{h^2 g(h)}{h} = \lim_{h \to 0} h \cdot g(h)$.
Since $g(x)$ is a continuous function,$\lim_{h \to 0} g(h) = g(0)$.
Therefore,$f'(x) = 0 \cdot g(0) = 0$.

(C) To find the derivative of $\log(\log x)$ with respect to $x$,we use the chain rule.
Let $y = \log(\log x)$.
Applying the chain rule,$\frac{dy}{dx} = \frac{d}{d(\log x)} \log(\log x) \cdot \frac{d}{dx} \log x$.
Since $\frac{d}{du} \log u = \frac{1}{u}$ and $\frac{d}{dx} \log x = \frac{1}{x}$,we get:
$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$.
This can be written as $(x \log x)^{-1}$.
Therefore,the correct option is $C$.

(A) Given expression: $y = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2$
Expand the square using the identity $(a+b)^2 = a^2 + b^2 + 2ab$:
$y = (\sqrt{x})^2 + \left( \frac{1}{\sqrt{x}} \right)^2 + 2(\sqrt{x}) \left( \frac{1}{\sqrt{x}} \right)$
$y = x + \frac{1}{x} + 2$
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(2)$
$\frac{dy}{dx} = 1 - x^{-2} + 0$
$\frac{dy}{dx} = 1 - \frac{1}{x^2}$
Thus,the correct option is $A$.

(C) Given $y = x + \frac{1}{x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(x^{-1})$
$\frac{dy}{dx} = 1 - x^{-2} = 1 - \frac{1}{x^2}$.
Now,consider the expression $x^2 \frac{dy}{dx} - xy + 2$:
Substitute $\frac{dy}{dx} = 1 - \frac{1}{x^2}$ and $y = x + \frac{1}{x}$ into the expression:
$= x^2 \left( 1 - \frac{1}{x^2} \right) - x \left( x + \frac{1}{x} \right) + 2$
$= (x^2 - 1) - (x^2 + 1) + 2$
$= x^2 - 1 - x^2 - 1 + 2$
$= 0$.
Thus,$x^2 \frac{dy}{dx} - xy + 2 = 0$.

(B) We need to find the derivative of $f(x) = \frac{1}{x^4 \sec x}$.
Since $\frac{1}{\sec x} = \cos x$,we can rewrite the function as $f(x) = \frac{\cos x}{x^4}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = \cos x$ and $v = x^4$:
$u' = -\sin x$ and $v' = 4x^3$.
Substituting these into the quotient rule:
$\frac{d}{dx} \left( \frac{\cos x}{x^4} \right) = \frac{x^4(-\sin x) - \cos x(4x^3)}{(x^4)^2}$
$= \frac{-x^4 \sin x - 4x^3 \cos x}{x^8}$
$= \frac{-x^3(x \sin x + 4 \cos x)}{x^8}$
$= \frac{-(x \sin x + 4 \cos x)}{x^5}$.
Thus,the correct option is $B$.

(C) Given the expression $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(\frac{x^2}{2!}) + \dots + \frac{d}{dx}(\frac{x^n}{n!})$
Since $\frac{d}{dx}(x^k) = k \cdot x^{k-1}$,we have:
$\frac{dy}{dx} = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots + \frac{nx^{n-1}}{n!}$
Simplifying the terms:
$\frac{dy}{dx} = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!}$
Now,observe that $y = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{x^n}{n!}$.
Therefore,$\frac{dy}{dx} = y - \frac{x^n}{n!}$.

(A) Given the series $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$.
We know that the Maclaurin series expansion for the exponential function is $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$.
Therefore,$y = e^x$.
Now,differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$.
Since $y = e^x$,we have $\frac{dy}{dx} = y$.

(A) Given the equation $y = \frac{1}{a - z}$.
Rearranging the equation to solve for $z$:
$a - z = \frac{1}{y}$
$z = a - \frac{1}{y}$
Now,differentiate $z$ with respect to $y$:
$\frac{dz}{dy} = \frac{d}{dy}(a - \frac{1}{y})$
Since the derivative of a constant $a$ is $0$ and the derivative of $-\frac{1}{y}$ is $\frac{1}{y^2}$:
$\frac{dz}{dy} = 0 - (-1) \times y^{-2} = \frac{1}{y^2}$
Substitute $y = \frac{1}{a - z}$ back into the expression:
$\frac{dz}{dy} = \frac{1}{(\frac{1}{a - z})^2} = (a - z)^2$
Since $(a - z)^2 = (z - a)^2$,the final result is $(z - a)^2$.

