Derivatives by definition Questions in English

(A) To evaluate the limit $\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h}$,we rationalize the numerator:
$\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h} \times \frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }} = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h) - x}}{{h(\sqrt {x + h} + \sqrt x )}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{h}{{h(\sqrt {x + h} + \sqrt x )}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {x + h} + \sqrt x }}$
$= \frac{1}{{\sqrt {x + 0} + \sqrt x }} = \frac{1}{{2\sqrt x }}$
Alternatively,using $L'\text{Hospital's rule}$ by differentiating with respect to $h$:
$\mathop {\lim }\limits_{h \to 0} \frac{{\frac{d}{{dh}}(\sqrt {x + h} - \sqrt x )}}{{\frac{d}{{dh}}(h)}} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{2\sqrt {x + h} }}}}{1} = \frac{1}{{2\sqrt x }}$.

(B) The given expression $\lim_{h \to 0} \frac{f(r + h) - f(r)}{h}$ is the definition of the derivative of $f(r)$ with respect to $r$,denoted as $f'(r)$ or $\frac{df}{dr}$.
Given $f(r) = \pi r^2$.
Applying the power rule for differentiation,$\frac{d}{dr}(\pi r^2) = \pi \cdot \frac{d}{dr}(r^2) = \pi \cdot (2r) = 2\pi r$.
Therefore,$\lim_{h \to 0} \frac{f(r + h) - f(r)}{h} = 2\pi r$.

(B) Let $L = \lim_{x \to a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$.
Adding and subtracting $g(a)f(a)$ in the numerator:
$L = \lim_{x \to a} \frac{g(x)f(a) - g(a)f(a) + g(a)f(a) - g(a)f(x)}{x - a}$
$L = \lim_{x \to a} \left[ f(a) \frac{g(x) - g(a)}{x - a} - g(a) \frac{f(x) - f(a)}{x - a} \right]$
Using the definition of the derivative,$\lim_{x \to a} \frac{g(x) - g(a)}{x - a} = g'(a)$ and $\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$.
$L = f(a)g'(a) - g(a)f'(a)$
Substituting the given values: $L = (2)(2) - (-1)(1) = 4 + 1 = 5$.

(C) The given limit is of the form $\mathop {\lim }\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$,which is the definition of the derivative $f'(a)$ for the function $f(x) = x^2 \sin x$.
First,find the derivative of $f(x) = x^2 \sin x$ with respect to $x$ using the product rule:
$f'(x) = \frac{d}{dx}(x^2) \sin x + x^2 \frac{d}{dx}(\sin x)$
$f'(x) = 2x \sin x + x^2 \cos x$
Evaluating at $x = a$ gives:
$f'(a) = 2a \sin a + a^2 \cos a$
Alternatively,applying $L$-Hospital's rule with respect to $h$:
$\mathop {\lim }\limits_{h \to 0} \frac{\frac{d}{dh}((a+h)^2 \sin(a+h) - a^2 \sin a)}{1}$
$= \mathop {\lim }\limits_{h \to 0} (2(a+h) \sin(a+h) + (a+h)^2 \cos(a+h))$
$= 2a \sin a + a^2 \cos a$

(A) Let $f(y) = (x + y)\sec(x + y)$. We need to find $\mathop {\lim }\limits_{y \to 0} \frac{f(y) - f(0)}{y}$,which is the definition of the derivative $f'(0)$ with respect to $y$.
$f'(y) = \frac{d}{dy} [(x + y)\sec(x + y)]$
Using the product rule: $f'(y) = 1 \cdot \sec(x + y) + (x + y) \cdot \sec(x + y)\tan(x + y)$.
Evaluating at $y = 0$:
$f'(0) = \sec(x) + x\sec(x)\tan(x)$
$f'(0) = \sec x(1 + x\tan x)$.

