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Aus Förberedande kurs i matematik 2

für 1.1 Inledning till derivata
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__NOTOC__
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" | &nbsp;
{{Vald flik|[[1.1 Inledning till derivata|Theory]]}}
{{Ej vald flik|[[1.1 Övningar|Exercises]]}}
| style="border-bottom:1px solid #797979"  width="100%"| &nbsp;
|}

{{Info|
'''Contents:'''

* Definition of the derivative (overview). 
* Derivative of <math>x^\alpha</math>, <math>\ln x</math>, <math>e^x</math>, <math>\cos x</math>, <math>\sin x</math> and <math>\tan x</math>.
*  Derivative of  sums and differences. 
* Tangents and normals to curves. 
}}

{{Info|
'''Learning outcomes:'''

After this section, you will have learned:
* That the first derivative <math>f^{\,\prime}(a)</math> is  the slope of the curve <math>y&#061;f(x)</math> at the  point <math>x&#061;a</math>.
* That the first derivative is the instantaneous rate of change of a quantity (such as speed, price increase, and so on.). 
* That there are functions that are not differentiable (such as <math>f(x)&#061;\vert x\vert</math> at <math>x&#061;0</math>). 
* To differentiate <math>x^\alpha</math>, <math>\ln x</math>, <math>e^x</math>, <math>\cos x</math>, <math>\sin x</math> and <math>\tan x</math> as well as the sums / differences of such terms. 
* To determine the tangent  and normal to the curve  <math>y&#061;f(x)</math>.
* That the derivative can be denoted by <math>f^{\,\prime}(x)</math> och <math>df/dx(x)</math>.
}}

== Introduction ==

When studying mathematical functions and their graphs  one of the main areas of study of a function is the way it  changes, i.e. whether a function is increasing or decreasing and the rate at which this is taking place. 


One makes use of the concept rate of change  (or speed of change), which is a measure of how the value of the function (<math>y</math>) is changing per unit increase in the variable (<math>x</math>). If one knows  two points on a graph of a  function one can get a measure of the rate of change of the function between these points by calculating the increment ratio 

{{Fristående formel||<math>\frac{\Delta y}{\Delta x}= \frac{\text{increment in @(i)y@(/i)}}{\text{increment in@(i)x@(/i)}}</math>}}
	 
<div class="exempel">
''' Example 1'''

The linear functions  <math>f(x)=x</math> and <math>g(x)=-2x</math> change by the same amount everywhere. Their rates of change are <math>1</math> and. <math>−2</math>, which are the slopes of these straight lines. 




<center>
{|
| align="center" |{{:1.1 - Figur - Grafen till f(x) = x}}
| width="30px" |
| align="center" |{{:1.1 - Figur - Grafen till f(x) = -2x}}
|-
| align="center" |<small>Graph of ''f''(''x'') = ''x'' has slope&nbsp;1.</small>
| width="30px" |
| align="center" |<small>Graph of ''g''(''x'') = - 2''x'' has slope&nbsp;-&nbsp;2.</small>
|}
</center>
    
                  	 
Thus for a linear function the rate of change  is the same as the slope. 

</div>

For  a function for which the value changes  with time, it is natural to use the term speed of change because rate of change specifies how the value of the functions is changing per unit time. 

If a car is moving at a speed of 80&nbsp;km/h then the distance traveled, ''s'' km,  after ''t'' hours is given by the function <math>s(t)=80 t</math>.
The rate of change of the function  indicates how the value of the function  is changing per hour, which of course is the same as the car's speed, 80&nbsp;km/h.

For non-linear functions however,  the slope of the curve of the function  is changing all the time and thus also the function's rate of change. In order to determine how such a function is changing, we can either give  the functions average change ( its mean) between two points on the curve of the function, or the instantaneous rate of change  at one point on the curve. 

<div class="exempel">
''' Example 2'''

For the function  <math>f(x)=4x-x^2</math> one has <math>f(1)=3</math>,  <math>f(2)=4</math> and <math>f(4)=0</math>.

