Derivative at a point, Standard differentiation Questions in English

(A) Let $y = e^{x^{3}}$.
To differentiate $y$ with respect to $x$,we use the chain rule.
$\frac{dy}{dx} = \frac{d}{dx}(e^{x^{3}})$.
Applying the chain rule,we treat $x^{3}$ as a function $u$,so $\frac{d}{dx}(e^{u}) = e^{u} \cdot \frac{du}{dx}$.
$\frac{dy}{dx} = e^{x^{3}} \cdot \frac{d}{dx}(x^{3})$.
Since $\frac{d}{dx}(x^{3}) = 3x^{2}$,we get:
$\frac{dy}{dx} = e^{x^{3}} \cdot 3x^{2} = 3x^{2}e^{x^{3}}$.

Let $y = \sin \left(\tan ^{-1} e^{-x}\right)$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left[ \sin \left( \tan^{-1} e^{-x} \right) \right]$
$= \cos \left( \tan^{-1} e^{-x} \right) \cdot \frac{d}{dx} \left( \tan^{-1} e^{-x} \right)$
$= \cos \left( \tan^{-1} e^{-x} \right) \cdot \frac{1}{1 + (e^{-x})^2} \cdot \frac{d}{dx} (e^{-x})$
$= \cos \left( \tan^{-1} e^{-x} \right) \cdot \frac{1}{1 + e^{-2x}} \cdot (e^{-x} \cdot -1)$
$= \frac{-e^{-x} \cos \left( \tan^{-1} e^{-x} \right)}{1 + e^{-2x}}$

Let $y = e^{x}+e^{x^{2}}+e^{x^{3}}+e^{x^{4}}+e^{x^{5}}$.
To find the derivative with respect to $x$,we apply the sum rule and the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}(e^{x}) + \frac{d}{dx}(e^{x^{2}}) + \frac{d}{dx}(e^{x^{3}}) + \frac{d}{dx}(e^{x^{4}}) + \frac{d}{dx}(e^{x^{5}})$
Using the chain rule $\frac{d}{dx}(e^{u}) = e^{u} \cdot \frac{du}{dx}$:
$= e^{x} + e^{x^{2}} \cdot \frac{d}{dx}(x^{2}) + e^{x^{3}} \cdot \frac{d}{dx}(x^{3}) + e^{x^{4}} \cdot \frac{d}{dx}(x^{4}) + e^{x^{5}} \cdot \frac{d}{dx}(x^{5})$
$= e^{x} + e^{x^{2}} \cdot (2x) + e^{x^{3}} \cdot (3x^{2}) + e^{x^{4}} \cdot (4x^{3}) + e^{x^{5}} \cdot (5x^{4})$
$= e^{x} + 2x e^{x^{2}} + 3x^{2} e^{x^{3}} + 4x^{3} e^{x^{4}} + 5x^{4} e^{x^{5}}$

Let $y = \sqrt{e^{\sqrt{x}}}$.
Then,$y^2 = e^{\sqrt{x}}$.
By differentiating this relationship with respect to $x$,we obtain:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(e^{\sqrt{x}})$.
Using the chain rule,we get:
$2y \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{d}{dx}(\sqrt{x})$.
Since $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$,we have:
$2y \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$.
$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4y\sqrt{x}}$.
Substituting $y = \sqrt{e^{\sqrt{x}}}$ back into the equation:
$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}$.
$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

Let $y = \frac{\cos x}{\log x}$.
By using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$,we obtain:
$\frac{dy}{dx} = \frac{(\log x) \frac{d}{dx}(\cos x) - (\cos x) \frac{d}{dx}(\log x)}{(\log x)^2}$
Since $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\log x) = \frac{1}{x}$,we have:
$\frac{dy}{dx} = \frac{(\log x)(-\sin x) - (\cos x)(\frac{1}{x})}{(\log x)^2}$
$\frac{dy}{dx} = \frac{-\sin x \log x - \frac{\cos x}{x}}{(\log x)^2}$
Multiplying the numerator and denominator by $x$,we get:
$\frac{dy}{dx} = \frac{-(x \sin x \log x + \cos x)}{x(\log x)^2}, x > 0$

Let $y = \cos (\log x + e^x)$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\cos (\log x + e^x)]$
Using the derivative of $\cos(u)$ which is $-\sin(u) \cdot \frac{du}{dx}$,we get:
$\frac{dy}{dx} = -\sin (\log x + e^x) \cdot \frac{d}{dx} (\log x + e^x)$
Now,differentiate the terms inside the bracket:
$\frac{d}{dx} (\log x) = \frac{1}{x}$ and $\frac{d}{dx} (e^x) = e^x$.
Substituting these back into the expression:
$\frac{dy}{dx} = -\sin (\log x + e^x) \cdot (\frac{1}{x} + e^x)$
Therefore,the final derivative is:
$\frac{dy}{dx} = -(\frac{1}{x} + e^x) \sin (\log x + e^x)$ for $x > 0$.