(B) Given that $f'(x) = \tan^4 x$ and $f(0) = 0$.
We need to evaluate $\lim_{x \to 0} \frac{f(x)}{x}$.
Since $f(0) = 0$,we can write this as:
$\lim_{x \to 0} \frac{f(x) - 0}{x - 0} = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$.
By the definition of the derivative,this limit is equal to $f'(0)$.
Since $f'(x) = \tan^4 x$,we have $f'(0) = \tan^4(0) = 0^4 = 0$.
Therefore,the limit is $0$.

(A) We have the function $f(x) = 3x$.
To find the derivative at $x=2$,we use the definition of the derivative:
$f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$
Substituting $f(x) = 3x$ into the formula:
$f'(2) = \lim_{h \to 0} \frac{3(2+h) - 3(2)}{h}$
$f'(2) = \lim_{h \to 0} \frac{6 + 3h - 6}{h}$
$f'(2) = \lim_{h \to 0} \frac{3h}{h}$
$f'(2) = \lim_{h \to 0} 3 = 3$
Thus,the derivative of the function $f(x) = 3x$ at $x=2$ is $3$.

(B) Let $f(x) = \sin x$.
Then,the derivative at $x=0$ is given by $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$.
Substituting $f(x) = \sin x$,we get $f'(0) = \lim_{h \to 0} \frac{\sin(0+h) - \sin(0)}{h}$.
Since $\sin(0) = 0$,this simplifies to $f'(0) = \lim_{h \to 0} \frac{\sin h}{h}$.
Using the standard limit $\lim_{h \to 0} \frac{\sin h}{h} = 1$,we find $f'(0) = 1$.

(A) The derivative of a function $f(x)$ is defined as $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Given $f(x) = 10x$,we have $f(x+h) = 10(x+h) = 10x + 10h$.
Substituting these into the limit definition:
$f'(x) = \lim_{h \to 0} \frac{(10x + 10h) - 10x}{h}$
$f'(x) = \lim_{h \to 0} \frac{10h}{h}$
$f'(x) = \lim_{h \to 0} (10) = 10$.

(A) We have,$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Substituting $f(x) = x^2$,we get:
$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{(x+h)^2 - x^2}{h}$.
Expanding the numerator:
$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h}$.
Simplifying the expression:
$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{2xh + h^2}{h} = \mathop {\lim }\limits_{h \to 0} (2x + h)$.
Evaluating the limit as $h \to 0$:
$f'(x) = 2x + 0 = 2x$.

(A) By the definition of the derivative,we have:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Since $f(x) = a$ is a constant function,$f(x+h) = a$ and $f(x) = a$.
Substituting these values,we get:
$f'(x) = \lim_{h \to 0} \frac{a - a}{h}$
$f'(x) = \lim_{h \to 0} \frac{0}{h}$
Since $h \neq 0$ as $h \to 0$,the expression $\frac{0}{h}$ is equal to $0$.
Therefore,$f'(x) = 0$.

(A) We have the definition of the derivative as $f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Substituting $f(x) = \frac{1}{x}$,we get:
$f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h}$
Taking the common denominator in the numerator:
$f^{\prime}(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x - (x+h)}{x(x+h)} \right]$
Simplifying the expression:
$f^{\prime}(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{-h}{x(x+h)} \right]$
Canceling $h$ and evaluating the limit as $h \to 0$:
$f^{\prime}(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^{2}}$.

