1.1 Introduction to derivatives

From Förberedande kurs i matematik 2

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__NOTOC__
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{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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{{Selected tab|[[1.1 Introduction to derivatives|Theory]]}}
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{{Not selected tab|[[1.1 Exercises|Exercises]]}}
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| style="border-bottom:1px solid #797979" width="100%"|  
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Contents:'''
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* Derivatans definition (översiktligt).
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* Definition of the derivative (overview).
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* Derivatan av <math>x^\alpha</math>, <math>\ln x</math>, <math>e^x</math>, <math>\cos x</math>, <math>\sin x</math> och <math>\tan x</math>.
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* 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>.
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* Derivata av summa och differens.
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* Derivative of sums and differences.
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* Tangent och normal till kurvor.
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* Tangents and normals to curves.
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned:
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* Förstå derivatan <math>f^{\,\prime}(a)</math> som lutningen av kurvan <math>y&#061;f(x)</math> i punkten <math>x&#061;a</math>.
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* That the first derivative <math>f^{\,\prime}(a)</math> is the gradient of the curve <math>y&#061;f(x)</math> at the point <math>x&#061;a</math>.
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* Förstå derivatan som den momentana ändringstakten av en storhet (exempelvis fart, prisökning, osv.).
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* That the first derivative is the instantaneous rate of change of a quantity (such as speed, price increase, and so on.).
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* Veta att det finns funktioner som inte är deriverbara (t.ex. <math>f(x)&#061;\vert x\vert</math> i <math>x&#061;0</math>).
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* That there are functions that are not differentiable (such as <math>f(x)&#061;\vert x\vert</math> at <math>x&#061;0</math>).
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* Kunna derivera <math>x^\alpha</math>, <math>\ln x</math>, <math>e^x</math>, <math>\cos x</math>, <math>\sin x</math> och <math>\tan x</math> samt summor/differenser av sådana termer.
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* 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.
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* Kunna bestämma tangent och normal till <math>y&#061;f(x)</math>.
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* To determine the tangent and normal to the curve <math>y&#061;f(x)</math>.
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* Veta att derivatan kan betecknas med <math>f^{\,\prime}(x)</math> och <math>df/dx(x)</math>.
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* That the derivative can be denoted by <math>f^{\,\prime}(x)</math> or <math>df/dx(x)</math>.
}}
}}
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== Inledning ==
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== Introduction ==
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När man studerar matematiska funktioner och deras grafer är ett av de viktigaste områdena studiet av en funktions förändring, dvs. om en funktion ökar eller minskar samt i vilken takt detta sker.
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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.
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Man använder sig här av begreppet förändringsgrad (eller förändringshastighet), vilket är ett mått på hur funktionens värde (<math>y</math>) ändras för varje enhets ökning av variabelvärdet (<math>x</math>). Om man känner till två punkter på en funktions graf kan man få ett mått på funktionens förändringsgrad mellan dessa punkter genom att beräkna ändringskvoten
 
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{{Fristående formel||<math>\frac{\Delta y}{\Delta x}= \frac{\text{skillnad i @(i)y@(/i)-led}}{\text{skillnad i @(i)x@(/i)-led}}</math>}}
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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
 +
 
 +
{{Displayed math||<math>\frac{\Delta y}{\Delta x}= \frac{\text{increment in @(i)y@(/i)}}{\text{increment in@(i)x@(/i)}}</math>}}
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' 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 gradients of these straight lines.
 +
 
 +
 
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De linjära funktionerna <math>f(x)=x</math> respektive <math>g(x)=-2x</math> förändras på samma sätt hela tiden. Deras förändringsgrad är <math>1</math> resp. <math>−2</math>, vilket vi känner till som linjernas respektive riktningskoefficient.
 
<center>
<center>
{|
{|
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| align="center" |{{:1.1 - Figur - Grafen till f(x) = x}}
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| align="center" |{{:1.1 - Figure - The graph of f(x) = x}}
| width="30px" |
| width="30px" |
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| align="center" |{{:1.1 - Figur - Grafen till f(x) = -2x}}
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| align="center" |{{:1.1 - Figure - The graph of f(x) = -2x}}
|-
|-
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| align="center" |<small>Grafen till ''f''(''x'') = ''x'' har riktningskoefficient&nbsp;1.</small>
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| align="center" |<small>Graph of ''f''(''x'') = ''x'' has gradient&nbsp;1.</small>
| width="30px" |
| width="30px" |
-
| align="center" |<small>Grafen till ''g''(''x'') = - 2''x'' har riktningskoefficient&nbsp;-&nbsp;2.</small>
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| align="center" |<small>Graph of ''g''(''x'') = - 2''x'' has gradient&nbsp;-&nbsp;2.</small>
|}
|}
</center>
</center>
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För en linjär funktion gäller alltså att funktionens förändringsgrad är samma som linjens riktningskoefficient.