(A) Let $y = a^{x}$.
Taking the natural logarithm on both sides,we get:
$\log y = x \log a$
Differentiating both sides with respect to $x$,we have:
$\frac{1}{y} \frac{dy}{dx} = \log a$
Multiplying both sides by $y$,we get:
$\frac{dy}{dx} = y \log a$
Substituting $y = a^{x}$ back into the equation:
$\frac{dy}{dx} = a^{x} \log a$
Alternatively,using the chain rule:
$\frac{d}{dx}(a^{x}) = \frac{d}{dx}(e^{x \log a}) = e^{x \log a} \cdot \frac{d}{dx}(x \log a) = a^{x} \cdot \log a$.

Let $y = (x^{2}-5x+8)(x^{3}+7x+9)$.
Let $u = x^{2}-5x+8$ and $v = x^{3}+7x+9$.
Then $y = uv$.
By the product rule,$\frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx}$.
$\frac{du}{dx} = \frac{d}{dx}(x^{2}-5x+8) = 2x-5$.
$\frac{dv}{dx} = \frac{d}{dx}(x^{3}+7x+9) = 3x^{2}+7$.
Substituting these into the formula:
$\frac{dy}{dx} = (2x-5)(x^{3}+7x+9) + (x^{2}-5x+8)(3x^{2}+7)$.
Expanding the terms:
$(2x-5)(x^{3}+7x+9) = 2x^{4} + 14x^{2} + 18x - 5x^{3} - 35x - 45 = 2x^{4} - 5x^{3} + 14x^{2} - 17x - 45$.
$(x^{2}-5x+8)(3x^{2}+7) = 3x^{4} + 7x^{2} - 15x^{3} - 35x + 24x^{2} + 56 = 3x^{4} - 15x^{3} + 31x^{2} - 35x + 56$.
Adding the two expressions:
$\frac{dy}{dx} = (2x^{4} - 5x^{3} + 14x^{2} - 17x - 45) + (3x^{4} - 15x^{3} + 31x^{2} - 35x + 56)$.
$\frac{dy}{dx} = 5x^{4} - 20x^{3} + 45x^{2} - 52x + 11$.

Let $y = (x^{2}-5x+8)(x^{3}+7x+9)$.
First,expand the product:
$y = x^{2}(x^{3}+7x+9) - 5x(x^{3}+7x+9) + 8(x^{3}+7x+9)$
$y = x^{5} + 7x^{3} + 9x^{2} - 5x^{4} - 35x^{2} - 45x + 8x^{3} + 56x + 72$
Combine like terms:
$y = x^{5} - 5x^{4} + (7+8)x^{3} + (9-35)x^{2} + (-45+56)x + 72$
$y = x^{5} - 5x^{4} + 15x^{3} - 26x^{2} + 11x + 72$
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^{5} - 5x^{4} + 15x^{3} - 26x^{2} + 11x + 72)$
Using the power rule $\frac{d}{dx}(x^{n}) = nx^{n-1}$:
$\frac{dy}{dx} = 5x^{4} - 5(4x^{3}) + 15(3x^{2}) - 26(2x) + 11(1) + 0$
$\frac{dy}{dx} = 5x^{4} - 20x^{3} + 45x^{2} - 52x + 11$

Given $f(x)=2x^{2}+3x-5$. The derivative $f^{\prime}(x)$ is given by $\frac{d}{dx}(2x^{2}+3x-5) = 4x+3$.
First,we find $f^{\prime}(-1)$:
$f^{\prime}(-1) = 4(-1)+3 = -4+3 = -1$.
Next,we find $f^{\prime}(0)$:
$f^{\prime}(0) = 4(0)+3 = 3$.
Now,we verify the expression $f^{\prime}(0)+3f^{\prime}(-1)$:
$f^{\prime}(0)+3f^{\prime}(-1) = 3 + 3(-1) = 3 - 3 = 0$.
Thus,it is proved that $f^{\prime}(0)+3f^{\prime}(-1)=0$.