(A) Let $f(x) = \sin x$. Then by the definition of the derivative:
$\frac{df(x)}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}$
Using the trigonometric identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$:
$= \lim_{h \to 0} \frac{2 \cos\left(\frac{2x+h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$
$= \lim_{h \to 0} \cos\left(x + \frac{h}{2}\right) \cdot \lim_{h \to 0} \frac{\sin(h/2)}{h/2}$
Since $\lim_{h \to 0} \cos(x + h/2) = \cos x$ and $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we get:
$= \cos x \cdot 1 = \cos x$

(B) Let $f(x) = \tan x$. Then by the definition of the derivative:
$\frac{df(x)}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{\tan(x + h) - \tan x}{h}$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{\sin(x + h)}{\cos(x + h)} - \frac{\sin x}{\cos x} \right]$
$= \lim_{h \to 0} \frac{\sin(x + h)\cos x - \cos(x + h)\sin x}{h \cos(x + h)\cos x}$
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \lim_{h \to 0} \frac{\sin(x + h - x)}{h \cos(x + h)\cos x}$
$= \lim_{h \to 0} \left( \frac{\sin h}{h} \right) \cdot \lim_{h \to 0} \left( \frac{1}{\cos(x + h)\cos x} \right)$
Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$:
$= 1 \cdot \frac{1}{\cos^2 x} = \sec^2 x$

(C) Let $f(x) = x^{2} - 2$.
To find the derivative at $x = 10$,we use the definition of the derivative:
$f'(10) = \lim_{h \to 0} \frac{f(10+h) - f(10)}{h}$
Substitute $f(x) = x^{2} - 2$ into the formula:
$f'(10) = \lim_{h \to 0} \frac{[(10+h)^{2} - 2] - [10^{2} - 2]}{h}$
Expand the terms:
$f'(10) = \lim_{h \to 0} \frac{100 + 20h + h^{2} - 2 - 100 + 2}{h}$
Simplify the expression:
$f'(10) = \lim_{h \to 0} \frac{20h + h^{2}}{h}$
Divide by $h$:
$f'(10) = \lim_{h \to 0} (20 + h)$
Evaluate the limit as $h \to 0$:
$f'(10) = 20 + 0 = 20$
Thus,the derivative of $x^{2} - 2$ at $x = 10$ is $20$.

(B) Let $f(x) = x$.
By the definition of the derivative,$f'(a) = \mathop{\lim}\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.
For $a = 1$,we have:
$f'(1) = \mathop{\lim}\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$
Since $f(x) = x$,we substitute $f(1+h) = 1+h$ and $f(1) = 1$:
$f'(1) = \mathop{\lim}\limits_{h \to 0} \frac{(1+h) - 1}{h}$
$f'(1) = \mathop{\lim}\limits_{h \to 0} \frac{h}{h}$
$f'(1) = \mathop{\lim}\limits_{h \to 0} (1) = 1$.
Thus,the derivative of $x$ at $x = 1$ is $1$.

(A) Let $f(x) = 99x$.
By the definition of the derivative,$f'(a) = \mathop{\lim}\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$.
Here,$a = 100$,so $f'(100) = \mathop{\lim}\limits_{h \to 0} \frac{f(100+h) - f(100)}{h}$.
Substituting $f(x) = 99x$:
$f'(100) = \mathop{\lim}\limits_{h \to 0} \frac{99(100+h) - 99(100)}{h}$.
$f'(100) = \mathop{\lim}\limits_{h \to 0} \frac{9900 + 99h - 9900}{h}$.
$f'(100) = \mathop{\lim}\limits_{h \to 0} \frac{99h}{h}$.
$f'(100) = \mathop{\lim}\limits_{h \to 0} (99) = 99$.
Thus,the derivative of $99x$ at $x=100$ is $99$.

(A) Let $f(x) = x^{3}-27$. By the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$f'(x) = \lim_{h \to 0} \frac{[(x+h)^{3} - 27] - [x^{3} - 27]}{h}$
$f'(x) = \lim_{h \to 0} \frac{x^{3} + 3x^{2}h + 3xh^{2} + h^{3} - 27 - x^{3} + 27}{h}$
$f'(x) = \lim_{h \to 0} \frac{3x^{2}h + 3xh^{2} + h^{3}}{h}$
$f'(x) = \lim_{h \to 0} (3x^{2} + 3xh + h^{2})$
As $h \to 0$,$f'(x) = 3x^{2} + 3x(0) + (0)^{2} = 3x^{2}$.