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Thus for a linear function the rate of change is the same as the gradient.
</div>
</div>
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Om man har en funktion där funktionsvärdet förändras med tiden är det naturligt att använda begreppet förändringshastighet, eftersom förändringsgraden här anger hur funktionsvärdet ändras per tidsenhet.
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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>.
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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.
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Om en bil rör sig med hastigheten 80&nbsp;km/h så kan den tillryggalagda sträckan, ''s'' km, efter ''t'' timmar beskrivas med funktionen <math>s(t)=80 t</math>.
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Funktionens förändringsgrad anger hur funktionsvärdet ändras per timme, vilket naturligtvis är detsamma som bilens hastighet, 80&nbsp;km/h.
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För icke-linjära funktioner gäller ju att lutningen på funktionskurvan ändras hela tiden och därmed också funktionens förändringsgrad. För att bestämma hur en sådan funktion förändras kan vi antingen ange funktionens genomsnittliga förändring (medelförändringen) mellan två punkter på funktionskurvan, eller den momentana förändringsgraden i en punkt på kurvan.
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For non-linear functions however, the gradient of the graph of the function (that is, the function's rate of change) varies from point to point. We can either give the function's average (mean) rate of change between two points on the curve of the function, or the instantaneous rate of change at one point on the curve. The mean rate of change is fairly easy to calculate; how to calculate the instantaneous rate of change forms the main focus of this section.
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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För funktionen <math>f(x)=4x-x^2</math> är <math>f(1)=3</math>, <math>f(2)=4</math> och <math>f(4)=0</math>.
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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">
<ol type="a">
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<li>Medelförändringen (medellutningen) från <math>x = 1</math>
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<li> Mean rate of change (mean gradient) from <math>x = 1</math>
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till <math>x = 2</math> är
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to <math>x = 2</math> is
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{{Fristående formel||<math>\frac{\Delta y}{\Delta x}= \frac{f(2)-f(1)}{2-1}
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{{Displayed math||<math>\frac{\Delta y}{\Delta x}= \frac{f(2)-f(1)}{2-1}
= \frac{4-3}{1}=1\,\mbox{,}</math>}}
= \frac{4-3}{1}=1\,\mbox{,}</math>}}
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och funktionen ökar i detta intervall.</li>
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and the function increases in this interval. </li>
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<li>Medelförändringen från <math>x = 2</math> till <math>x = 4</math> är
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<li> Mean rate of change from <math>x = 2</math> to <math>x = 4</math> is
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{{Fristående formel||<math>\frac{\Delta y}{\Delta x}= \frac{f(4)-f(2)}{4-2}
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{{Displayed math||<math>\frac{\Delta y}{\Delta x}= \frac{f(4)-f(2)}{4-2}
= \frac{0-4}{2}=-2\,\mbox{,}</math>}}
= \frac{0-4}{2}=-2\,\mbox{,}</math>}}
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och funktionen avtar i detta intervall.</li>
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and the function decreases in this interval.</li>
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<li>Mellan <math>x = 1</math> och <math>x = 4</math> är medelförändringen
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<li> Between <math>x = 1</math> and <math>x = 4</math> the mean rate of change is
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{{Fristående formel||<math>\frac{\Delta y}{\Delta x} = \frac{f(4)-f(1)}{4-1}
+
{{Displayed math||<math>\frac{\Delta y}{\Delta x} = \frac{f(4)-f(1)}{4-1}
= \frac{0-3}{3} = -1\,\mbox{.}</math>}}
= \frac{0-3}{3} = -1\,\mbox{.}</math>}}
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I genomsnitt är funktionen avtagande i detta intervall, även om funktionen både växer och avtar i intervallet.</li>
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On average, the function decreases in this interval, although the function both increases and decreases within this interval.t.</li>
</ol>
</ol>
<center>
<center>
{| align="center"
{| align="center"
-
| align="center" |{{:1.1 - Figur - Medelförändring för f(x) = x(4 - x) mellan x = 1 och x = 2}}
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| align="center" |{{:1.1 - Figure - Mean change of f(x) = x(4 - x) between x = 1 and x = 2}}
| width="30px" |
| width="30px" |
-
| align="center" |{{:1.1 - Figur - Medelförändring för f(x) = x(4 - x) mellan x = 1 och x = 4}}