<ol type="a">
  <li> Mean change (mean slope) from <math>x = 1</math>
    to <math>x = 2</math> is
{{Fristående formel||<math>\frac{\Delta y}{\Delta x}= \frac{f(2)-f(1)}{2-1}
  = \frac{4-3}{1}=1\,\mbox{,}</math>}}
and the function increases in this interval. </li>

  <li> Mean change  from <math>x = 2</math> to <math>x = 4</math> is
{{Fristående formel||<math>\frac{\Delta y}{\Delta x}= \frac{f(4)-f(2)}{4-2}
  = \frac{0-4}{2}=-2\,\mbox{,}</math>}}
and the function decreases in this interval.</li>

  <li> Between <math>x = 1</math> and <math>x = 4</math>  the mean change is
{{Fristående formel||<math>\frac{\Delta y}{\Delta x} = \frac{f(4)-f(1)}{4-1}
  = \frac{0-3}{3} = -1\,\mbox{.}</math>}}
On average, the function decreases in this interval, although the function  both increases and decreases within this interval.t.</li>
</ol>

<center>
{| align="center"
| align="center" |{{:1.1 - Figur - Medelförändring för f(x) = x(4 - x) mellan x = 1 och x = 2}}
| width="30px" |
| align="center" |{{:1.1 - Figur - Medelförändring för f(x) = x(4 - x) mellan x = 1 och x = 4}}
|-
| align="center" |<small>Between  ''x'' = 1 and ''x'' = 2 the function has the mean change 1/1 = 1.</small>
| width="30px" |
| align="center" |<small>Between  ''x'' = 1 and ''x'' = 4  the function has the mean change (-3)/3 = -1.</small>
|}
</center>



</div>


==Definition of the derivative ==

To calculate the instantaneous rate of change of a function, that is, the slope of its curve at a point ''P'', we  temporarily use an additional point ''Q'' in the vicinity of ''P'' and construct the increment ratio between ''P'' and ''Q'':

<center>{{:1.1 - Figur - Differenskvot mellan P och Q}}</center>

'''Increment ratio'''
{{Fristående formel||<math>\frac{\Delta y}{\Delta x}
  = \frac{f(x+h)-f(x)}{(x+h)-x}= \frac{f(x+h)-f(x)}{h}\,\mbox{.}</math>}}


If we allow ''Q''  to approach ''P''  (that is allow <math>h \rightarrow 0</math>) we can work out what the value would be  if the points coincided, and thus obtain the slope at the point ''P''. We call this value ''the derivative'' of <math>f(x)</math> at the point ''P'', and can be interpreted as the instantaneous rate of change of <math>f(x)</math> at the point ''P''.

The derivative of a function <math>f(x)</math> is written as <math>f^{\,\prime}(x)</math>  and may be formally defined as follows: 

<div class="regel">
''The derivative'' of a function <math>f(x)</math>, is defined as 
{{Fristående formel||<math>f^{\,\prime}(x)
  = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \,\mbox{.}</math>}}
If <math>f^{\,\prime}(x_0)</math> exists, one says  that  <math>f(x)</math> is ''differentiable'' at the point <math>x=x_0</math>.
</div>

Different notations for the derivative are used, for example, 

{| border="0" cellpadding="5" cellspacing="0" align="center"
!width="200" style="border-bottom:2px solid grey;" align="center" |  Function
!width="200" style="border-bottom:2px solid grey;" align="center" |  derivative
|-
| align="center" | <math>f(x)</math>
| align="center" | <math>f^{\,\prime}(x)</math>	   
|-
| align="center" | <math>y</math>
| align="center" | <math>y^{\,\prime}</math>
|-
| align="center" | <math>y</math>
| align="center" | <math>Dy</math>
|-
| align="center" | <math>y</math>
| align="center" | <math>\dfrac{dy}{dx}</math>
|-
| align="center" | <math>s(t)</math>
| align="center" | <math>\dot s(t)</math>
|}


== The sign of the derivative ==

The sign of the derivative (+/−) tells us if the functions graph slopes upwards or downwards, that is, if the function is increasing or decreasing: 

* <math>f^{\,\prime}(x) > 0</math> (positive slope) means that <math>f(x)</math> is increasing.
* <math>f^{\,\prime}(x) < 0</math>  (negative slope) means that <math>f(x)</math> is decreasing. 
* <math>f^{\,\prime}(x) = 0</math> (no slope) means that <math>f(x)</math> is stationary (horisontal). 


