Derivatives

On the curve defined by $y = f(x)$ , as $x$ changes the value from $x_{0}$ to $x_{0}+h$ , the value of $y$ changes from $f(x_{0})$ to $f(x_{0}+h)$ . Then the difference

$\displaystyle \frac{f(x_{0}+h) - f(x_{0})}{h}$

is called difference quotient. Now think of this using the graph. Then a line goes through two points $(x_{0},f(x_{0}))$ , $(x_{0} + h, f(x_{0} + h))$ is called secant line.

**Figure 2.1:** secant line
$\includegraphics[width=11cm]{CALCFIG/Fig2-1-1.eps}$

Difference Quotient $f(x_0 + h) - f(x_0)$ is called increment of $y$ and denoted by $\Delta y$ , $x_0 + h - x_0$ is denoted by $\Delta x$ . Then

$\displaystyle \frac{\Delta y}{\Delta x} = \frac{f(x_0 + h) - f(x_0)}{h}$

which is the ratio between the increment of $y$

and the increment of $x$

. Right-Hand Limit $h$

approaches 0 from the right means that the value of $h$

is always positive and approaches 0.

As $h$ approaches 0 from the right, the secant line is getting close to the red line. Similarly, as $h$ approaches 0 from the left, the secant line is getting close to the same red line. This red line is called tangent line at $(x_{0},f(x_{0}))$ . Differential Coefficient Suppose that $f(x)$ is defined on an interval containing $x_{0}$ . If

$\displaystyle \lim_{h \rightarrow 0} \frac{f(x_{0}+h) - f(x_{0})}{h} = A {\rm provided} A \neq \pm \infty$

exists, then we say $f(x)$

is differentiable at $x = x_{0}$ . The number $A$

is called differntiable coefficient at $x_{0}$ and denoted by $f^{\prime}(x_{0})$ . To be differentiable, the limit $A$

must be real number.

NOTE The slope of the secant line is given by

$\displaystyle \frac{f(x_{0}+h) - f(x_{0})}{h}$

Thus the slope of the tangent line is

$\displaystyle \lim_{h \to 0}\frac{f(x_{0} +h) - f(x_{0})}{h}$

Then the differential coefficient at $x = x_{0}$ can be thought as the slope tangent line.

Example 2..1 Using the definition of differential coefficient, find a differential coefficient of $f(x) = \sin{x}$ at $x = 0$

Definition Let $x = x_{0} + h$ . Then we can express the differential constant in the following way.
$f'(x_{0}) = \lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}.$

SOLUTION By the definition of differential coefficient, we have

$\displaystyle \lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \rightarrow 0}\frac{\sin{h} - 0}{h} = 1$

Thus, $f^{\prime}(0) = 1$ $\blacksquare$

Exercise 2..1 For $f(x) = x^2$

, using the definition of differential coefficient, find $f^{\prime}(-3)$

Check
.

SOLUTION By the definition of differential coefficient with $x_{0} = -3$ , we have

$\displaystyle \lim_{h \to 0}\frac{f(-3+h) - f(-3)}{h}$	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{(-3+h)^2 - (-3)^2}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{9 -6h + h^2 - 9}{h} = \lim_{h \to 0}\frac{h(h-6)}{h}$
	$\displaystyle =$	$\displaystyle \lim_{ h\to 0}(h-6) = -6\ensuremath{ \blacksquare}$

Equation of Tangent Line The equation of a tangent line for $y = f(x)$ at $(x_{0},y_{0})$ is given by

$\displaystyle y - y_{0} = f'(x_{0})(x - x_{0}).$

NOTE Since $f'(x_0)$ is the same as the slope of the tangent line, the slope of the line connecting two points $(x,y)$ and $(x_0, y_0)$ on the tangent line is equal to . Equation of Normal Line A line perpendicular to a tangent line is called normal line. The equation of a normal line to a function $y = f(x)$ at $(x_{0},y_{0})$ is given by

$\displaystyle y - y_{0} = -\frac{1}{f'(x_{0})}(x - x_{0})$

**Figure 2.2:** Tangent and Normal Line
$\includegraphics[width=3cm]{SOFTFIG-2/normal_gr1.eps}$

Linear Approximation The equation of a tangent line of $y = f(x)$ is a linear approximation. In other words, for $x \fallingdotseq a$ , we can approximate the value of $f(x)$ by the tangent line. $f(x) \fallingdotseq f(x_{0}) + f'(x_{0})(x - x_{0})$ .

