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

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

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

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

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

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 $f^{\prime}(x_{0})$. ^2.1

NOTE The slope of the secant line is given by

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

Thus the slope of the tangent line is

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

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 $x = 0$.

SOLUTION By the definition of differential coefficient, we have

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

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

Exercise 2..1   For $f(x) = x^2$, using the definition of differential coefficient, find $f^{\prime}(-3)$

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

$$\lim_{h \to 0}\frac{f(-3+h) - f(-3)}{h} = \lim_{h \to 0}\frac{(-3+h)^2 - (-3)^2}{h}$$

$$= \lim_{h \to 0}\frac{9 -6h + h^2 - 9}{h} = \lim_{h \to 0}\frac{h(h-6)}{h}$$

$$= \lim_{ h\to 0}(h-6) = -6\ensuremath{\ \blacksquare}$$

Equation of Tangent Line

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

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

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 $f'(x_0)$.

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

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

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 $(-3,9)$ is $f'(-3) = -6$. Thus the equation of a tangent line is

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

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

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

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

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 $(0,0)$ is $f'(0) = 1$. Thus, the equation of the tangent line is

$$y = x$$

Similarly, we have the equation of the normal line

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

Left-Hand Differential Coefficient

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

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

Right-Hand Differential Coefficient

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

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 $x = x_{0}$.

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 $f_{-}^{\prime}(0)$.

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

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

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

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

Exercise 2..3   Determine whether the following function is differentiable at $x = 0$. $${f(x) = \left\{\begin{array}{ll} x^2 \sin{\frac{1}{x}}, & x \neq 0\\ 0, & x = 0 \end{array}\right.}$$

SOLUTION

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

Now we have

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

Let $h \to 0$. Then

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

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

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 $x = x_{0}$.

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

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

and $x$ approaches $x_{0}$. Then

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

Note that $f(x)$ is differentiable at $x = x_{0}$. Thus

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

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

This implies that

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

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

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

NOTE The symbols of derivatives are

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

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.
$${1. \ f(x) = x^{n},\ n \mbox{integer}}$$ $${2. \ f(x) = \sin{x}}$$ $${3. \ f(x) = e^x}$$

SOLUTION 1.

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

$$= \lim_{h \rightarrow 0}\frac{nx^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n}{h} \ \$$

$$= nx^{n-1} + \lim_{h \rightarrow 0}[\binom{n}{2}x^{n-2}h + \binom{n}{3}x^{n-3}h^2 + \cdots + h^{n-1}]$$

$$= nx^{n-1} \ensuremath{\ \blacksquare}$$

2.

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

$$= \lim_{h \rightarrow 0}\frac{\sin{x}\cos{h} + \cos{x}\sin{h} - \sin{x}}{h} \ \$$

$$= \lim_{h \rightarrow 0}[\sin{x}\frac{\cos{h} - 1}{h} + \cos{x}\frac{\sin{h}}{h}]$$

$$= \sin{x} \lim_{h \rightarrow 0}\frac{-\sin^{2}{h}}{h(\cos{h} + 1)} + \cos{x}$$

$$= \sin{x} \lim_{h \rightarrow 0}\frac{\sin{h}}{h}\cdot\frac{-\sin{h}}{\cos{h} + 1} + \cos{x} \ \$$

$$= \sin{x}(1 \cdot - \frac{0}{2}) + \cos{x} = \cos{x}\ensuremath{\ \blacksquare}$$

3.

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

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

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

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

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

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

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

SOLUTION 1.

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

$$= \lim_{h \rightarrow 0}\frac{\cos{x}\cos{h} - \sin{x}\sin{h} - \cos{x}}{h} \ \$$

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

$$= -\sin{x}$$

3. By the definition of the derivative, we have

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

Now using the law of logarithm, to transform the right-hand side,

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

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

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

$$\lim_{h \to 0}\log \big(1 + \frac{h}{x}\big)^{\frac{1}{h}} = \lim_{t \to \pm \infty}\log \big(1 + \frac{1}{t}\big)^{\frac{t}{x}} = \frac{1}{x}\lim_{t \to \pm \infty}\log \big(1+ \frac{1}{t}\big)^{t}.$$

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

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

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

Differentiation Formula

Theorem 2..2   Let $f(x), g(x), h(x)$ be differentiable and $c$ be constant

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

Proof $3.$ .

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

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

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

$$= f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$$

Similarly for other cases $\ \blacksquare$

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

SOLUTION1.

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

2.

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

$$= e^x \sin{x} + e^x \cos{x} = e^x (\sin{x} + \cos{x}) \ensuremath{\ \blacksquare}$$

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

SOLUTION1.

$$(\tan{x})^{\prime} = (\frac{\sin{x}}{\cos{x}})^{\prime}$$

$$\underbrace{=}_{quotient rule} \frac{(\sin{x})^{\prime}\cos{x} - \sin{x}(\cos{x})^{\prime}}{\cos^{2}{x}}$$

$$= \frac{\cos^{2}{x} + \sin^{2}{x}}{\cos^{2}{x}}$$

$$= \frac{1}{\cos^{2}{x}} = \sec^{2}{x} \ensuremath{\ \blacksquare}$$

2.

$$y^{\prime} = (x^2 \log{x})^{\prime} \underbrace{=}_{product rule} (x^2)^{\prime}\log{x} + x^2(\log{x})'$$

$$= 2x\log{x} + x^2 \cdot \frac{1}{x} = x (2\log{x} + x) \ensuremath{\ \blacksquare}$$

Exercise A

1.

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

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

2.

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

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

3.

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

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

4.

Find the derivative of the following functions:?D

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

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

Exercise B

1.

Find the derivative of the following functions:?D

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

2.

Find the differential of the following functions:?D

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

3.

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

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

4.

Find the derivative of the following functions:?D

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

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