<div class="exempel">
''' Example 3'''

<ol type="a">
  <li><math>f(2)=3\ </math> means that '''the value of the function''' is 
    <math>3</math> at <math>x=2</math>.</li> 
  <li><math>f^{\,\prime}(2)=3\ </math>means that '''the value of the derivative''' 
    is <math>3</math> when <math>x=2</math>, which in turn means that the
    functions graph has a slope <math>3</math> at <math>x=2</math>.</li>
</ol>
</div>

<div class="exempel">
''' Example 4'''

From  the figure one can obtain that 
<center>
{| align="center"
|- 
| valign="center" |
<math>\begin{align*}
f^{\,\prime}(a) &> 0\\[4pt]
f(b) &= 0\\[4pt]
f^{\,\prime}(c) &= 0\\[4pt]
f(d) &= 0\\[4pt]
f^{\,\prime}(e) &= 0\\[4pt]
f(e) &< 0\\[4pt]
f^{\,\prime}(g) &> 0
\end{align*}</math>
| width="30px" |
| valign="center" |
{{:1.1 - Figur - Grafen y = f(x) med punkter x = a, b, c, d, e och g}}
|}
</center>

Note the different meanings of <math>f(x)</math> and <math>f^{\,\prime}(x)</math>.

</div>

<div class="exempel">
''' Example 5'''

The temperature in a thermos is given by a function, where  <math>T(t)</math> is the temperature of the thermos after <math>t</math>  minutes. Interpret  the following using mathematical symbols:

<ol type="a">
  <li> After 10&nbsp;minutes the temperature is 80°.<br><br>
      <math>T(10)=80</math><br><br></li>
  <li> After 2&nbsp;minutes the temperature is dropping in the thermos by 3° per minute <br><br>
      <math>T'(2)=-3</math> (the temperature is decreasing, which is why the derivative is   negative)<br><br></li> 
</ol>
</div>

<div class="exempel">
''' Example 6'''

The function <math>f(x)=|x|</math> does not have  a  derivative at <math>x=0</math>. One cannot determine how the graph  of the function slopes at the point  <math>(0,0)</math> (see figure below). 

One can express this, for example, in one of the following ways:"<math>f^{\,\prime}(0)</math> does not exist", "<math>f^{\,\prime}(0)</math> is not defined " or "<math>f(x)</math> is not differentiable at <math>x=0</math>". 

<center>{{:1.1 - Figur - Grafen till f(x) = beloppet av x}}</center>
<center><small> graph of the function ''f''(''x'') = |''x''|</small></center>

</div>


== Differentiation rules  ==

Using the definition of differentiation one can  determine the derivatives for the standard types of functions. 


<div class="exempel">
''' Example 7'''

If <math>f(x)=x^2</math>  then, according to the definition of the increment ratio

{{Fristående formel||<math>\frac{(x+h)^2-x^2}{h}=\frac{x^2+2hx+h^2-x^2}{h}
  = \frac{h(2x+h)}{h} = 2x + h\,\mbox{.}</math>}}
	 
If we then let <math>h</math> go to zero, we see that the slope at the point becomes <math>2x</math>. We have thus shown that the slope of an arbitrary point on the curve <math>y=x^2</math> is <math>2x</math>. That is, the derivative of <math>x^2</math> is <math>2x</math>.
</div>

In a similar way, we can deduce general differentiation rules: 

{| border="0" cellpadding="5" cellspacing="0" align="center"
!width="200" style="border-bottom:2px solid grey;" align="center" | Function
!width="200" style="border-bottom:2px solid grey;" align="center" | Derivative
|-
| align="center" |<math>x^n</math>
| align="center" |<math>nx^{n-1}</math>	   
|-
| align="center" |<math>\ln x</math>
| align="center" |<math>1/x</math>
|-
| align="center" |<math>e^x</math>
| align="center" |<math>e^x</math>
|-
| align="center" |<math>\sin x</math>
| align="center" |<math>\cos x</math>
|-
| align="center" |<math>\cos x</math>
| align="center" |<math>-\sin x</math>
|-
| align="center" |<math>\tan x</math>
| align="center" |<math>1/\cos^2 x</math>
|}


In addition, for sums and differences of expressions of functions  one has
{{Fristående formel||<math>D(f(x) +g(x))
  = f^{\,\prime}(x) + g'(x)\,\mbox{.}</math>}}

Additionally, if ''k'' is a constant, then 
{{Fristående formel||<math>D(k \cdot f(x))
  = k \cdot f^{\,\prime}(x)\,\mbox{.}</math>}}