Example 2..2 Find the tangent line and the normal line of $f(x) = x^{2}$ at $(-3,9)$

SOLUTION Since the slope of the tangent line at $(-3,9)$

. Thus the equation of a tangent line is

$\displaystyle y - 9 = -6(x - (-3))$

$\displaystyle y - 9 = -6(x + 3).$

On the other hand, the equation of the normal line is

$\displaystyle y - 9 = \frac{1}{6}(x + 3)\ensuremath{ \blacksquare}$

Equation of Line Consider an equation of line goes through $(a,b)$ with the slope $m$ . Now take any point $(x,y)$ different from $(a,b)$ on this line. Then the slope of the line is $m = \frac{y-b}{x-a}$ . Thus, we have $y-b = m(x-a)$ .

Exercise 2..2 Find the equation of a tangent line and the normal line of $f(x) = \sin{x}$ .

SOLUTION By Example2.1, the slope of the tangent line at $(0,0)$

. Thus, the equation of the tangent line is

$\displaystyle y = x$

Similarly, we have the equation of the normal line

$\displaystyle y = -x \ensuremath{ \blacksquare}$

Left-Hand Differential Coefficient

$\displaystyle \lim_{h \rightarrow 0-} \frac{f(x_{0}+h) - f(x_{0})}{h}$

exists. Then this value is called left-hand differential coefficient and denoted by $f_{-}^{\prime}(x_{0})$ .

Left-Hand A left-hand differential coefficient is the same as the slope of tangent line as $x$ approaches from the left at $x_0$ .
Right-Hand The right-hand differential coefficient is the same as the slope of tangent line as $x$ approaches from the right at $x_0$ .

Right-Hand Differential Coefficient

$\displaystyle \lim_{h \rightarrow 0+} \frac{f(x_{0}+h) - f(x_{0})}{h}$

exists. Then this value is called right-hand differential coefficient and denoted by $f_{+}^{\prime}(x_{0})$ .

NOTE By the definition of differentiable function, if $f_{-}^{\prime}(x_{0})$ and $f_{+}^{\prime}(x_{0})$ exist and their values are equal, then $f(x)$ is differentiable at $x = x_{0}$ .

Differentiable Fcts If $f_{-}^{\prime}(x_{0}) = f_{+}^{\prime}(x_{0})$ , then $f(x)$ is differentiable at $x_0$ and denoted by $f^{\prime}(x_{0})$ . From this, if the graph of function has sharp edge, then the function is not differentiable.

Example 2..3 Determine whether the function $f(x) = \vert x\vert$ is differentiable at $x = 0$

SOLUTION We need to check the left-hand differential coefficient $f_{-}^\prime (0)$ and the right-hand differential coefficient $f_{+}^\prime(0)$ . We first find $f_{-}^{\prime}(0)$ .

$\displaystyle f_{-}^{\prime}(0) = \lim_{h \rightarrow 0-}\frac{f(0+h) - f(0)}{h... ...ightarrow 0-}\frac{\vert h\vert}{h} = \lim_{h \rightarrow 0-}\frac{-h}{h} = -1$

We next find $f_{+}^{\prime}(0)$ .