(A) Since the derivative measures the rate of change of a function,it is intuitive that the derivative of a constant function is zero at every point.
For $f(x) = 3$,the derivative at any point $x = a$ is given by:
$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0} \frac{3 - 3}{h} = \lim_{h \to 0} \frac{0}{h} = 0$.
Thus,at $x = 0$,$f'(0) = 0$.
At $x = 3$,$f'(3) = 0$.

(A) To find the derivative of the function $f(x) = 6 x^{100}-x^{55}+x$,we use the power rule $\frac{d}{dx}(x^n) = n x^{n-1}$.
Applying this to each term:
$\frac{d}{dx}(6 x^{100}) = 6 \times 100 x^{99} = 600 x^{99}$.
$\frac{d}{dx}(-x^{55}) = -55 x^{54}$.
$\frac{d}{dx}(x) = 1$.
Combining these,the derivative is $600 x^{99}-55 x^{54}+1$.

(A) The given function is $f(x) = 1 + x + x^{2} + x^{3} + \dots + x^{50}$.
Taking the derivative with respect to $x$,we get $f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(x^{2}) + \dots + \frac{d}{dx}(x^{50})$.
Using the power rule $\frac{d}{dx}(x^{n}) = nx^{n-1}$,we have $f'(x) = 0 + 1 + 2x + 3x^{2} + \dots + 50x^{49}$.
At $x = 1$,the value is $f'(1) = 1 + 2(1) + 3(1)^{2} + \dots + 50(1)^{49}$.
This simplifies to the sum of the first $50$ natural numbers: $1 + 2 + 3 + \dots + 50$.
The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.
For $n = 50$,the sum is $\frac{50 \times 51}{2} = 25 \times 51 = 1275$.

(B) Given the function $f(x) = \frac{x+1}{x}$.
We can simplify the function as $f(x) = \frac{x}{x} + \frac{1}{x} = 1 + x^{-1}$.
Now,differentiate with respect to $x$:
$\frac{d}{dx} f(x) = \frac{d}{dx} (1 + x^{-1})$.
Using the power rule $\frac{d}{dx} x^n = nx^{n-1}$:
$\frac{d}{dx} (1) + \frac{d}{dx} (x^{-1}) = 0 + (-1)x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$.

(B) To find the derivative of $f(x) = \sin^{2} x$,we use the chain rule.
Let $u = \sin x$,then $f(x) = u^{2}$.
$\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$
$\frac{df}{dx} = 2u \cdot \cos x$
Substituting $u = \sin x$ back into the equation:
$\frac{df}{dx} = 2 \sin x \cos x$
Using the trigonometric identity $\sin 2x = 2 \sin x \cos x$,we get:
$\frac{df}{dx} = \sin 2x$

(N/A) The given function is $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$.
Taking the derivative with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx} \left( \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1 \right)$.
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f^{\prime}(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + \dots + \frac{2x}{2} + 1 + 0$.
$f^{\prime}(x) = x^{99} + x^{98} + \dots + x + 1$.
At $x = 0$,$f^{\prime}(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1$.
At $x = 1$,$f^{\prime}(1) = 1^{99} + 1^{98} + \dots + 1 + 1$.
Since there are $100$ terms in the sum,$f^{\prime}(1) = 1 \times 100 = 100$.
Thus,$f^{\prime}(1) = 100 \times 1 = 100 f^{\prime}(0)$.
Hence,$f^{\prime}(1) = 100 f^{\prime}(0)$ is proved.

Let $f(x) = x^{n} + a x^{n-1} + a^{2} x^{n-2} + \dots + a^{n-1} x + a^{n}$.
$\frac{d}{dx} f(x) = \frac{d}{dx} (x^{n} + a x^{n-1} + a^{2} x^{n-2} + \dots + a^{n-1} x + a^{n})$.
Using the linearity property of derivatives:
$= \frac{d}{dx}(x^{n}) + a \frac{d}{dx}(x^{n-1}) + a^{2} \frac{d}{dx}(x^{n-2}) + \dots + a^{n-1} \frac{d}{dx}(x) + \frac{d}{dx}(a^{n})$.
Applying the power rule $\frac{d}{dx}(x^{k}) = k x^{k-1}$ and noting that $\frac{d}{dx}(a^{n}) = 0$ since $a$ is a constant:
$= n x^{n-1} + a(n-1) x^{n-2} + a^{2}(n-2) x^{n-3} + \dots + a^{n-1}(1) + 0$.
Thus,$f'(x) = n x^{n-1} + a(n-1) x^{n-2} + a^{2}(n-2) x^{n-3} + \dots + a^{n-1}$.