(A) Let $f(x) = (x-1)(x-2) = x^2 - 3x + 2$.
By the first principle of derivatives,$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Substituting $f(x+h) = (x+h)^2 - 3(x+h) + 2$:
$f'(x) = \lim_{h \to 0} \frac{[(x+h)^2 - 3(x+h) + 2] - [x^2 - 3x + 2]}{h}$
$f'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2}{h}$
$f'(x) = \lim_{h \to 0} \frac{2xh + h^2 - 3h}{h}$
$f'(x) = \lim_{h \to 0} (2x + h - 3)$
As $h \to 0$,$f'(x) = 2x - 3$.

(A) Let $f(x) = \frac{1}{x^{2}}$.
By the first principle of derivatives,$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Substituting $f(x) = \frac{1}{x^{2}}$,we get:
$f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^{2}} - \frac{1}{x^{2}}}{h}$
$f'(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x^{2} - (x+h)^{2}}{x^{2}(x+h)^{2}} \right]$
$f'(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x^{2} - (x^{2} + 2xh + h^{2})}{x^{2}(x+h)^{2}} \right]$
$f'(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{-2xh - h^{2}}{x^{2}(x+h)^{2}} \right]$
$f'(x) = \lim_{h \to 0} \frac{-h(2x + h)}{h \cdot x^{2}(x+h)^{2}}$
$f'(x) = \lim_{h \to 0} \frac{-(2x + h)}{x^{2}(x+h)^{2}}$
Applying the limit $h \to 0$:
$f'(x) = \frac{-(2x + 0)}{x^{2}(x+0)^{2}} = \frac{-2x}{x^{2} \cdot x^{2}} = \frac{-2x}{x^{4}} = \frac{-2}{x^{3}}$.

(A) Let $f(x) = \frac{x+1}{x-1}$. By the first principle of derivatives,$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Substituting $f(x+h) = \frac{x+h+1}{x+h-1}$ and $f(x) = \frac{x+1}{x-1}$:
$f^{\prime}(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{x+h+1}{x+h-1} - \frac{x+1}{x-1} \right]$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{(x-1)(x+h+1) - (x+1)(x+h-1)}{(x-1)(x+h-1)} \right]$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{(x^2 + xh + x - x - h - 1) - (x^2 + xh - x + x + h - 1)}{(x-1)(x+h-1)} \right]$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{-2h}{(x-1)(x+h-1)} \right]$
$= \lim_{h \to 0} \frac{-2}{(x-1)(x+h-1)} = \frac{-2}{(x-1)^2}$.

(A) Let $f(x) = \cos x$. According to the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}$
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$= \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$
$= \lim_{h \to 0} \left[ \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right) \right]$
$= \cos x \cdot (0) - \sin x \cdot (1)$
$= -\sin x$

(A) Let $f(x) = \sec x$. Accordingly,from the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{\sec(x+h) - \sec x}{h}$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{1}{\cos(x+h)} - \frac{1}{\cos x} \right]$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{\cos x - \cos(x+h)}{\cos x \cos(x+h)} \right]$
Using the identity $\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$:
$= \frac{1}{\cos x} \cdot \lim_{h \to 0} \frac{-2 \sin \left( \frac{2x+h}{2} \right) \sin \left( \frac{-h}{2} \right)}{h \cos(x+h)}$
$= \frac{1}{\cos x} \cdot \lim_{h \to 0} \left[ \frac{\sin \left( \frac{2x+h}{2} \right)}{\cos(x+h)} \cdot \frac{\sin(h/2)}{h/2} \right]$
$= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1$
$= \sec x \tan x$