$\displaystyle f_{+}^{\prime}(0) = \lim_{h \rightarrow 0+}\frac{f(0+h) - f(0)}{h... ...\rightarrow 0+}\frac{\vert h\vert}{h} = \lim_{h \rightarrow 0+}\frac{h}{h} = 1$

Thus $f(x) = \vert x\vert$ is not differentiable at $x = 0$

$\blacksquare$

Exercise 2..3 Determine whether the following function is differentiable at $x = 0$

. $\displaystyle{f(x) = \left\{\begin{array}{ll} x^2 \sin{\frac{1}{x}}, & x \neq 0\\ 0, & x = 0 \end{array}\right.}$

SOLUTION

$\displaystyle f^{\prime}(0) = \lim_{h \rightarrow 0}\frac{f(0+h) - f(0)}{h} = \... ... 0}\frac{h^2 \sin{\frac{1}{h}}}{h} = \lim_{h \rightarrow 0}h\sin{\frac{1}{h}}.$

Now we have

$\displaystyle 0 \leq \vert h\sin{\frac{1}{h}}\vert \leq \vert h\vert$

Let $h \to 0$ . Then

$\displaystyle 0 \leq \lim_{h \rightarrow 0}\vert h\sin{\frac{1}{h}}\vert \leq 0.$

Thus, we have $f^{\prime}(0) = 0$ and the function is differentiable at $x = 0$

$\blacksquare$

**Figure 2.3:** Exercise2-3
$\includegraphics[width=3.5cm]{SOFTFIG-2/x^2sin1overx_gr1.eps}$

Since $\sin{\frac{1}{h}}$ bibrates, we squeeze from the both sides. As $0 \leq \vert\sin{\frac{1}{h}}\vert \leq 1$ , we multiply the both sides by $\vert h\vert$ .

Note that $f(x) = \vert x\vert$ is not differentiable at $x = 0$ but continuous at $x = 0$ . What kind of relation can we find between differentiablility and continuity.

Differentiability implies Continuity

Theorem 2..1 Suppse that $f(x)$

is differentiable at $x = x_{0}$ . Then $f(x)$

is continuous at $x = x_{0}$ .

Continuity
is continuous at $x = x_{0}$ if and only if $\lim_{x \to x_{0}}f(x) = f(x_{0})$ .

Proof Need to show $\lim_{x \rightarrow x_{0}} f(x) = f(x_{0})$ . Rewrite

$\displaystyle f(x) - f(x_{0}) = \frac{f(x) - f(x_{0})}{x - x_{0}}(x - x_{0})$

and

approaches $x_{0}$ . Then

$\displaystyle \lim_{x \rightarrow x_{0}}(f(x) - f(x_{0})) = \lim_{x \rightarrow x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}(x - x_{0})$

Note that

is differentiable at $x = x_{0}$ . Thus

$\displaystyle \lim_{x \rightarrow x_{0}}(f(x) - f(x_{0}))$	$\displaystyle =$	$\displaystyle \lim_{x \rightarrow x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}\cdot \lim_{x \rightarrow x_{0}}(x - x_{0})$
	$\displaystyle =$	$\displaystyle f^{\prime}(x_{0}) \cdot \lim_{x \rightarrow x_{0}}(x - x_{0}) = 0$

This implies that

$\displaystyle \lim_{x \rightarrow x_{0}} f(x) = f(x_{0}) \ensuremath{ \blacksquare}$

Differentiability
is differentiable at $x = x_{0}$ means that $\lim_{x \to x_{0}}\frac{f(x) - f(x_{0})}{x-x_{0}} = f'(x_{0})$ .

The converse of this statement is not true. In other words, continuity does not imply differentiability. see Example2.3.

Derivatives If $f(x)$ is differentiable at each point on some interval $I$ , then we say $f(x)$ is differentiable on $I$ . In this case, we associate the value of $f(x)$ to each point in $I$ to get Derivative which is define by

$\displaystyle f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$

Derivatives The differentiable constant of $f(x)$ can be thought of the instantaneous rate of change at fixed point. On the other hand, the derivative of $f(x)$ can be thought of the instantaneous rate of change at arbitrary point. Prime Notation When we write $(f(x))'$ , we must differentiate with respect to $x$ .

NOTE The symbols of derivatives are

$\displaystyle \frac{df(x)}{dx}, Df(x), (f(x))'$

To find the derivative of $f(x)$

, we say differentiate.

Example 2..4 Find the derivative of the following function using the definition of the derivative.
$\displaystyle{1. f(x) = x^{n}, n \mbox{integer}}$ $\displaystyle{2. f(x) = \sin{x}}$ $\displaystyle{3. f(x) = e^x}$

SOLUTION 1.