(A) Let $f(x) = (x-a)(x-b)$.
Expanding the expression,we get:
$f(x) = x^2 - ax - bx + ab = x^2 - (a+b)x + ab$.
Now,differentiate with respect to $x$:
$f'(x) = \frac{d}{dx}(x^2 - (a+b)x + ab)$.
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the fact that the derivative of a constant is $0$:
$f'(x) = 2x - (a+b) + 0$.
Therefore,$f'(x) = 2x - a - b$.

(A) Let $f(x) = (ax^2 + b)^2$.
Using the chain rule,where $\frac{d}{dx}[u(x)^n] = n \cdot u(x)^{n-1} \cdot u'(x)$:
$f'(x) = 2(ax^2 + b) \cdot \frac{d}{dx}(ax^2 + b)$
$f'(x) = 2(ax^2 + b) \cdot (2ax + 0)$
$f'(x) = 2(ax^2 + b) \cdot (2ax)$
$f'(x) = 4ax(ax^2 + b)$.

(A) Let $f(x) = \frac{x-a}{x-b}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = x-a$ and $v = x-b$.
$\frac{du}{dx} = 1$ and $\frac{dv}{dx} = 1$.
$f'(x) = \frac{(x-b)(1) - (x-a)(1)}{(x-b)^2}$.
$f'(x) = \frac{x - b - x + a}{(x-b)^2}$.
$f'(x) = \frac{a-b}{(x-b)^2}$.

(A) Let $f(x) = \frac{x^n - a^n}{x - a}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = x^n - a^n$ and $v = x - a$.
$u' = n x^{n-1}$ and $v' = 1$.
$f'(x) = \frac{(x - a)(n x^{n-1}) - (x^n - a^n)(1)}{(x - a)^2}$.
$f'(x) = \frac{n x^n - n a x^{n-1} - x^n + a^n}{(x - a)^2}$.
$f'(x) = \frac{(n - 1) x^n - n a x^{n-1} + a^n}{(x - a)^2}$.

(A) Let $f(x) = 2x - \frac{3}{4}$.
To find the derivative $f'(x)$,we differentiate with respect to $x$:
$f'(x) = \frac{d}{dx}(2x - \frac{3}{4})$
Using the linearity property of the derivative:
$f'(x) = 2 \frac{d}{dx}(x) - \frac{d}{dx}(\frac{3}{4})$
Since $\frac{d}{dx}(x) = 1$ and the derivative of a constant $\frac{3}{4}$ is $0$:
$f'(x) = 2(1) - 0$
$f'(x) = 2$.

(A) Let $f(x) = (5x^{3} + 3x - 1)(x - 1)$.
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (5x^{3} + 3x - 1)\frac{d}{dx}(x - 1) + (x - 1)\frac{d}{dx}(5x^{3} + 3x - 1)$
$f'(x) = (5x^{3} + 3x - 1)(1) + (x - 1)(15x^{2} + 3)$
$f'(x) = 5x^{3} + 3x - 1 + (15x^{3} + 3x - 15x^{2} - 3)$
$f'(x) = 5x^{3} + 15x^{3} - 15x^{2} + 3x + 3x - 1 - 3$
$f'(x) = 20x^{3} - 15x^{2} + 6x - 4$.

(C) Let $f(x) = x^{-3}(5+3x) = 5x^{-3} + 3x^{-2}$.
Now,differentiate with respect to $x$:
$f'(x) = \frac{d}{dx}(5x^{-3} + 3x^{-2})$
$= 5(-3)x^{-4} + 3(-2)x^{-3}$
$= -15x^{-4} - 6x^{-3}$
$= -3x^{-4}(5 + 2x)$
Alternatively,factoring out $-3x^{-3}$:
$= -3x^{-3}(2 + 5x^{-1}) = -3x^{-3}(\frac{2x+5}{x}) = \frac{-3x^{-3}}{x}(2x+5)$.

(A) Let $f(x) = x^{5}(3-6x^{-9})$.
First,simplify the expression:
$f(x) = 3x^{5} - 6x^{5-9} = 3x^{5} - 6x^{-4}$.
Now,differentiate with respect to $x$ using the power rule $\frac{d}{dx}(x^{n}) = nx^{n-1}$:
$f'(x) = \frac{d}{dx}(3x^{5}) - \frac{d}{dx}(6x^{-4})$.
$f'(x) = 3(5x^{4}) - 6(-4x^{-4-1})$.
$f'(x) = 15x^{4} + 24x^{-5}$.
$f'(x) = 15x^{4} + \frac{24}{x^{5}}$.