(A) Let $y = x^2 e^x \sin x$. Applying the product rule for three functions,$\frac{d}{dx}(uvw) = u'vw + uv'w + uvw'$.
Here,$u = x^2$,$v = e^x$,and $w = \sin x$.
$\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot e^x \sin x + x^2 \cdot \frac{d}{dx}(e^x) \cdot \sin x + x^2 \cdot e^x \cdot \frac{d}{dx}(\sin x)$
$= 2x e^x \sin x + x^2 e^x \sin x + x^2 e^x \cos x$
$= x e^x (2 \sin x + x \sin x + x \cos x)$.

(C) To find the derivative of $\cos((1 - x^2)^2)$,we use the chain rule.
Let $y = \cos((1 - x^2)^2)$.
Applying the chain rule: $\frac{dy}{dx} = -\sin((1 - x^2)^2) \cdot \frac{d}{dx}((1 - x^2)^2)$.
Now,differentiate $(1 - x^2)^2$ using the power rule and chain rule:
$\frac{d}{dx}((1 - x^2)^2) = 2(1 - x^2) \cdot \frac{d}{dx}(1 - x^2) = 2(1 - x^2) \cdot (-2x) = -4x(1 - x^2)$.
Substituting this back into the expression:
$\frac{dy}{dx} = -\sin((1 - x^2)^2) \cdot (-4x(1 - x^2)) = 4x(1 - x^2)\sin((1 - x^2)^2)$.
Thus,the correct option is $C$.

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

(D) Given the function $y = \cos (\sin {x^2}).$
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 \frac{d}{dx}(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} $
Therefore,$\cos {x^2} = \cos \frac{\pi }{2} = 0 $
Substituting this value into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\sin (\sin \frac{\pi }{2}) \cdot 0 \cdot 2\sqrt {\frac{\pi }{2}} = 0 $
Thus,the correct option is $D$.

(A) We know that $\log |x|$ is defined as:
$\log |x| = \begin{cases} \log x, & \text{if } x > 0 \\ \log(-x), & \text{if } x < 0 \end{cases}$
Case $1$: If $x > 0$,then $\frac{d}{dx}(\log x) = \frac{1}{x}$.
Case $2$: If $x < 0$,then using the chain rule,$\frac{d}{dx}(\log(-x)) = \frac{1}{-x} \cdot \frac{d}{dx}(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.
Since the derivative is $\frac{1}{x}$ in both cases,we conclude that $\frac{d}{dx} \log |x| = \frac{1}{x}$ for all $x \ne 0$.

(D) Given $f(x) = x^2 - 3x$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3$.
We are given the condition $f(x) = f'(x)$:
$x^2 - 3x = 2x - 3$.
Rearranging the terms to form a quadratic equation:
$x^2 - 5x + 3 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-5, c=3$:
$x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(3)}}{2(1)} = \frac{5 \pm \sqrt{25 - 12}}{2} = \frac{5 \pm \sqrt{13}}{2}$.
Thus,the points are $\frac{5 + \sqrt{13}}{2}$ and $\frac{5 - \sqrt{13}}{2}$,which are not listed in the options.
Therefore,the correct option is $D$.

(C) Given the function $f(x) = mx + c$.
Taking the derivative with respect to $x$,we get $f'(x) = m$.
Given $f'(0) = 1$,substituting $x = 0$ into $f'(x) = m$ gives $m = 1$.
Given $f(0) = 1$,substituting $x = 0$ into $f(x) = mx + c$ gives $f(0) = m(0) + c = 1$,so $c = 1$.
Thus,the function is $f(x) = 1x + 1 = x + 1$.
To find $f(2)$,substitute $x = 2$ into the function: $f(2) = 2 + 1 = 3$.