<div class="exempel">
''' Example 8'''
<ol type="a">
  <li><math>D(2x^3 - 4x + 10 - \sin x)
        = 2\,D x^3 - 4\,D x + D 10 - D \sin x</math><br>
      <math>\phantom{D(2x^3 - 4x + 10 - \sin x)}{}
        = 2\cdot 3x^2 - 4\cdot 1 + 0 - \cos x</math></li> 
  <li><math> y= 3 \ln x + 2e^x \quad</math>  gives that
      <math>\quad y'= 3 \cdot\frac{1}{x} + 2 e^x
        = \frac{3}{x} + 2 e^x\,</math>.</li>
  <li><math>\frac{d}{dx}\Bigl(\frac{3x^2}{5} - \frac{x^3}{2}\Bigr)
        = \frac{d}{dx}\bigl(\tfrac{3}{5}x^2 - \tfrac{1}{2}x^3\bigr)
        = \tfrac{3}{5}\cdot 2x - \tfrac{1}{2}\cdot 3x^2
        = \tfrac{6}{5}x - \tfrac{3}{2}x^2\,</math>.</li>
  <li><math>s(t)= v_0t + \frac{at^2}{2} \quad</math>  gives that
      <math>\quad s'(t)=v_0 + \frac{2at}{2} = v_0 + at\,</math>.</li>
</ol>

</div> 

<div class="exempel">
''' Example 9'''
<ol type="a">
  <li><math>f(x) = \frac{1}{x} = x^{-1} \quad</math>  gives that
      <math>\quad f^{\,\prime}(x) = -1 \cdot x^{-2}
        = -\frac{1}{x^2}\,</math>.</li>
  <li><math>f(x)= \frac{1}{3x^2} = \tfrac{1}{3}\,x^{-2} \quad</math> gives that <math>\quad f^{\,\prime}(x) = \tfrac{1}{3}\cdot(-2)x^{-3}
      = -\tfrac{2}{3} \cdot x^{-3} = -\frac{2}{3x^3}\,</math>.</li>
  <li><math>g(t) = \frac{t^2 - 2t + 1}{t} = t -2 + \frac{1}{t} \quad</math>
     gives that<math>\quad g'(t) = 1 - \frac{1}{t^2}\,</math>.</li>
  <li><math>y = \Bigl( x^2 + \frac{1}{x} \Bigr)^2
        = (x^2)^2 + 2 \cdot x^2 \cdot \frac{1}{x} + \Bigl(\frac{1}{x} \Bigr)^2
        = x^4 + 2x + x^{-2}</math><br>
<math>\qquad\quad</math> gives that<math>\quad y' =4x^3 + 2 -2x^{-3}
        = 4x^3 + 2 - \frac{2}{x^3}\,</math>.</li>
</ol>

</div>

<div class="exempel">
''' Example 10'''

The function <math>f(x)=x^2 + x^{-2}</math> has the derivative 

{{Fristående formel||<math>f^{\,\prime}(x) = 2x^1 -2x^{-3}
  = 2x - \frac{2}{x^3}\,\mbox{.}</math>}}

This means, for example, that <math>f^{\,\prime}(2) = 2\cdot 2 - 2/2^3= 4- \frac{1}{4} = \frac{15}{4}</math>  and that <math>f^{\,\prime}(-1) = 2 \cdot (-1) - 2/(-1)^3 = -2 + 2 = 0</math>. However, the derivative <math>f'(0)</math> is not defined. 

</div>

<div class="exempel">
''' Example 11'''

An object moves according to <math>s(t) = t^3 -4t^2 +5t</math>, where <math>s(t)</math> km is the distance from the starting point after <math>t</math> hours. Calculate <math>s'(3)</math> and explain what the value stands for. 
<br>
<br>
Differentiating with respect to the time  

{{Fristående formel||<math>s'(t) = 3t^2 - 8t +5\qquad
  \text{which gives that}\qquad s'(3) = 3 \cdot 3^2 - 8 \cdot 3 + 5
  = 8\,\mbox{.}</math>}}

This might suggest that after 3 hours the object's speed is 8 km/h. 