(A) Let $f(x) = \operatorname{cosec} x$. Using the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{\operatorname{cosec}(x+h) - \operatorname{cosec} x}{h}$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{1}{\sin(x+h)} - \frac{1}{\sin x} \right]$
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{\sin x - \sin(x+h)}{\sin x \sin(x+h)} \right]$
Using the identity $\sin A - \sin B = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$:
$= \lim_{h \to 0} \frac{1}{h} \left[ \frac{2 \cos(\frac{2x+h}{2}) \sin(-\frac{h}{2})}{\sin x \sin(x+h)} \right]$
$= \lim_{h \to 0} \left[ -\cos(\frac{2x+h}{2}) \cdot \frac{\sin(h/2)}{h/2} \cdot \frac{1}{\sin x \sin(x+h)} \right]$
Since $\lim_{h \to 0} \frac{\sin(h/2)}{h/2} = 1$:
$= -\cos x \cdot 1 \cdot \frac{1}{\sin^2 x} = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = -\cot x \operatorname{cosec} x$

The derivative of a function $f(x)$ by the first principle is given by $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Given $f(x) = \frac{2x+3}{x-2}$,we have:
$f'(x) = \lim_{h \to 0} \frac{\frac{2(x+h)+3}{x+h-2} - \frac{2x+3}{x-2}}{h}$
Taking the common denominator in the numerator:
$f'(x) = \lim_{h \to 0} \frac{(2x+2h+3)(x-2) - (2x+3)(x+h-2)}{h(x-2)(x+h-2)}$
Expanding the terms:
$f'(x) = \lim_{h \to 0} \frac{(2x^2 - 4x + 2xh - 4h + 3x - 6) - (2x^2 + 2xh - 4x + 3x + 3h - 6)}{h(x-2)(x+h-2)}$
Simplifying the numerator:
$f'(x) = \lim_{h \to 0} \frac{2x^2 - x + 2xh - 4h - 6 - 2x^2 - 2xh + x - 3h + 6}{h(x-2)(x+h-2)}$
$f'(x) = \lim_{h \to 0} \frac{-7h}{h(x-2)(x+h-2)}$
Canceling $h$ and taking the limit as $h \to 0$:
$f'(x) = \frac{-7}{(x-2)(x-2)} = -\frac{7}{(x-2)^2}$.

The function is not defined at $x = 0$. Using the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{(x+h + \frac{1}{x+h}) - (x + \frac{1}{x})}{h}$
$= \lim_{h \to 0} \frac{1}{h} [h + \frac{1}{x+h} - \frac{1}{x}]$
$= \lim_{h \to 0} \frac{1}{h} [h + \frac{x - (x+h)}{x(x+h)}]$
$= \lim_{h \to 0} \frac{1}{h} [h - \frac{h}{x(x+h)}]$
$= \lim_{h \to 0} [1 - \frac{1}{x(x+h)}]$
$= 1 - \frac{1}{x^2}$
Thus,the derivative is $1 - \frac{1}{x^2}$ for $x \neq 0$.

(A) By the first principle,the derivative $f'(x)$ is defined as:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substituting $f(x) = \sin x + \cos x$:
$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) + \cos(x+h) - (\sin x + \cos x)}{h}$
Using the sum formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$f'(x) = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h + \cos x \cos h - \sin x \sin h - \sin x - \cos x}{h}$
Rearranging terms:
$f'(x) = \lim_{h \to 0} \left[ \sin x \left( \frac{\cos h - 1}{h} \right) + \cos x \left( \frac{\sin h}{h} \right) + \cos x \left( \frac{\cos h - 1}{h} \right) - \sin x \left( \frac{\sin h}{h} \right) \right]$
Using standard limits $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$:
$f'(x) = \sin x(0) + \cos x(1) + \cos x(0) - \sin x(1)$
$f'(x) = \cos x - \sin x$