(A) Given $y = x \left[ \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) + \sin x \right] + \frac{1}{2\sqrt{x}}$.
Using the identity $(a+b)(a-b) = a^2 - b^2$,we have $\left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \cos x$.
Substituting this into the expression for $y$:
$y = x(\cos x + \sin x) + \frac{1}{2}x^{-1/2}$.
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}[x(\cos x + \sin x)] + \frac{d}{dx}(\frac{1}{2}x^{-1/2})$.
$\frac{dy}{dx} = x(-\sin x + \cos x) + 1(\cos x + \sin x) + \frac{1}{2} \cdot (-\frac{1}{2})x^{-3/2}$.
$\frac{dy}{dx} = -x\sin x + x\cos x + \cos x + \sin x - \frac{1}{4x\sqrt{x}}$.
$\frac{dy}{dx} = (1 + x)\cos x + (1 - x)\sin x - \frac{1}{4x\sqrt{x}}$.

(A) Let $y = a^x + \log x \cdot \sin x$.
Differentiating with respect to $x$,we apply the sum rule and the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(a^x) + \frac{d}{dx}(\log x \cdot \sin x)$.
Using the standard derivative formula $\frac{d}{dx}(a^x) = a^x \log_e a$ and the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:
$\frac{dy}{dx} = a^x \log_e a + \left( \log x \cdot \frac{d}{dx}(\sin x) + \sin x \cdot \frac{d}{dx}(\log x) \right)$.
$\frac{dy}{dx} = a^x \log_e a + \log x \cdot \cos x + \sin x \cdot \frac{1}{x}$.
Thus,$\frac{dy}{dx} = a^x \log_e a + \frac{\sin x}{x} + \log x \cdot \cos x$.
Therefore,the correct option is $A$.

(D) Let $y = \tan^{-1} \left( \frac{ax - b}{bx + a} \right)$.
We can rewrite the argument of the inverse tangent function by dividing the numerator and denominator by $a$:
$y = \tan^{-1} \left( \frac{x - \frac{b}{a}}{1 + \frac{b}{a}x} \right)$.
Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A - B}{1 + AB} \right)$,we get:
$y = \tan^{-1}(x) - \tan^{-1} \left( \frac{b}{a} \right)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(x) \right) - \frac{d}{dx} \left( \tan^{-1} \left( \frac{b}{a} \right) \right)$.
Since $\tan^{-1} \left( \frac{b}{a} \right)$ is a constant,its derivative is $0$.
Therefore,$\frac{dy}{dx} = \frac{1}{1 + x^2} - 0 = \frac{1}{1 + x^2}$.
Comparing this with the given options,the correct answer is not listed.

(B) Let $y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}}$.
Using the identities $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:
$y = \sqrt{\frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x + \sin^2 x + 2 \sin x \cos x}} = \sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} = \frac{\cos x - \sin x}{\cos x + \sin x}$.
Dividing numerator and denominator by $\cos x$:
$y = \frac{1 - \tan x}{1 + \tan x} = \tan \left( \frac{\pi}{4} - x \right)$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \sec^2 \left( \frac{\pi}{4} - x \right) \cdot \frac{d}{dx} \left( \frac{\pi}{4} - x \right) = \sec^2 \left( \frac{\pi}{4} - x \right) \cdot (-1) = -\sec^2 \left( \frac{\pi}{4} - x \right)$.

(D) We are given the expression $y = \frac{\cot^2 x - 1}{\cot^2 x + 1}$.
First,we simplify the expression using trigonometric identities:
$\frac{\cot^2 x - 1}{\cot^2 x + 1} = \frac{\frac{\cos^2 x}{\sin^2 x} - 1}{\frac{\cos^2 x}{\sin^2 x} + 1} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x}$.
Since $\cos^2 x + \sin^2 x = 1$ and $\cos^2 x - \sin^2 x = \cos 2x$,the expression simplifies to $\cos 2x$.
Now,we differentiate with respect to $x$:
$\frac{d}{dx}(\cos 2x) = -\sin 2x \cdot \frac{d}{dx}(2x) = -2\sin 2x$.
Thus,the correct option is $D$.