</div> 

<div class="exempel">
''' Example 12'''

The total cost  <math>T</math>  dollars for the manufacture of <math>x</math> objects is given by the function 

{{Fristående formel||<math> T(x) = 40000 + 370x -0{,}09x^2, \quad
  \text{for} \ 0 \le x \le 200\,\mbox{.}</math>}}

Calculate and explain the meaning of the following expressions. 
<br>
<br>
<ol type="a">
  <li><math>T(120)</math><br><br>
      <math>T(120)=40000 + 370 \cdot 120 - 0{,}09 \cdot 120^2
        = 83104\,</math>.<br> 
      The total cost to manufacture 120 objects is 83104 dollars.</li>
  <li><math>T'(120)</math><br><br>
        The derivative is given  by <math>T^{\,\prime}(x)= 370 - 0\textrm{.}18x</math> and 
       therefore, is 
      {{Fristående formel||<math>T^{\,\prime}(120) = 370 - 0\textrm{.}18 \cdot 120
        \approx 348\textrm{.}</math>}}
 Marginal costs ("the cost to produce an additional 1 object") of 120 manufactured objects is approximately 348 dollars.  </li>
</ol>

</div>


== Tangents and normals ==

A  ''tangent'' to a curve is a straight line tangential to the curve. 

A ''normal'' to a curve at a point on the curve is a straight line that is perpendicular to the curve at the
point (and hence perpendicular to the curve's tangent at this point). 

For perpendicular lines, the product of their slopes is  <math>–1</math>, i.e. if the tangents slope is  <math>k_T</math>  and the normals is <math>k_N</math> then <math>k_T \cdot k_N = -1</math>. Since we can determine the slope of a curve with the help of the derivative, we can also determine the equation of a tangent or a normal, if we know the equation for the curve. 


<div class="exempel">
''' Example 13'''

Determine the equation for the tangent and the normal to the curve  <math>y=x^2 + 1</math> at the point <math>(1,2)</math>.
<br>
<br>
We write the tangents equation as <math>y = kx + m</math>. Since it is to tangent (touch) the curve at  <math>x=1</math> it must have a slope of <math>k= y'(1)</math>,  i.e.

{{Fristående formel||<math>y' = 2x,\qquad y'(1) = 2\cdot 1 = 2</math>.}}

The tangent  also passes through the point <math>(1,2)</math> and therefore  <math>(1,2)</math>  must satisfy the tangents equation

{{Fristående formel||<math>2 = 2 \cdot 1 + m \quad \Leftrightarrow \quad
  m = 0</math>.}}

The tangents  equation is  thus <math>y=2x</math>.


The slope of the normal is  <math>k_N = -\frac{1}{k_T} = -\frac{1}{2}</math> .

In addition, the normal also passes through the point <math>(1, 2)</math> ,  i.e. 


{{Fristående formel||<math>2= -\frac{1}{2}\cdot 1 + m
  \quad \Leftrightarrow \quad m = \frac{5}{2}</math>.}}

The normal has the equation <math>y= -\frac{x}{2} + \frac{5}{2} = \frac{5-x}{2}</math>.



<center>
{| align="center"
| align="center" |{{:1.1 - Figur - Tangentlinjen y = 2x}}
| width="30px" |
| align="center" |{{:1.1 - Figur - Normallinjen y = (5 - x)/2}}
|-
| align="center" |<small>Tangent <math>y=2x</math></small>
| width="30px" |
| align="center" |<small>Normal <math>y=(5-x)/2</math></small>
|}
</center>



</div>

<div class="exempel">
''' Example 14'''

The curve <math>y = 2 \, e^x - 3x</math>  has a tangent with a slope of  <math>–1</math>. Determine the point of tangency (where the tangent touches the curve).
<br>
<br>
{| width="100%"
| width="90%" |
The derivative of the right-hand side is  <math>y' = 2 \, e^x -3</math> and at the point of tangency the derivative must  be equal to <math>-1</math>, that is,
<math>y' = -1</math>,  and this gives us the equation
{{Fristående formel||<math>2 \, e^x - 3=-1</math>}}
which has a solution <math> x=0</math>. At the point <math>x=0</math> the curve has <math>y</math>-value <math>y(0) = 2 \, e^0 - 3 \cdot 0 = 2</math> and therefore  the point of tangency is <math>(0,2)</math>.
| width="10%" |
||{{:1.1 - Figur - Kurvan y = 2e^x - 3x och tangentlinjen genom (0,2)}}
|}

</div>
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Aus Förberedande kurs i matematik 2