Let $y = \sqrt{3x+2} + \frac{1}{\sqrt{2x^2+4}} = (3x+2)^{\frac{1}{2}} + (2x^2+4)^{-\frac{1}{2}}$.
Applying the chain rule,we differentiate each term with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}[(3x+2)^{\frac{1}{2}}] + \frac{d}{dx}[(2x^2+4)^{-\frac{1}{2}}]$
$= \frac{1}{2}(3x+2)^{-\frac{1}{2}} \cdot \frac{d}{dx}(3x+2) + \left(-\frac{1}{2}\right)(2x^2+4)^{-\frac{3}{2}} \cdot \frac{d}{dx}(2x^2+4)$
$= \frac{1}{2}(3x+2)^{-\frac{1}{2}} \cdot (3) - \frac{1}{2}(2x^2+4)^{-\frac{3}{2}} \cdot (4x)$
$= \frac{3}{2\sqrt{3x+2}} - \frac{2x}{(2x^2+4)^{\frac{3}{2}}}$

(N/A) Let $y = e^{\sec ^{2} x} + 3 \cos ^{-1} x$.
This function is defined for all $x \in [-1, 1]$.
To find the derivative,we use the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}(e^{\sec ^{2} x}) + \frac{d}{dx}(3 \cos ^{-1} x)$
$= e^{\sec ^{2} x} \cdot \frac{d}{dx}(\sec ^{2} x) + 3 \cdot \left( -\frac{1}{\sqrt{1 - x^{2}}} \right)$
$= e^{\sec ^{2} x} \cdot (2 \sec x \cdot \frac{d}{dx}(\sec x)) - \frac{3}{\sqrt{1 - x^{2}}}$
$= e^{\sec ^{2} x} \cdot (2 \sec x \cdot \sec x \tan x) - \frac{3}{\sqrt{1 - x^{2}}}$
$= 2 \sec ^{2} x \tan x e^{\sec ^{2} x} - \frac{3}{\sqrt{1 - x^{2}}}$
Note that the derivative is valid for $x \in (-1, 1)$ because the derivative of $\cos ^{-1} x$ is only defined in the open interval $(-1, 1)$.

(A) Let $y = (3x^{2}-9x+5)^{9}$.
Using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [(3x^{2}-9x+5)^{9}]$
$= 9(3x^{2}-9x+5)^{8} \cdot \frac{d}{dx}(3x^{2}-9x+5)$
$= 9(3x^{2}-9x+5)^{8} \cdot (6x-9)$
$= 9(3x^{2}-9x+5)^{8} \cdot 3(2x-3)$
$= 27(3x^{2}-9x+5)^{8}(2x-3)$.

(N/A) Let $y = \sin^{3} x + \cos^{6} x$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sin^{3} x) + \frac{d}{dx}(\cos^{6} x)$.
Using the power rule and chain rule:
$\frac{dy}{dx} = 3 \sin^{2} x \cdot \frac{d}{dx}(\sin x) + 6 \cos^{5} x \cdot \frac{d}{dx}(\cos x)$.
Substituting the derivatives of $\sin x$ and $\cos x$:
$\frac{dy}{dx} = 3 \sin^{2} x \cdot \cos x + 6 \cos^{5} x \cdot (-\sin x)$.
Factoring out $3 \sin x \cos x$:
$\frac{dy}{dx} = 3 \sin x \cos x (\sin x - 2 \cos^{4} x)$.

Let $y = \sin^{-1}(x\sqrt{x})$.
Using the chain rule,we have:
$\frac{dy}{dx} = \frac{d}{dx} \sin^{-1}(x\sqrt{x})$
$= \frac{1}{\sqrt{1 - (x\sqrt{x})^2}} \cdot \frac{d}{dx}(x \cdot x^{1/2})$
$= \frac{1}{\sqrt{1 - x^3}} \cdot \frac{d}{dx}(x^{3/2})$
$= \frac{1}{\sqrt{1 - x^3}} \cdot \left(\frac{3}{2} x^{1/2}\right)$
$= \frac{3\sqrt{x}}{2\sqrt{1 - x^3}}$
$= \frac{3}{2} \sqrt{\frac{x}{1 - x^3}}$