$\displaystyle f^{\prime}(x)$	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0}\frac{(x+h)^{n} - x^n}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n}{h}$
	$\displaystyle =$	$\displaystyle nx^{n-1} + \lim_{h \rightarrow 0}[\binom{n}{2}x^{n-2}h + \binom{n}{3}x^{n-3}h^2 + \cdots + h^{n-1}]$
	$\displaystyle =$	$\displaystyle nx^{n-1} \ensuremath{ \blacksquare}$

Check
To find , replace by . Then we have .

Check
Using the addition formula for sine,

		$\displaystyle \sin(x+h)$
	$\displaystyle =$	$\displaystyle \sin{x}\cos{h}+\cos{x}\sin{h}$

$\displaystyle f^{\prime}(x)$	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0}\frac{\sin{(x+h)} - \sin{x}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{\sin{x}\cos{h} + \cos{x}\sin{h} - \sin{x}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}[\sin{x}\frac{\cos{h} - 1}{h} + \cos{x}\frac{\sin{h}}{h}]$
	$\displaystyle =$	$\displaystyle \sin{x} \lim_{h \rightarrow 0}\frac{-\sin^{2}{h}}{h(\cos{h} + 1)} + \cos{x}$
	$\displaystyle =$	$\displaystyle \sin{x} \lim_{h \rightarrow 0}\frac{\sin{h}}{h}\cdot\frac{-\sin{h}}{\cos{h} + 1} + \cos{x}$
	$\displaystyle =$	$\displaystyle \sin{x}(1 \cdot - \frac{0}{2}) + \cos{x} = \cos{x}\ensuremath{ \blacksquare}$

$\displaystyle f^{\prime}(x)$	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0}\frac{e^{x+h} - e^{x}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{e^{x}(e^{h} - 1)}{h} = e^{x}\lim_{h \rightarrow 0}\frac{e^{h} - 1}{h}$

Check
$e^{x+h} = e^{x}\cdot e^{h}$ .

Now let $t = e^h - 1$ . Then since $h = \log(t+1)$ , we have

$\displaystyle \lim_{h \to 0}\frac{e^h - 1}{h}$	$\displaystyle =$	$\displaystyle \lim_{t \to 0}\frac{t}{\log(t+1)} = \lim_{t \to 0}\frac{1}{\frac{1}{t}\log(t+1)}$
	$\displaystyle =$	$\displaystyle \lim_{t \to 0}\frac{1}{\log(t+1)^{\frac{1}{t}}} = \frac{1}{\log{e}} = 1.$

Thus , $f'(x) = e^{x} \ensuremath{ \blacksquare}$

Check
Let $u = \frac{1}{t}$ . Then $\lim_{t \to 0}(1 + t)^{\frac{1}{t}} = \lim_{u \to \pm \infty}(1 + \frac{1}{u})^{u} = e$ .

Exercise 2..4 Find the derivative of the following function using the definition.
$\displaystyle{1. f(x) = \cos{x}}$ $\displaystyle{2. f(x) = \log{x}}$

SOLUTION 1.

$\displaystyle f^{\prime}(x)$	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0}\frac{\cos{(x+h)} - \cos{x}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{\cos{x}\cos{h} - \sin{x}\sin{h} - \cos{x}}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}[\cos{x}\frac{\cos{h} - 1}{h} - \sin{x}\frac{\sin{h}}{h}] \left(\frac{\cos{h} - 1}{h} \to 0, \frac{\sin{h}}{h} \to 1\right)$
	$\displaystyle =$	$\displaystyle -\sin{x}$

Check

		$\displaystyle \cos(a+b)$
	$\displaystyle =$	$\displaystyle \cos{a}\cos{b}-\sin{a}\sin{b}$

3. By the definition of the derivative, we have

$\displaystyle \frac{d(\log{x})}{dx} = \lim_{h \to 0}\frac{\log(x+h) - \log{x}}{h}$

Now using the law of logarithm,

law of logarithm
$\log{p} -\log{q} = \log{\frac{p}{q}}$ $r\log{p} = \log{p}^r$ .