(B) Given $f(x) = x \tan^{-1} x$.
Applying the product rule for differentiation,$(uv)' = u'v + uv'$,where $u = x$ and $v = \tan^{-1} x$.
$f'(x) = \frac{d}{dx}(x) \cdot \tan^{-1} x + x \cdot \frac{d}{dx}(\tan^{-1} x)$.
$f'(x) = 1 \cdot \tan^{-1} x + x \cdot \frac{1}{1 + x^2}$.
$f'(x) = \tan^{-1} x + \frac{x}{1 + x^2}$.
Now,substitute $x = 1$ into the derivative:
$f'(1) = \tan^{-1}(1) + \frac{1}{1 + 1^2}$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$,we have:
$f'(1) = \frac{\pi}{4} + \frac{1}{2}$.

(C) Given $y = b \cos \log \left( \frac{x}{n} \right)^n$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = -b \sin \left( \log \left( \frac{x}{n} \right)^n \right) \cdot \frac{d}{dx} \left( \log \left( \frac{x}{n} \right)^n \right)$.
Using the property of logarithms,$\log \left( \frac{x}{n} \right)^n = n \log \left( \frac{x}{n} \right)$.
So,$\frac{d}{dx} \left( n \log \left( \frac{x}{n} \right) \right) = n \cdot \frac{1}{x/n} \cdot \frac{1}{n} = \frac{n}{x}$.
Therefore,$\frac{dy}{dx} = -b \sin \left( \log \left( \frac{x}{n} \right)^n \right) \cdot \frac{n}{x} = -\frac{nb}{x} \sin \log \left( \frac{x}{n} \right)^n$.

(A) Let $y = \sin^n x \cos nx$.
Using 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 $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$\frac{dy}{dx} = n \sin^{n-1} x \cos(nx + x) = n \sin^{n-1} x \cos(n+1)x$.

(A) Given $y = \log {\left( \frac{1 + x}{1 - x} \right)^{1/4}} - \frac{1}{2}{\tan ^{ - 1}}x$.
Using properties of logarithms,$y = \frac{1}{4} [\log(1 + x) - \log(1 - x)] - \frac{1}{2}{\tan ^{ - 1}}x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1}{1 + x} - \frac{1}{1 - x}(-1) \right] - \frac{1}{2} \left( \frac{1}{1 + x^2} \right)$
$\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1}{1 + x} + \frac{1}{1 - x} \right] - \frac{1}{2(1 + x^2)}$
$\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1 - x + 1 + x}{(1 + x)(1 - x)} \right] - \frac{1}{2(1 + x^2)}$
$\frac{dy}{dx} = \frac{1}{4} \left[ \frac{2}{1 - x^2} \right] - \frac{1}{2(1 + x^2)}$
$\frac{dy}{dx} = \frac{1}{2(1 - x^2)} - \frac{1}{2(1 + x^2)}$
$\frac{dy}{dx} = \frac{1}{2} \left[ \frac{(1 + x^2) - (1 - x^2)}{(1 - x^2)(1 + x^2)} \right]$
$\frac{dy}{dx} = \frac{1}{2} \left[ \frac{2x^2}{1 - x^4} \right] = \frac{x^2}{1 - x^4}$.

(B) Given $f(x) = \sqrt{1 + \cos^2(x^2)}$.
Using the chain rule,$f'(x) = \frac{1}{2\sqrt{1 + \cos^2(x^2)}} \cdot \frac{d}{dx}(1 + \cos^2(x^2))$.
$f'(x) = \frac{1}{2\sqrt{1 + \cos^2(x^2)}} \cdot (2 \cos(x^2) \cdot (-\sin(x^2)) \cdot 2x)$.
$f'(x) = \frac{-2x \cos(x^2) \sin(x^2)}{\sqrt{1 + \cos^2(x^2)}} = \frac{-x \sin(2x^2)}{\sqrt{1 + \cos^2(x^2)}}$.
At $x = \frac{\sqrt{\pi}}{2}$,$x^2 = \frac{\pi}{4}$.
$f'\left(\frac{\sqrt{\pi}}{2}\right) = \frac{-\frac{\sqrt{\pi}}{2} \sin(2 \cdot \frac{\pi}{4})}{\sqrt{1 + \cos^2(\frac{\pi}{4})}} = \frac{-\frac{\sqrt{\pi}}{2} \sin(\frac{\pi}{2})}{\sqrt{1 + (\frac{1}{\sqrt{2}})^2}}$.
Since $\sin(\frac{\pi}{2}) = 1$ and $\cos^2(\frac{\pi}{4}) = \frac{1}{2}$,
$f'\left(\frac{\sqrt{\pi}}{2}\right) = \frac{-\frac{\sqrt{\pi}}{2} \cdot 1}{\sqrt{1 + \frac{1}{2}}} = \frac{-\frac{\sqrt{\pi}}{2}}{\sqrt{\frac{3}{2}}} = -\frac{\sqrt{\pi}}{2} \cdot \sqrt{\frac{2}{3}} = -\sqrt{\frac{\pi}{4} \cdot \frac{2}{3}} = -\sqrt{\frac{\pi}{6}}$.