+
| align="center" |{{:1.1 - Figure - Mean change of f(x) = x(4 - x) between x = 1 and x = 4}}
|-
|-
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| align="center" |<small>Mellan ''x'' = 1 och ''x'' = 2 har funktionen medelförändringen 1/1 = 1.</small>
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| align="center" |<small>Between ''x'' = 1 and ''x'' = 2 the function has the mean rate of change 1/1 = 1.</small>
| width="30px" |
| width="30px" |
-
| align="center" |<small>Mellan ''x'' = 1 och ''x'' = 4 har funktionen medelförändringen (-3)/3 = -1.</small>
+
| align="center" |<small>Between ''x'' = 1 and ''x'' = 4 the function has the mean rate of change (-3)/3 = -1.</small>
|}
|}
</center>
</center>
 +
 +
</div>
</div>
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== Derivatans definition ==
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==Definition of the derivative ==
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För att beräkna den momentana förändringsgraden hos en funktion, dvs. funktionskurvans lutning i en punkt ''P'', tar vi temporärt hjälp av ytterligare en punkt ''Q'' i närheten av ''P'' och bildar ändringskvoten mellan ''P'' och ''Q'':
+
To calculate the instantaneous rate of change of a function, that is, the gradient 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>
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<center>{{:1.1 - Figure - Difference quotient of P and Q}}</center>
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'''Ändringskvoten'''
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'''Increment ratio'''
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{{Fristående formel||<math>\frac{\Delta y}{\Delta x}
+
{{Displayed math||<math>\frac{\Delta y}{\Delta x}
= \frac{f(x+h)-f(x)}{(x+h)-x}= \frac{f(x+h)-f(x)}{h}\,\mbox{.}</math>}}
= \frac{f(x+h)-f(x)}{(x+h)-x}= \frac{f(x+h)-f(x)}{h}\,\mbox{.}</math>}}
-
Om vi låter ''Q'' närma sig ''P'' (dvs. låter <math>h \rightarrow 0</math>) så kan vi lista ut vad värdet blir om punkterna sammanfaller och därmed få fram lutningen i punkten ''P''. Vi kallar detta värde för ''derivatan'' av <math>f(x)</math> i punkten ''P'', vilket kan tolkas som den momentana förändringsgraden av <math>f(x)</math> i punkten ''P''.
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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 gradient at the point ''P''. We call this value ''the derivative'' of <math>f(x)</math> at the point ''P''; it can be interpreted as the instantaneous rate of change of <math>f(x)</math> at the point ''P''.
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Derivatan av en funktion <math>f(x)</math> betecknas <math>f^{\,\prime}(x)</math> och kan formellt definieras så här:
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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">
<div class="regel">
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''Derivatan'' av en funktion <math>f(x)</math>, definieras som
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''The derivative'' of a function <math>f(x)</math>, is defined as
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{{Fristående formel||<math>f^{\,\prime}(x)
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{{Displayed math||<math>f^{\,\prime}(x)
= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \,\mbox{.}</math>}}
= \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \,\mbox{.}</math>}}
-
Om <math>f^{\,\prime}(x_0)</math> existerar, säger man att <math>f(x)</math> är ''deriverbar'' i punkten <math>x=x_0</math>.
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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>
</div>
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Olika symboler för derivatan förekommer, t.ex.
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Different notations for the derivative are used, for example,
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{| border="0" cellpadding="5" cellspacing="0" align="center"
{| border="0" cellpadding="5" cellspacing="0" align="center"
-
!width="200" style="border-bottom:2px solid grey;" align="center" | Funktion
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!width="200" style="border-bottom:2px solid grey;" align="center" | Function
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!width="200" style="border-bottom:2px solid grey;" align="center" | Derivata
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!width="200" style="border-bottom:2px solid grey;" align="center" | derivative
|-
|-
| align="center" | <math>f(x)</math>
| align="center" | <math>f(x)</math>
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== Derivatans tecken ==
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== The sign of the derivative ==
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Derivatans tecken (+/−) visar oss om funktionens graf lutar uppåt eller nedåt, dvs. om funktionen är växande eller avtagande:
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The sign of the derivative (+/−) tells us if the function's graph slopes upwards or downwards, that is, if the function is increasing or decreasing:
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* <math>f^{\,\prime}(x) > 0</math> (positiv lutning) medför att <math>f(x)</math> är växande.
