next up previous contents index
Next: Differentiation Formnulas Up: Differentiation Previous: Differentiation   Contents   Index

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]{SOFTFIG-2/Fig2-1.eps}
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 $f^{\prime}(x_{0})$. 2.1
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 $x = 0$.

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 $f^{\prime}(-3)$

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 $f'(x_0)$.
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

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

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

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

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

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

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

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

$\displaystyle y = x$

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

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 $x = 0$ $\ \blacksquare$

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

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

and $x$ 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 $f(x)$ 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} $

The converse of this statement is not true. In other words, continuity does not imply differentiability. see Example. 2.3

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} $

The symbols of derivatives are NOTE

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

To find the derivative of , we say . $f(x)$ differentiate

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

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

2.

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

3.
$\displaystyle f^{\prime}(x)$ $\displaystyle =$ $\displaystyle \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0}\frac{e^{x+h} - e^{x}}{h}$  
  $\displaystyle =$ $\displaystyle \lim_{h \rightarrow 0}\frac{e^{x}(e^{h} - 1)}{h} = e^{x}\lim_{h \rightarrow 0}\frac{e^{h} - 1}{h}$  

Now let . Then since , we have $t = e^h - 1$$h = \log(t+1)$
$\displaystyle \lim_{h \to 0}\frac{e^h - 1}{h}$ $\displaystyle =$ $\displaystyle \lim_{t \to 0}\frac{t}{\log(t+1)} = \lim_{t \to 0}\frac{1}{\frac{1}{t}\log(t+1)}$  
  $\displaystyle =$ $\displaystyle \lim_{t \to 0}\frac{1}{\log(t+1)^{\frac{1}{t}}} = \frac{1}{\log{e}} = 1.$  

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

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

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

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, 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 =$$\displaystyle \lim_{h \to 0}\log \big(1 + \frac{h}{x}\big)^{\frac{1}{h}}.$  
Set . Then as implies . Thus $\frac{x}{h} = t$$h \to 0$$t \to \pm\infty$
$\displaystyle \lim_{h \to 0}\log \big(1 + \frac{h}{x}\big)^{\frac{1}{h}}$$\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 , we have $\lim_{t \to \pm \infty}(1 + \frac{1}{t})^{t} = e$

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

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

  Let be differentiable and be constant
Theorem ..22$f(x), g(x), h(x)$$c$
    • $(f(x) \pm g(x))^{\prime} = f^{\prime}(x) \pm g^{\prime}(x)$
    • $(cf(x))^{\prime} = cf^{\prime}(x)$
    • $(f(x)g(x))^{\prime} = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$
    • $\left(\frac{f(x)}{g(x)}\right)^{\prime} = \frac{f^{\prime}(x)g(x) - f(x)g^{\prime}(x)}{g^{2}(x)}$

    . 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 $\ \blacksquare$

    Example ..25   Find the derivative of the followings.
    $\displaystyle{1. \ y = 3x^{3} + 2x +3}$$\displaystyle{2. \ y = e^{x}\sin{x}}$

    1. SOLUTION
    $\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}$  
    2.
    $\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}$  

      Find the derivative of the followings.
    Exercise ..25
    $\displaystyle{1. \ \tan{x}}$$\displaystyle{2. \ x^2 \log{x}}$

    1. SOLUTION
    $\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 =$$\displaystyle \frac{1}{\cos^{2}{x}} = \sec^{2}{x} \ensuremath{\ \blacksquare}$  
    2.
    $\displaystyle y^{\prime}$$\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?D

    (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:?D

    (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:?D

    (a)$\displaystyle{f(x) = \cos{3x}}$(b)$\displaystyle{f(x) = (x + 2)^{n} \ (n : \mbox{integer})}$
    2.
    Find the differential of the following functions:?D

    (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:?D

    (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:?D

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



       
    next up previous contents index
    Next:Differentiation Formnulas Up:Differentiation Previous:Differentiation Contents Index