(B) Let $y = \sqrt{\frac{1 + \cos 2x}{1 - \cos 2x}}$.
Using trigonometric identities $1 + \cos 2x = 2 \cos^2 x$ and $1 - \cos 2x = 2 \sin^2 x$,we get:
$y = \sqrt{\frac{2 \cos^2 x}{2 \sin^2 x}} = \sqrt{\cot^2 x} = |\cot x|$.
Assuming the expression is defined in the first quadrant,$y = \cot x$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cot x) = -\csc^2 x$.
Thus,the correct option is $B$.

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

(B) Given $y = \sin [\cos (\sin x)]$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\sin [\cos (\sin x)]) = \cos [\cos (\sin x)] \cdot \frac{d}{dx} (\cos (\sin x))$.
Now,differentiate $\cos (\sin x)$:
$\frac{d}{dx} (\cos (\sin x)) = -\sin (\sin x) \cdot \frac{d}{dx} (\sin x)$.
Finally,differentiate $\sin x$:
$\frac{d}{dx} (\sin x) = \cos x$.
Combining all these parts:
$\frac{dy}{dx} = \cos [\cos (\sin x)] \cdot [-\sin (\sin x)] \cdot \cos x = -\cos [\cos (\sin x)] \sin (\sin x) \cos x$.

(D) Given $y = t^{4/3} - 3t^{-2/3}$.
Applying the power rule $\frac{d}{dt}(t^n) = nt^{n-1}$:
$\frac{dy}{dt} = \frac{4}{3}t^{(4/3 - 1)} - 3 \times (-2/3)t^{(-2/3 - 1)}$
$\frac{dy}{dt} = \frac{4}{3}t^{1/3} + 2t^{-5/3}$
To simplify,find a common denominator of $3t^{5/3}$:
$\frac{dy}{dt} = \frac{4}{3}t^{1/3} \times \frac{t^{5/3}}{t^{5/3}} + 2t^{-5/3} = \frac{4t^{6/3}}{3t^{5/3}} + \frac{6}{3t^{5/3}}$
$\frac{dy}{dt} = \frac{4t^2 + 6}{3t^{5/3}} = \frac{2(2t^2 + 3)}{3t^{5/3}}$.

(B) Given $y = x^2 \log x + 2x^{-1/2}$.
Applying the product rule to $x^2 \log x$ and the power rule to $2x^{-1/2}$:
$\frac{dy}{dx} = \frac{d}{dx}(x^2 \log x) + \frac{d}{dx}(2x^{-1/2})$
$\frac{dy}{dx} = (x^2 \cdot \frac{1}{x} + \log x \cdot 2x) + 2(-\frac{1}{2}x^{-3/2})$
$\frac{dy}{dx} = x + 2x \log x - x^{-3/2}$
$\frac{dy}{dx} = x + 2x \log x - \frac{1}{x^{3/2}}$
Thus,the correct option is $B$.

(B) To find the derivative of $\frac{e^x}{1 + x^2}$,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 $u = e^x$ and $v = 1 + x^2$.
Then $\frac{du}{dx} = e^x$ and $\frac{dv}{dx} = 2x$.
Applying the quotient rule:
$\frac{d}{dx} \left( \frac{e^x}{1 + x^2} \right) = \frac{(1 + x^2)e^x - e^x(2x)}{(1 + x^2)^2}$.
Factor out $e^x$ from the numerator:
$= \frac{e^x(1 + x^2 - 2x)}{(1 + x^2)^2}$.
Recognizing that $1 + x^2 - 2x = (1 - x)^2$:
$= \frac{e^x(1 - x)^2}{(1 + x^2)^2}$.