Let $y = \frac{\cos^{-1}(\frac{x}{2})}{\sqrt{2x+7}}$.
Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,we have:
$\frac{dy}{dx} = \frac{\sqrt{2x+7} \cdot \frac{d}{dx}(\cos^{-1}(\frac{x}{2})) - \cos^{-1}(\frac{x}{2}) \cdot \frac{d}{dx}(\sqrt{2x+7})}{2x+7}$
Since $\frac{d}{dx}(\cos^{-1}(\frac{x}{2})) = \frac{-1}{\sqrt{1-(\frac{x}{2})^2}} \cdot \frac{1}{2} = \frac{-1}{\sqrt{4-x^2}}$ and $\frac{d}{dx}(\sqrt{2x+7}) = \frac{1}{2\sqrt{2x+7}} \cdot 2 = \frac{1}{\sqrt{2x+7}}$,we get:
$\frac{dy}{dx} = \frac{\sqrt{2x+7} \cdot (\frac{-1}{\sqrt{4-x^2}}) - \cos^{-1}(\frac{x}{2}) \cdot (\frac{1}{\sqrt{2x+7}})}{2x+7}$
$\frac{dy}{dx} = -\left[ \frac{1}{\sqrt{4-x^2}\sqrt{2x+7}} + \frac{\cos^{-1}(\frac{x}{2})}{(2x+7)^{3/2}} \right]$

(B) Let $y = \cos (a \cos x + b \sin x)$.
By using the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\cos (a \cos x + b \sin x)]$
Using the derivative of $\cos(u)$ which is $-\sin(u) \cdot \frac{du}{dx}$:
$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot \frac{d}{dx} (a \cos x + b \sin x)$
Now,differentiate the inner function:
$\frac{d}{dx} (a \cos x + b \sin x) = a(-\sin x) + b(\cos x) = b \cos x - a \sin x$
Substituting this back into the expression:
$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot (b \cos x - a \sin x)$
Rearranging the terms:
$\frac{dy}{dx} = (a \sin x - b \cos x) \sin (a \cos x + b \sin x)$

(N/A) Let $y = x^{x} + x^{a} + a^{x} + a^{a}$.
Let $u = x^{x}$,$v = x^{a}$,$w = a^{x}$,and $s = a^{a}$.
Then $y = u + v + w + s$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} + \frac{ds}{dx} \dots (1)$.
For $u = x^{x}$,taking log on both sides: $\log u = x \log x$. Differentiating w.r.t $x$: $\frac{1}{u} \frac{du}{dx} = \log x + x(\frac{1}{x}) = \log x + 1$. Thus,$\frac{du}{dx} = x^{x}(1 + \log x) \dots (2)$.
For $v = x^{a}$,using the power rule: $\frac{dv}{dx} = a x^{a-1} \dots (3)$.
For $w = a^{x}$,using the exponential derivative rule: $\frac{dw}{dx} = a^{x} \log a \dots (4)$.
For $s = a^{a}$,since $a$ is a constant,$s$ is a constant,so $\frac{ds}{dx} = 0 \dots (5)$.
Substituting $(2), (3), (4),$ and $(5)$ into $(1)$,we get $\frac{dy}{dx} = x^{x}(1 + \log x) + a x^{a-1} + a^{x} \log a$.

(A) Let $f(x) = x^{-4}(3 - 4x^{-5}) = 3x^{-4} - 4x^{-9}$.
Now,differentiate with respect to $x$:
$f'(x) = \frac{d}{dx}(3x^{-4} - 4x^{-9})$
Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f'(x) = 3(-4)x^{-4-1} - 4(-9)x^{-9-1}$
$f'(x) = -12x^{-5} + 36x^{-10}$
$f'(x) = -\frac{12}{x^5} + \frac{36}{x^{10}}$

Let $f(x) = \frac{2}{x+1} - \frac{x^2}{3x-1}$.
Using the linearity of the derivative:
$f'(x) = \frac{d}{dx}\left(\frac{2}{x+1}\right) - \frac{d}{dx}\left(\frac{x^2}{3x-1}\right)$.
Applying the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}$:
For the first term: $\frac{d}{dx}\left(\frac{2}{x+1}\right) = \frac{(x+1)(0) - 2(1)}{(x+1)^2} = \frac{-2}{(x+1)^2}$.
For the second term: $\frac{d}{dx}\left(\frac{x^2}{3x-1}\right) = \frac{(3x-1)(2x) - x^2(3)}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2}$.
Combining these results:
$f'(x) = -\frac{2}{(x+1)^2} - \frac{3x^2 - 2x}{(3x-1)^2}$.