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* <math>f^{\,\prime}(x) > 0</math> (positive gradient) means that <math>f(x)</math> is increasing.
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* <math>f^{\,\prime}(x) < 0</math> (negativ lutning) medför att <math>f(x)</math> är avtagande.
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* <math>f^{\,\prime}(x) < 0</math> (negative gradient) means that <math>f(x)</math> is decreasing.
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* <math>f^{\,\prime}(x) = 0</math> (ingen lutning) medför att <math>f(x)</math> är stationär (horisontell).
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* <math>f^{\,\prime}(x) = 0</math> (gradient zero) means that <math>f(x)</math> is stationary (horizontal).
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
-
<li><math>f(2)=3\ </math> betyder att '''funktionens värde''' är
+
<li><math>f(2)=3\ </math> means that '''the value of the function''' is
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<math>3</math> när <math>x=2</math>.</li>
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<math>3</math> at <math>x=2</math>.</li>
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<li><math>f^{\,\prime}(2)=3\ </math> betyder att '''derivatans värde'''
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<li><math>f^{\,\prime}(2)=3\ </math>means that '''the value of the derivative'''
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är <math>3</math> när <math>x=2</math>, vilket i sin tur betyder att
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is <math>3</math> when <math>x=2</math>, which in turn means that the
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funktionens graf har lutningen <math>3</math> när <math>x=2</math>.</li>
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function's graph has a gradient <math>3</math> at <math>x=2</math>.</li>
</ol>
</ol>
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 4'''
+
''' Example 4'''
-
 
+
-
I figuren kan man utläsa att
+
 +
From the figure one can obtain that
<center>
<center>
{| align="center"
{| align="center"
Line 183: Line 192:
| width="30px" |
| width="30px" |
| valign="center" |
| valign="center" |
-
{{:1.1 - Figur - Grafen y = f(x) med punkter x = a, b, c, d, e och g}}
+
{{:1.1 - Figure - The graph of y = f(x) with points x = a, b, c, d, e and g}}
|}
|}
</center>
</center>
-
Notera betydelsen av <math>f(x)</math> respektive <math>f^{\,\prime}(x)</math>.
+
Note the different meanings of <math>f(x)</math> and <math>f^{\,\prime}(x)</math>.
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 5'''
+
''' Example 5'''
-
Temperaturen i en termos beskrivs av en funktion, där <math>T(t)</math> är temperaturen i termosen efter <math>t</math> minuter. Skriv följande med matematiska symboler:
+
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">
<ol type="a">
-
<li>Efter 10&nbsp;minuter är temperaturen 80°.<br><br>
+
<li> After 10&nbsp;minutes the temperature is 80°.<br><br>
<math>T(10)=80</math><br><br></li>
<math>T(10)=80</math><br><br></li>
-
<li>Efter 2&nbsp;minuter sjunker temperaturen i termosen med 3° per
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<li> After 2&nbsp;minutes the temperature is dropping in the thermos by 3° per minute <br><br>
-
minut.<br><br>
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<math>T'(2)=-3</math> (the temperature is decreasing, which is why the derivative is negative)<br><br></li>
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<math>T'(2)=-3</math> (funktionen är avtagande, varför derivatan är
+
-
negativ)<br><br></li>
+
</ol>
</ol>
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 6'''
+
''' Example 6'''
-
Funktionen <math>f(x)=|x|</math> saknar derivata då <math>x=0</math>. Man kan nämligen inte bestämma hur funktionens graf lutar i punkten <math>(0,0)</math> (se figuren nedan).
+
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).
-
Man kan uttrycka detta på exempelvis något av följande sätt: "<math>f^{\,\prime}(0)</math> existerar inte" , "<math>f^{\,\prime}(0)</math> är ej definerad" eller "<math>f(x)</math> är inte deriverbar i <math>x=0</math>".