(D) Given $y = \frac{\tan x + \cot x}{\tan x - \cot x}$.
We can simplify the expression by converting it to $\sin x$ and $\cos x$ or using trigonometric identities.
$y = \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}} = \frac{\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}}{\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}} = \frac{1}{-(\cos^2 x - \sin^2 x)} = -\frac{1}{\cos 2x} = -\sec 2x$.
Now,differentiate $y = -\sec 2x$ with respect to $x$ using the chain rule:
$\frac{dy}{dx} = -(\sec 2x \tan 2x) \cdot \frac{d}{dx}(2x) = -2 \sec 2x \tan 2x$.
Thus,the correct option is $D$.

(A) Given $A = \frac{2^x \cot x}{\sqrt{x}}$.
Applying the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
$\frac{dA}{dx} = \frac{\sqrt{x} \frac{d}{dx}(2^x \cot x) - 2^x \cot x \frac{d}{dx}(\sqrt{x})}{x}$
$= \frac{\sqrt{x} [2^x \ln 2 \cot x - 2^x \csc^2 x] - 2^x \cot x \frac{1}{2\sqrt{x}}}{x}$
$= \frac{2^x [\sqrt{x} \ln 2 \cot x - \sqrt{x} \csc^2 x - \frac{\cot x}{2\sqrt{x}}]}{x}$
$= \frac{2^x [2x \ln 2 \cot x - 2x \csc^2 x - \cot x]}{2x^{3/2}}$
$= \frac{2^x [2x \ln 2 \cot x - \cot x - 2x \csc^2 x]}{2x^{3/2}}$
$= \frac{2^{x-1} [\cot x (2x \ln 2 - 1) - 2x \csc^2 x]}{x^{3/2}}$
Since $\ln(4^x/e) = \ln(4^x) - \ln e = x \ln 4 - 1 = 2x \ln 2 - 1$,the expression becomes:
$= \frac{2^{x-1} \{ -2x \csc^2 x + \cot x \cdot \log(\frac{4^x}{e}) \}}{x^{3/2}}$.

(B) To find the derivative of $\frac{\log x}{\sin x}$,we use the quotient rule: $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = \log x$ and $v = \sin x$.
Then,$\frac{du}{dx} = \frac{1}{x}$ and $\frac{dv}{dx} = \cos x$.
Applying the quotient rule:
$\frac{d}{dx} \left( \frac{\log x}{\sin x} \right) = \frac{\sin x \cdot \frac{1}{x} - \log x \cdot \cos x}{(\sin x)^2}$.
This simplifies to $\frac{\frac{\sin x}{x} - \log x \cdot \cos x}{\sin^2 x}$.

(A) Given $y = \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$.
Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})$:
$y = \frac{(\sqrt{x^2 + 1} + \sqrt{x^2 - 1})^2}{(\sqrt{x^2 + 1})^2 - (\sqrt{x^2 - 1})^2}$
$y = \frac{(x^2 + 1) + (x^2 - 1) + 2\sqrt{(x^2 + 1)(x^2 - 1)}}{(x^2 + 1) - (x^2 - 1)}$
$y = \frac{2x^2 + 2\sqrt{x^4 - 1}}{2}$
$y = x^2 + \sqrt{x^4 - 1}$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}((x^4 - 1)^{1/2})$
$\frac{dy}{dx} = 2x + \frac{1}{2}(x^4 - 1)^{-1/2} \cdot (4x^3)$
$\frac{dy}{dx} = 2x + \frac{2x^3}{\sqrt{x^4 - 1}}$.