(A) Let $f(x) = \sin x \cos x$. We can rewrite the function as $f(x) = \frac{1}{2} \sin 2x$.
Using the chain rule,the derivative is:
$f'(x) = \frac{d}{dx} \left( \frac{1}{2} \sin 2x \right)$
$f'(x) = \frac{1}{2} \cdot \cos 2x \cdot \frac{d}{dx}(2x)$
$f'(x) = \frac{1}{2} \cdot \cos 2x \cdot 2$
$f'(x) = \cos 2x$

(A) Let $f(x) = 5 \sec x + 4 \cos x$.
Using the derivative rules for trigonometric functions:
$\frac{d}{dx} (\sec x) = \sec x \tan x$
$\frac{d}{dx} (\cos x) = -\sin x$
Applying the linearity property of derivatives:
$f'(x) = \frac{d}{dx} (5 \sec x + 4 \cos x)$
$f'(x) = 5 \frac{d}{dx} (\sec x) + 4 \frac{d}{dx} (\cos x)$
$f'(x) = 5 (\sec x \tan x) + 4 (-\sin x)$
$f'(x) = 5 \sec x \tan x - 4 \sin x$

(A) Let $f(x) = 3 \cot x + 5 \operatorname{cosec} x$. Using the derivative rules:
$\frac{d}{dx} [3 \cot x + 5 \operatorname{cosec} x] = 3 \frac{d}{dx} (\cot x) + 5 \frac{d}{dx} (\operatorname{cosec} x)$
We know that $\frac{d}{dx} (\cot x) = -\operatorname{cosec}^{2} x$ and $\frac{d}{dx} (\operatorname{cosec} x) = -\operatorname{cosec} x \cot x$.
Substituting these values:
$f'(x) = 3(-\operatorname{cosec}^{2} x) + 5(-\operatorname{cosec} x \cot x)$
$f'(x) = -3 \operatorname{cosec}^{2} x - 5 \operatorname{cosec} x \cot x$

(A) Given the function $f(x) = 5 \sin x - 6 \cos x + 7$.
To find the derivative $f'(x)$,we use the standard derivative rules:
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\text{constant}) = 0$
Applying these rules:
$f'(x) = \frac{d}{dx}(5 \sin x) - \frac{d}{dx}(6 \cos x) + \frac{d}{dx}(7)$
$f'(x) = 5 \cos x - 6(-\sin x) + 0$
$f'(x) = 5 \cos x + 6 \sin x$.

(A) Let $f(x) = 2 \tan x - 7 \sec x$.
Using the standard derivative formulas:
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(\sec x) = \sec x \tan x$
Applying the linearity property of derivatives:
$f'(x) = \frac{d}{dx}(2 \tan x - 7 \sec x)$
$f'(x) = 2 \frac{d}{dx}(\tan x) - 7 \frac{d}{dx}(\sec x)$
$f'(x) = 2 \sec^2 x - 7 \sec x \tan x$

(A) To find the derivative of $f(x) = \sin 2x$,we use the chain rule of differentiation.
Let $u = 2x$,then $f(x) = \sin u$.
By the chain rule,$\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.
$\frac{df}{du} = \cos u = \cos 2x$.
$\frac{du}{dx} = \frac{d}{dx}(2x) = 2$.
Therefore,$\frac{df}{dx} = \cos 2x \cdot 2 = 2 \cos 2x$.

(A) By definition,$g(x) = \cot x = \frac{\cos x}{\sin x}$. We use the quotient rule on this function wherever it is defined.
$\frac{dg}{dx} = \frac{d}{dx}(\cot x) = \frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)$
$= \frac{(\cos x)'(\sin x) - (\cos x)(\sin x)'}{(\sin x)^2}$
$= \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}$
$= -\frac{\sin^2 x + \cos^2 x}{\sin^2 x} = -\csc^2 x$

Let $h(x) = \frac{x^{5}-\cos x}{\sin x}$. We use the quotient rule,$\left(\frac{u}{v}\right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^{2}}$,where $u = x^{5}-\cos x$ and $v = \sin x$.
$h^{\prime}(x) = \frac{\frac{d}{dx}(x^{5}-\cos x) \cdot \sin x - (x^{5}-\cos x) \cdot \frac{d}{dx}(\sin x)}{(\sin x)^{2}}$
$h^{\prime}(x) = \frac{(5x^{4} + \sin x) \sin x - (x^{5}-\cos x) \cos x}{\sin^{2} x}$
$h^{\prime}(x) = \frac{5x^{4} \sin x + \sin^{2} x - x^{5} \cos x + \cos^{2} x}{\sin^{2} x}$
Since $\sin^{2} x + \cos^{2} x = 1$,we have:
$h^{\prime}(x) = \frac{5x^{4} \sin x - x^{5} \cos x + 1}{\sin^{2} x}$

(A) Let $f(x) = \frac{x+\cos x}{\tan x}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$,where $u = x+\cos x$ and $v = \tan x$.
$u' = \frac{d}{dx}(x+\cos x) = 1 - \sin x$.
$v' = \frac{d}{dx}(\tan x) = \sec^2 x$.
Substituting these into the formula:
$f'(x) = \frac{(1-\sin x) \tan x - (x+\cos x) \sec^2 x}{(\tan x)^2}$.
Thus,the derivative is $\frac{(1-\sin x) \tan x - (x+\cos x) \sec^2 x}{\tan^2 x}$.