To transform the right-hand side,

$\displaystyle \lim_{h \to 0}\frac{1}{h}(\log(x+h) - \log{x})$	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\log \big(\frac{x+h}{x}\big)^{\frac{1}{h}}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\log \big(1 + \frac{h}{x}\big)^{\frac{1}{h}}.$

Set $\frac{x}{h} = t$ . Then as $h \to 0$ implies $t \to \pm\infty$ . Thus

$\displaystyle \lim_{h \to 0}\log \big(1 + \frac{h}{x}\big)^{\frac{1}{h}}$

$\displaystyle =$

$\displaystyle \lim_{t \to \pm \infty}\log \big(1 + \frac{1}{t}\big)^{\frac{t}{x}} = \frac{1}{x}\lim_{t \to \pm \infty}\log \big(1+ \frac{1}{t}\big)^{t}.$

Since $\lim_{t \to \pm \infty}(1 + \frac{1}{t})^{t} = e$ , we have

$\displaystyle \frac{d(\log{x})}{dx} = \frac{1}{x} \ensuremath{ \blacksquare}$

$\log{e}$
Note that $\log{e}$ means $\log_{e}{e}$ . Thus $\log{e} = 1$ .

As you saw, finding the derivative of a function by the definition is not easy. So we show useful derivative formulas.

Basic Formulas $\begin{displaymath}\begin{array}{ll} 1. & (x^n)' = nx^{n-1}\\ 2. &(e^{x})' = e^... ...2}{x} \\ & \hskip 1.3cm = \frac{1}{\cos^{2}{x}}\\ \end{array}\end{displaymath}$

Differentiation Formula

Theorem 2..2 Let $f(x), g(x), h(x)$

be differentiable and $c$

be constant

$(f(x) \pm g(x))^{\prime} = f^{\prime}(x) \pm g^{\prime}(x)$
$(cf(x))^{\prime} = cf^{\prime}(x)$
$(f(x)g(x))^{\prime} = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$
$\left(\frac{f(x)}{g(x)}\right)^{\prime} = \frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{g^{2}(x)}$

Proof $3. $ .

$\displaystyle (f(x)g(x))^{\prime}$	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \rightarrow 0}\frac{(f(x+h)- f(x))g(x+h) + f(x)(g(x+h) - g(x))}{h}$
	$\displaystyle =$	$\displaystyle f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$

Similarly for other cases $\blacksquare$ Basic Technique Subtract $f(x)g(x+h)$

then add the same.

Check
$\footnotesize {\displaystyle{\lim_{h \to 0}}\frac{f(x+h) - f(x)}{h} = f'(x)}$ $\footnotesize {\displaystyle{\lim_{h \to 0}}\frac{g(x+h) - g(x)}{h} = g'(x)}$ .

Example 2..5 Find the derivative of the followings.
$\displaystyle{1. y = 3x^{3} + 2x +3}$ $\displaystyle{2. y = e^{x}\sin{x}}$

SOLUTION1.

$\displaystyle y^{\prime}$

$\displaystyle =$

$\displaystyle (3x^{3} + 2x +3)^{\prime} \underbrace{=}_{sum rule} (3x^3)^{\prim... ...^{\prime} + 3^{\prime} = 3\cdot 3x^2 + 2 = 9x^2 + 2 \ensuremath{ \blacksquare}$

sum rule $(f(x) + g(x))' = f'(x) + g'(x)$

$\displaystyle y^{\prime}$	$\displaystyle =$	$\displaystyle (e^x \sin{x})^{\prime} \underbrace{=}_{product rule} (e^x)^{\prime}\sin{x} + e^{x}(\sin{x})'$
	$\displaystyle =$	$\displaystyle e^x \sin{x} + e^x \cos{x} = e^x (\sin{x} + \cos{x}) \ensuremath{ \blacksquare}$

product rule $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$

Exercise 2..5 Find the derivative of the followings.
$\displaystyle{1. \tan{x}}$ $\displaystyle{2. x^2 \log{x}}$

SOLUTION1.