(A) Given $y = (x \cot^3 x)^{3/2}$.
Applying the chain rule,$\frac{dy}{dx} = \frac{3}{2}(x \cot^3 x)^{1/2} \cdot \frac{d}{dx}(x \cot^3 x)$.
Using the product rule for the derivative of $(x \cot^3 x)$:
$\frac{d}{dx}(x \cot^3 x) = \frac{d}{dx}(x) \cdot \cot^3 x + x \cdot \frac{d}{dx}(\cot^3 x)$.
$= 1 \cdot \cot^3 x + x \cdot (3 \cot^2 x \cdot \frac{d}{dx}(\cot x))$.
$= \cot^3 x + x \cdot (3 \cot^2 x \cdot (-\csc^2 x))$.
$= \cot^3 x - 3x \cot^2 x \csc^2 x$.
Therefore,$\frac{dy}{dx} = \frac{3}{2}(x \cot^3 x)^{1/2} [\cot^3 x - 3x \cot^2 x \csc^2 x]$.

(B) To find the derivative of $y = \cos(\sin(x^2))$,we use the chain rule.
Let $u = \sin(x^2)$ and $v = x^2$.
Then $y = \cos(u)$,$u = \sin(v)$,and $v = x^2$.
Applying the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.
$\frac{dy}{du} = -\sin(u) = -\sin(\sin(x^2))$.
$\frac{du}{dv} = \cos(v) = \cos(x^2)$.
$\frac{dv}{dx} = 2x$.
Multiplying these together: $\frac{dy}{dx} = -\sin(\sin(x^2)) \cdot \cos(x^2) \cdot 2x$.
Thus,the correct option is $B$.

(C) Given $y = \sin (\sqrt {\sin x + \cos x} )$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \cos (\sqrt {\sin x + \cos x}) \cdot \frac{d}{dx} (\sqrt {\sin x + \cos x})$
Using the derivative of $\sqrt{u}$ which is $\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$:
$\frac{dy}{dx} = \cos (\sqrt {\sin x + \cos x}) \cdot \frac{1}{2\sqrt{\sin x + \cos x}} \cdot \frac{d}{dx} (\sin x + \cos x)$
Since $\frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x$:
$\frac{dy}{dx} = \frac{1}{2} \frac{\cos (\sqrt {\sin x + \cos x})}{\sqrt {\sin x + \cos x}} (\cos x - \sin x)$.

(D) Given $y = \sin \left( \frac{1 + x^2}{1 - x^2} \right)$.
Using the chain rule,$\frac{dy}{dx} = \cos \left( \frac{1 + x^2}{1 - x^2} \right) \cdot \frac{d}{dx} \left( \frac{1 + x^2}{1 - x^2} \right)$.
Applying the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$\frac{d}{dx} \left( \frac{1 + x^2}{1 - x^2} \right) = \frac{(1 - x^2)(2x) - (1 + x^2)(-2x)}{(1 - x^2)^2}$
$= \frac{2x - 2x^3 + 2x + 2x^3}{(1 - x^2)^2} = \frac{4x}{(1 - x^2)^2}$.
Therefore,$\frac{dy}{dx} = \frac{4x}{(1 - x^2)^2} \cos \left( \frac{1 + x^2}{1 - x^2} \right)$.

(A) Given $y = \sqrt{\frac{1 + \tan x}{1 - \tan x}}$.
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we know that $\tan\left(\frac{\pi}{4} + x\right) = \frac{\tan(\pi/4) + \tan x}{1 - \tan(\pi/4)\tan x} = \frac{1 + \tan x}{1 - \tan x}$.
So,$y = \sqrt{\tan\left(\frac{\pi}{4} + x\right)}$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan\left(\frac{\pi}{4} + x\right)}} \cdot \frac{d}{dx}\left(\tan\left(\frac{\pi}{4} + x\right)\right)$.
$\frac{dy}{dx} = \frac{1}{2\sqrt{\tan\left(\frac{\pi}{4} + x\right)}} \cdot \sec^2\left(\frac{\pi}{4} + x\right)$.
Since $\frac{1}{\sqrt{\tan(\pi/4 + x)}} = \sqrt{\frac{1 - \tan x}{1 + \tan x}}$,we get:
$\frac{dy}{dx} = \frac{1}{2}\sqrt{\frac{1 - \tan x}{1 + \tan x}} \cdot \sec^2\left(\frac{\pi}{4} + x\right)$.