+
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>{{:1.1 - Figure - The graph of f(x) = ∣x∣}}</center>
-
<center><small>Grafen till funktionen ''f''(''x'') = |''x''|</small></center>
+
<center><small> graph of the function ''f''(''x'') = |''x''|</small></center>
</div>
</div>
-
== Deriveringsregler ==
+
== Differentiation rules ==
 +
 
 +
Using the definition of differentiation one can determine the derivatives for the standard types of functions.
-
Med hjälp av derivatans definition kan man bestämma derivatan för de vanliga funktionstyperna.
 
<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
+
''' Example 7'''
-
Om <math>f(x)=x^2</math> så får vi enligt definitionen av derivata ändringskvoten
+
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}
+
{{Displayed math||<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>}}
= \frac{h(2x+h)}{h} = 2x + h\,\mbox{.}</math>}}
-
Om vi sedan låter <math>h</math> gå mot noll så ser vi att lutningen i punkten blir <math>2x</math>. Vi har därmed visat att lutningen i en godtycklig punkt på kurvan <math>y=x^2</math> är <math>2x</math>, dvs. derivatan av <math>x^2</math> är <math>2x</math>.
+
If we then let <math>h</math> go to zero, we see that the gradient at the point becomes <math>2x</math>. We have thus shown that the gradient 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>
</div>
-
På liknande sätt kan man härleda allmänna deriveringsregler:
+
In a similar way, we can deduce general differentiation rules:
{| border="0" cellpadding="5" cellspacing="0" align="center"
{| border="0" cellpadding="5" cellspacing="0" align="center"
-
!width="200" style="border-bottom:2px solid grey;" align="center" |Funktion
+
!width="200" style="border-bottom:2px solid grey;" align="center" | Function
-
!width="200" style="border-bottom:2px solid grey;" align="center" |Derivata
+
!width="200" style="border-bottom:2px solid grey;" align="center" | Derivative
|-
|-
| align="center" |<math>x^n</math>
| align="center" |<math>x^n</math>
Line 260: Line 268:
-
Dessutom gäller för summor och differenser av funktionsuttryck att
+
In addition, for sums and differences of expressions of functions one has
-
{{Fristående formel||<math>D(f(x) +g(x))
+
{{Displayed math||<math>D(f(x) +g(x))
= f^{\,\prime}(x) + g'(x)\,\mbox{.}</math>}}
= f^{\,\prime}(x) + g'(x)\,\mbox{.}</math>}}
-
Samt, om ''k'' är en konstant, att
+
Additionally, if ''k'' is a constant, then
-
{{Fristående formel||<math>D(k \cdot f(x))
+
{{Displayed math||<math>D(k \, f(x))
-
= k \cdot f^{\,\prime}(x)\,\mbox{.}</math>}}
+
= k \, f^{\,\prime}(x)\,\mbox{.}</math>}}
<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
+
''' Example 8'''
<ol type="a">
<ol type="a">
<li><math>D(2x^3 - 4x + 10 - \sin x)
<li><math>D(2x^3 - 4x + 10 - \sin x)
= 2\,D x^3 - 4\,D x + D 10 - D \sin x</math><br>
= 2\,D x^3 - 4\,D x + D 10 - D \sin x</math><br>
<math>\phantom{D(2x^3 - 4x + 10 - \sin x)}{}
<math>\phantom{D(2x^3 - 4x + 10 - \sin x)}{}
-
= 2\cdot 3x^2 - 4\cdot 1 + 0 - \cos x</math></li>
+
= 2\times 3x^2 - 4\times 1 + 0 - \cos x</math></li>
-
<li><math> y= 3 \ln x + 2e^x \quad</math> ger att
+
<li><math> y= 3 \ln x + 2e^x \quad</math> gives that
-
<math>\quad y'= 3 \cdot\frac{1}{x} + 2 e^x
+