$\displaystyle (\tan{x})^{\prime}$	$\displaystyle =$	$\displaystyle (\frac{\sin{x}}{\cos{x}})^{\prime}$
	$\displaystyle \underbrace{=}_{quotient rule}$	$\displaystyle \frac{(\sin{x})^{\prime}\cos{x} - \sin{x}(\cos{x})^{\prime}}{\cos^{2}{x}}$
	$\displaystyle =$	$\displaystyle \frac{\cos^{2}{x} + \sin^{2}{x}}{\cos^{2}{x}}$
	$\displaystyle =$	$\displaystyle \frac{1}{\cos^{2}{x}} = \sec^{2}{x} \ensuremath{ \blacksquare}$

Quotient rule $(\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$

Derivative of $\log{x}$
$(\log{x})' = \frac{1}{x}$ .

$\displaystyle y^{\prime}$	$\displaystyle =$	$\displaystyle (x^2 \log{x})^{\prime} \underbrace{=}_{product rule} (x^2)^{\prime}\log{x} + x^2(\log{x})'$
	$\displaystyle =$	$\displaystyle 2x\log{x} + x^2 \cdot \frac{1}{x} = x (2\log{x} + x) \ensuremath{ \blacksquare}$

Exercise A

1.

Find the derivative of the following functions by using the definition of derivatives

(a) $\displaystyle{f(x) = c}$ (b) $\displaystyle{f(x) = \sqrt{x-1}}$ (c) $\displaystyle{f(x) = \frac{1}{x^{2}}}$

2.

Find the differential coefficient of the following function by using the definition of the derivative at $x = 2$

(a) $\displaystyle{f(x) = 5x - x^{2}}$ (b) $\displaystyle{f(x) = (3x - 7)^{2}}$

3.

Find the equation of tangent line to the graph of the function at the given point.

(a) $\displaystyle{f(x) = x^{2} - 5x + 3, a = 2}$ (b) $\displaystyle{f(x) = 5 - x^{3}, a = 2}$ (c) $\displaystyle{f(x) = \sqrt{x}, a = 4}$

4.

Find the derivative of the following functions:

(a) $\displaystyle{y = 11x^{5} - 6x^{3} + 8}$ (b) $\displaystyle{y = -\frac{1}{x^{2}}}$ (c) $\displaystyle{y = (x^{2} - 1)(x-3)}$

(d) $\displaystyle{y = \frac{x-1}{x-2}}$ (e) $\displaystyle{y = \frac{x^{2}-1}{2x+3}}$ (f) $\displaystyle{y = \frac{6 - 1/x}{x-2}}$ (g) $\displaystyle{y = \frac{1 + x^{4}}{x^{2}}}$

Exercise B

1.

Find the derivative of the following functions:

(a) $\displaystyle{f(x) = \cos{3x}}$ (b) $\displaystyle{f(x) = (x + 2)^{n} (n : \mbox{integer})}$

2.

Find the differential of the following functions:

(a) $\displaystyle{f(x) = x^4}$ (b) $\displaystyle{f(x) = e^{x}}$

3.

Find the left-hand and right-hand differential coefficient of the following functions:

(a) $\displaystyle{f(x) = \vert x^2 + x\vert}$ (b) $\displaystyle{f(x) = \left\{\begin{array}{ll} x^2 \sin{\frac{1}{x}}, & x \neq 0\\ 0, & x = 0 \end{array}\right.}$ (c) $\displaystyle{f(x) = \sqrt{x^3 + x^2}}$

4.

Find the derivative of the following functions:

(a) $\displaystyle{y = \frac{3x-1}{x^{2} + 1}}$ (b) $\displaystyle{y = \sec{x}}$ (c) $\displaystyle{y = {\rm cosec}{x}}$ (d) $\displaystyle{y = \cot{x}}$

(e) $\displaystyle{y = x^{2}e^{x}}$ (f) $\displaystyle{y = e^{x}\sin{x}}$ (g) $\displaystyle{y = \frac{e^{x}}{\sin{x}}}$