(C) To find the derivative of $({x^2} + \cos x)^4$,we use the chain rule.
Let $u = {x^2} + \cos x$.
Then the expression becomes $u^4$.
The derivative with respect to $x$ is given by $\frac{d}{dx}(u^4) = \frac{d}{du}(u^4) \cdot \frac{du}{dx}$.
$\frac{d}{du}(u^4) = 4u^3$.
$\frac{du}{dx} = \frac{d}{dx}({x^2} + \cos x) = 2x - \sin x$.
Combining these,we get $4({x^2} + \cos x)^3 \cdot (2x - \sin x)$.

(A) Let $y = \sqrt{x \sin x}$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{2 \sqrt{x \sin x}} \cdot \frac{d}{dx}(x \sin x)$.
Applying the product rule to $x \sin x$,we get $\frac{d}{dx}(x \sin x) = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x$.
Therefore,$\frac{dy}{dx} = \frac{\sin x + x \cos x}{2 \sqrt{x \sin x}}$.

(B) Let $y = \sqrt{\sec^2 x + \text{cosec}^2 x}$.
First,simplify the expression inside the square root:
$y = \sqrt{\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}} = \sqrt{\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}} = \sqrt{\frac{1}{\sin^2 x \cos^2 x}}$
Using the identity $\sin 2x = 2 \sin x \cos x$,we have $\sin x \cos x = \frac{\sin 2x}{2}$.
Thus,$y = \sqrt{\frac{1}{(\frac{\sin 2x}{2})^2}} = \sqrt{\frac{4}{\sin^2 2x}} = \frac{2}{\sin 2x} = 2 \text{cosec } 2x$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(2 \text{cosec } 2x) = 2 \cdot (-\text{cosec } 2x \cot 2x) \cdot \frac{d}{dx}(2x)$
$\frac{dy}{dx} = 2 \cdot (-\text{cosec } 2x \cot 2x) \cdot 2 = -4 \text{cosec } 2x \cot 2x$.

(A) Let $y = \frac{\sec x + \tan x}{\sec x - \tan x}$.
Simplifying the expression: $y = \frac{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}} = \frac{1 + \sin x}{1 - \sin x}$.
Now,differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$u = 1 + \sin x \implies u' = \cos x$
$v = 1 - \sin x \implies v' = -\cos x$
$\frac{dy}{dx} = \frac{(1 - \sin x)(\cos x) - (1 + \sin x)(-\cos x)}{(1 - \sin x)^2}$
$\frac{dy}{dx} = \frac{\cos x - \sin x \cos x + \cos x + \sin x \cos x}{(1 - \sin x)^2}$
$\frac{dy}{dx} = \frac{2\cos x}{(1 - \sin x)^2}$.

(B) To find the derivative of $f(x) = x^3 \tan^2 \frac{x}{2}$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.
Let $u = x^3$ and $v = \tan^2 \frac{x}{2}$.
Then $\frac{du}{dx} = 3x^2$.
Using the chain rule for $v$,$\frac{dv}{dx} = 2 \tan \frac{x}{2} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \tan \frac{x}{2} \sec^2 \frac{x}{2}$.
Now,applying the product rule:
$\frac{d}{dx} \left( x^3 \tan^2 \frac{x}{2} \right) = x^3 \left( \tan \frac{x}{2} \sec^2 \frac{x}{2} \right) + \tan^2 \frac{x}{2} (3x^2)$.
$= x^3 \tan \frac{x}{2} \sec^2 \frac{x}{2} + 3x^2 \tan^2 \frac{x}{2}$.

(B) Given $f(x) = \tan^{-1}\left( \frac{\sin x}{1 + \cos x} \right)$.
Using trigonometric identities,$\sin x = 2 \sin\left( \frac{x}{2} \right) \cos\left( \frac{x}{2} \right)$ and $1 + \cos x = 2 \cos^2\left( \frac{x}{2} \right)$.
Substituting these into the function,we get $f(x) = \tan^{-1}\left( \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} \right) = \tan^{-1}\left( \tan\left( \frac{x}{2} \right) \right)$.
Thus,$f(x) = \frac{x}{2}$.
Differentiating with respect to $x$,we get $f'(x) = \frac{1}{2}$.
Therefore,$f'\left( \frac{\pi}{3} \right) = \frac{1}{2}$.