<math>\quad y'= 3 \times\frac{1}{x} + 2 e^x
= \frac{3}{x} + 2 e^x\,</math>.</li>
= \frac{3}{x} + 2 e^x\,</math>.</li>
<li><math>\frac{d}{dx}\Bigl(\frac{3x^2}{5} - \frac{x^3}{2}\Bigr)
<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)
= \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{3}{5}\times 2x - \tfrac{1}{2}\times 3x^2
= \tfrac{6}{5}x - \tfrac{3}{2}x^2\,</math>.</li>
= \tfrac{6}{5}x - \tfrac{3}{2}x^2\,</math>.</li>
-
<li><math>s(t)= v_0t + \frac{at^2}{2} \quad</math> ger att
+
<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>
<math>\quad s'(t)=v_0 + \frac{2at}{2} = v_0 + at\,</math>.</li>
</ol>
</ol>
Line 290: Line 298:
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
+
''' Example 9'''
<ol type="a">
<ol type="a">
-
<li><math>f(x) = \frac{1}{x} = x^{-1} \quad</math> ger att
+
<li><math>f(x) = \frac{1}{x} = x^{-1} \quad</math> gives that
-
<math>\quad f^{\,\prime}(x) = -1 \cdot x^{-2}
+
<math>\quad f^{\,\prime}(x) = -1 \times x^{-2}
= -\frac{1}{x^2}\,</math>.</li>
= -\frac{1}{x^2}\,</math>.</li>
-
<li><math>f(x)= \frac{1}{3x^2} = \tfrac{1}{3}\,x^{-2} \quad</math> ger
+
<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}\times(-2)x^{-3}
-
att <math>\quad f^{\,\prime}(x) = \tfrac{1}{3}\cdot(-2)x^{-3}
+
= -\tfrac{2}{3} \times x^{-3} = -\frac{2}{3x^3}\,</math>.</li>
-
= -\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>
<li><math>g(t) = \frac{t^2 - 2t + 1}{t} = t -2 + \frac{1}{t} \quad</math>
-
ger att <math>\quad g'(t) = 1 - \frac{1}{t^2}\,</math>.</li>
+
gives that<math>\quad g'(t) = 1 - \frac{1}{t^2}\,</math>.</li>
<li><math>y = \Bigl( x^2 + \frac{1}{x} \Bigr)^2
<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^2)^2 + 2 \, x^2 \times \frac{1}{x} + \Bigl(\frac{1}{x} \Bigr)^2
= x^4 + 2x + x^{-2}</math><br>
= x^4 + 2x + x^{-2}</math><br>
-
<math>\qquad\quad</math> ger att <math>\quad y' =4x^3 + 2 -2x^{-3}
+
<math>\qquad\quad</math> gives that<math>\quad y' =4x^3 + 2 -2x^{-3}
= 4x^3 + 2 - \frac{2}{x^3}\,</math>.</li>
= 4x^3 + 2 - \frac{2}{x^3}\,</math>.</li>
</ol>
</ol>
Line 310: Line 317:
<div class="exempel">
<div class="exempel">
-
'''Exempel 10'''
+
''' Example 10'''
-
Funktionen <math>f(x)=x^2 + x^{-2}</math> har derivatan
+
The function <math>f(x)=x^2 + x^{-2}</math> has the derivative
-
{{Fristående formel||<math>f^{\,\prime}(x) = 2x^1 -2x^{-3}
+
{{Displayed math||<math>f^{\,\prime}(x) = 2x^1 -2x^{-3}
= 2x - \frac{2}{x^3}\,\mbox{.}</math>}}
= 2x - \frac{2}{x^3}\,\mbox{.}</math>}}
-
Detta betyder exempelvis att <math>f^{\,\prime}(2) = 2\cdot 2 - 2/2^3= 4- \frac{1}{4} = \frac{15}{4}</math> och att <math>f^{\,\prime}(-1) = 2 \cdot (-1) - 2/(-1)^3 = -2 + 2 = 0</math>. Däremot är derivatan <math>f'(0)</math> inte definierad.
+
This means, for example, that <math>f^{\,\prime}(2) = 2\times 2 - 2/2^3= 4- \frac{1}{4} = \frac{15}{4}</math> and that <math>f^{\,\prime}(-1) = 2 \times (-1) - 2/(-1)^3 = -2 + 2 = 0</math>. However, the derivative <math>f'(0)</math> is not defined.
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 11'''
+
''' Example 11'''
-
Ett föremål rör sig enligt <math>s(t) = t^3 -4t^2 +5t</math>, där <math>s(t)</math> km är avståndet från startpunkten efter <math>t</math> timmar. Beräkna <math>s'(3)</math> och förklara vad värdet står för.