(A) By the first principle of derivatives,$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Given $f(x) = x \sin x$,we have $f(x+h) = (x+h) \sin(x+h)$.
$f'(x) = \lim_{h \to 0} \frac{(x+h) \sin(x+h) - x \sin x}{h}$.
Using the identity $\sin(x+h) = \sin x \cos h + \cos x \sin h$:
$f'(x) = \lim_{h \to 0} \frac{(x+h)(\sin x \cos h + \cos x \sin h) - x \sin x}{h}$.
$f'(x) = \lim_{h \to 0} \frac{x \sin x \cos h + x \cos x \sin h + h \sin x \cos h + h \cos x \sin h - x \sin x}{h}$.
$f'(x) = \lim_{h \to 0} \left[ x \sin x \left( \frac{\cos h - 1}{h} \right) + x \cos x \left( \frac{\sin h}{h} \right) + \sin x \cos h + \cos x \sin h \right]$.
As $h \to 0$,$\frac{\cos h - 1}{h} \to 0$,$\frac{\sin h}{h} \to 1$,$\cos h \to 1$,and $\sin h \to 0$.
$f'(x) = x \sin x (0) + x \cos x (1) + \sin x (1) + \cos x (0) = x \cos x + \sin x$.

(A) Let $f(x) = -x$. Accordingly,$f(x+h) = -(x+h)$.
By the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substituting the values:
$f'(x) = \lim_{h \to 0} \frac{-(x+h) - (-x)}{h}$
Simplifying the expression:
$f'(x) = \lim_{h \to 0} \frac{-x - h + x}{h}$
$f'(x) = \lim_{h \to 0} \frac{-h}{h}$
$f'(x) = \lim_{h \to 0} (-1)$
Therefore,$f'(x) = -1$.

(A) Let $f(x) = -x^{-1} = -\frac{1}{x}$.
By the first principle of derivatives,$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Substituting $f(x+h) = -\frac{1}{x+h}$ and $f(x) = -\frac{1}{x}$:
$f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x+h} - (-\frac{1}{x})}{h}$
$f'(x) = \lim_{h \to 0} \frac{1}{h} \left( \frac{1}{x} - \frac{1}{x+h} \right)$
$f'(x) = \lim_{h \to 0} \frac{1}{h} \left( \frac{x+h - x}{x(x+h)} \right)$
$f'(x) = \lim_{h \to 0} \frac{1}{h} \left( \frac{h}{x(x+h)} \right)$
$f'(x) = \lim_{h \to 0} \frac{1}{x(x+h)}$
As $h \to 0$,$f'(x) = \frac{1}{x(x+0)} = \frac{1}{x^2}$.

(A) Let $f(x) = \sin(x+1)$. Accordingly,$f(x+h) = \sin(x+h+1)$.
By the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{\sin(x+h+1) - \sin(x+1)}{h}$
Using the identity $\sin A - \sin B = 2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})$:
$= \lim_{h \to 0} \frac{2 \cos(\frac{x+h+1+x+1}{2}) \sin(\frac{x+h+1-x-1}{2})}{h}$
$= \lim_{h \to 0} \frac{2 \cos(\frac{2x+h+2}{2}) \sin(\frac{h}{2})}{h}$
$= \lim_{h \to 0} [\cos(\frac{2x+h+2}{2}) \cdot \frac{\sin(h/2)}{h/2}]$
$= \cos(\frac{2x+2}{2}) \cdot 1$
$= \cos(x+1)$

(A) Let $f(x) = \cos \left(x-\frac{\pi}{8}\right)$. Accordingly,$f(x+h) = \cos \left(x+h-\frac{\pi}{8}\right)$.
By the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
$= \lim_{h \to 0} \frac{1}{h} \left[ \cos \left(x+h-\frac{\pi}{8}\right) - \cos \left(x-\frac{\pi}{8}\right) \right]$
Using the identity $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$:
$= \lim_{h \to 0} \frac{1}{h} \left[ -2 \sin \left(\frac{2x+h-\frac{\pi}{4}}{2}\right) \sin \left(\frac{h}{2}\right) \right]$
$= \lim_{h \to 0} \left[ -\sin \left(x + \frac{h}{2} - \frac{\pi}{8}\right) \cdot \frac{\sin(h/2)}{h/2} \right]$
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ as $h \to 0 \Rightarrow \frac{h}{2} \to 0$:
$= -\sin \left(x - \frac{\pi}{8}\right) \cdot 1 = -\sin \left(x - \frac{\pi}{8}\right)$.