Let $f(x) = (p x+q)\left(\frac{r}{x}+s\right)$.
Using the product rule,$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (p x+q) \frac{d}{dx}\left(\frac{r}{x}+s\right) + \left(\frac{r}{x}+s\right) \frac{d}{dx}(p x+q)$
$f'(x) = (p x+q) \left(-\frac{r}{x^2}\right) + \left(\frac{r}{x}+s\right) (p)$
$f'(x) = -\frac{p r x}{x^2} - \frac{q r}{x^2} + \frac{p r}{x} + p s$
$f'(x) = -\frac{p r}{x} - \frac{q r}{x^2} + \frac{p r}{x} + p s$
$f'(x) = p s - \frac{q r}{x^2}$

Let $f(x) = (ax+b)(cx+d)^{2}$.
Using the product rule $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$:
$f'(x) = (ax+b) \frac{d}{dx}(cx+d)^{2} + (cx+d)^{2} \frac{d}{dx}(ax+b)$
Applying the chain rule to $\frac{d}{dx}(cx+d)^{2} = 2(cx+d) \cdot c = 2c(cx+d)$:
$f'(x) = (ax+b)[2c(cx+d)] + (cx+d)^{2}(a)$
Factoring out $(cx+d)$:
$f'(x) = (cx+d)[2c(ax+b) + a(cx+d)]$
$f'(x) = (cx+d)[2acx + 2bc + acx + ad]$
$f'(x) = (cx+d)(3acx + 2bc + ad)$

Let $f(x) = \frac{a x+b}{c x+d}$.
Using the quotient rule,$\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$.
Here,$u(x) = ax+b$ and $v(x) = cx+d$.
Then $u'(x) = a$ and $v'(x) = c$.
Substituting these into the formula:
$f'(x) = \frac{(cx+d)(a) - (ax+b)(c)}{(cx+d)^2}$
$f'(x) = \frac{acx + ad - acx - bc}{(cx+d)^2}$
$f'(x) = \frac{ad - bc}{(cx+d)^2}$

Let $f(x) = \frac{1+\frac{1}{x}}{1-\frac{1}{x}} = \frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}$,where $x \neq 0$ and $x \neq 1$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$f'(x) = \frac{(x-1) \frac{d}{dx}(x+1) - (x+1) \frac{d}{dx}(x-1)}{(x-1)^2}$
$f'(x) = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2}$
$f'(x) = \frac{x - 1 - x - 1}{(x-1)^2}$
$f'(x) = \frac{-2}{(x-1)^2}$

Let $f(x) = \frac{1}{ax^{2}+bx+c}$.
Using the quotient rule,where $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}}$:
$f^{\prime}(x) = \frac{(ax^{2}+bx+c) \frac{d}{dx}(1) - (1) \frac{d}{dx}(ax^{2}+bx+c)}{(ax^{2}+bx+c)^{2}}$
Since $\frac{d}{dx}(1) = 0$ and $\frac{d}{dx}(ax^{2}+bx+c) = 2ax+b$:
$f^{\prime}(x) = \frac{(ax^{2}+bx+c)(0) - (2ax+b)}{(ax^{2}+bx+c)^{2}}$
$f^{\prime}(x) = \frac{-(2ax+b)}{(ax^{2}+bx+c)^{2}}$

Let $f(x) = \frac{ax+b}{px^{2}+qx+r}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}}$:
$f'(x) = \frac{(px^{2}+qx+r) \frac{d}{dx}(ax+b) - (ax+b) \frac{d}{dx}(px^{2}+qx+r)}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{(px^{2}+qx+r)(a) - (ax+b)(2px+q)}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{apx^{2} + aqx + ar - (2apx^{2} + aqx + 2bpx + bq)}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{apx^{2} + aqx + ar - 2apx^{2} - aqx - 2bpx - bq}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{-apx^{2} - 2bpx + ar - bq}{(px^{2}+qx+r)^{2}}$