(A) Let $f(x) = x + a$.
Accordingly,$f(x + h) = x + h + a$.
By the first principle of derivatives:
$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
$= \lim_{h \to 0} \frac{(x + h + a) - (x + a)}{h}$
$= \lim_{h \to 0} \frac{h}{h}$
$= \lim_{h \to 0} (1)$
$= 1$.

Let $f(x) = (ax + b)^n$.
By the first principle of derivatives,$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
$f'(x) = \lim_{h \to 0} \frac{(a(x+h) + b)^n - (ax + b)^n}{h}$.
$f'(x) = \lim_{h \to 0} \frac{(ax + b + ah)^n - (ax + b)^n}{h}$.
$f'(x) = \lim_{h \to 0} \frac{(ax + b)^n (1 + \frac{ah}{ax+b})^n - (ax + b)^n}{h}$.
Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2}z^2 + \dots$,we get:
$f'(x) = \lim_{h \to 0} \frac{(ax + b)^n [1 + n(\frac{ah}{ax+b}) + \dots] - (ax + b)^n}{h}$.
$f'(x) = (ax + b)^n \lim_{h \to 0} \frac{n(\frac{ah}{ax+b}) + \dots}{h}$.
$f'(x) = (ax + b)^n \cdot \frac{na}{ax+b} = na(ax + b)^{n-1}$.

Let $f(x) = \sin (x+a)$.
By the first principle of derivatives:
$f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
Substituting the function:
$f^{\prime}(x) = \lim_{h \to 0} \frac{\sin (x+h+a) - \sin (x+a)}{h}$
Using the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$:
$f^{\prime}(x) = \lim_{h \to 0} \frac{2 \cos \left(\frac{x+h+a+x+a}{2}\right) \sin \left(\frac{x+h+a-x-a}{2}\right)}{h}$
$f^{\prime}(x) = \lim_{h \to 0} \frac{2 \cos \left(\frac{2x+2a+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$f^{\prime}(x) = \lim_{h \to 0} \cos \left(x+a+\frac{h}{2}\right) \cdot \frac{\sin (h/2)}{h/2}$
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$f^{\prime}(x) = \cos (x+a) \cdot 1 = \cos (x+a)$.

(C) Given $f(x) = 3x^{10} - 7x^{8} + 5x^{6} - 21x^{3} + 3x^{2} - 7$.
We need to evaluate the limit $L = \lim_{h \rightarrow 0} \frac{f(1-h) - f(1)}{h^{3} + 3h}$.
We can rewrite the expression as:
$L = \lim_{h \rightarrow 0} \left( \frac{f(1-h) - f(1)}{-h} \cdot \frac{-h}{h(h^{2} + 3)} \right)$.
Since $f'(1) = \lim_{h \rightarrow 0} \frac{f(1-h) - f(1)}{-h}$,we have:
$L = f'(1) \cdot \lim_{h \rightarrow 0} \frac{-1}{h^{2} + 3}$.
First,calculate the derivative $f'(x) = 30x^{9} - 56x^{7} + 30x^{5} - 63x^{2} + 6x$.
Evaluating at $x = 1$:
$f'(1) = 30(1)^{9} - 56(1)^{7} + 30(1)^{5} - 63(1)^{2} + 6(1) = 30 - 56 + 30 - 63 + 6 = -53$.
Now substitute this into the limit expression:
$L = (-53) \cdot \left( \frac{-1}{0^{2} + 3} \right) = (-53) \cdot \left( -\frac{1}{3} \right) = \frac{53}{3}$.