+
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>
<br>
<br>
-
Tidsderivatan ges av
+
Differentiating with respect to the time
-
{{Fristående formel||<math>s'(t) = 3t^2 - 8t +5\qquad
+
{{Displayed math||<math>s'(t) = 3t^2 - 8t +5\qquad
-
\text{vilket ger att}\qquad s'(3) = 3 \cdot 3^2 - 8 \cdot 3 + 5
+
\text{which gives that}\qquad s'(3) = 3 \times 3^2 - 8 \times 3 + 5
= 8\,\mbox{.}</math>}}
= 8\,\mbox{.}</math>}}
-
Detta kan tolkas som att efter 3 timmar är föremålets hastighet 8 km/h.
+
This might suggest that after 3 hours the object's speed is 8 km/h.
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 12'''
+
''' Example 12'''
-
Totalkostnaden <math>T</math> kr för tillverkning av <math>x</math> gummidräkter ges av funktionen
+
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
+
{{Displayed math||<math> T(x) = 40000 + 370x -0{,}09x^2, \quad
-
\text{för} \ 0 \le x \le 200\,\mbox{.}</math>}}
+
\text{for} \ 0 \le x \le 200\,\mbox{.}</math>}}
-
Beräkna och förklara innebörden av nedanstående uttryck.
+
Calculate and explain the meaning of the following expressions.
<br>
<br>
<br>
<br>
<ol type="a">
<ol type="a">
<li><math>T(120)</math><br><br>
<li><math>T(120)</math><br><br>
-
<math>T(120)=40000 + 370 \cdot 120 - 0{,}09 \cdot 120^2
+
<math>T(120)=40000 + 370 \times 120 - 0{,}09 \times 120^2
= 83104\,</math>.<br>
= 83104\,</math>.<br>
-
Totalkostnaden för att tillverka 120 gummidräkter är 83104 kr.</li>
+
The total cost to manufacture 120 objects is 83104 dollars.</li>
<li><math>T'(120)</math><br><br>
<li><math>T'(120)</math><br><br>
-
Derivatan ges av <math>T^{\,\prime}(x)= 370 - 0{,}18x</math> och
+
The derivative is given by <math>T^{\,\prime}(x)= 370 - 0\textrm{.}18x</math> and
-
därför är
+
therefore, is
-
{{Fristående formel||<math>T^{\,\prime}(120) = 370 - 0{,}18 \cdot 120
+
{{Displayed math||<math>T^{\,\prime}(120) = 370 - 0\textrm{.}18 \times 120
-
\approx 348</math>.}}
+
\approx 348\textrm{.}</math>}}
-
Marginalkostnaden ("kostnaden för att tillverka ytterligare 1 enhet") vid 120 tillverkade gummidräkter är approximativt 348 kr.</li>
+
Marginal costs ("the cost to produce an additional 1 object") of 120 manufactured objects is approximately 348 dollars. </li>
</ol>
</ol>
Line 364: Line 371:
-
== Tangenter och normaler ==
+
== Tangents and normals ==
-
En ''tangent'' till en kurva är en rät linje som tangerar kurvan.
+
A ''tangent'' to a curve is a straight line tangential to the curve.
-
En ''normal'' till en kurva är en rät linje som är vinkelrät mot kurvan i en viss punkt på kurvan (och därmed också vinkelrät mot kurvans tangent i denna punkt).
+
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).
-
För vinkelräta linjer gäller att produkten av deras riktningskoefficienter är <math>–1</math>, dvs. om tangentens riktningskoefficient betecknas <math>k_T</math> och normalens <math>k_N</math> så är <math>k_T \cdot k_N = -1</math>. Eftersom vi kan bestämma lutningen på en kurva med hjälp av derivatan så kan vi också bestämma ekvationen för en tangent eller en normal om vi känner till funktionsuttrycket för kurvan.
+
For perpendicular lines, the product of their gradients is <math>–1</math>, i.e. if the tangents gradient is <math>k_T</math> and the normals is <math>k_N</math> then <math>k_T \, k_N = -1</math>. Since we can determine the gradient 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">
<div class="exempel">
-
'''Exempel 13'''
+
''' Example 13'''
-
Bestäm ekvationen för tangenten respektive normalen till kurvan <math>y=x^2 + 1</math> i punkten <math>(1,2)</math>.
+
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>
<br>
<br>
-
Vi skriver tangentens ekvation som <math>y = kx + m</math>. Eftersom den ska tangera kurvan i <math>x=1</math> har vi att dess lutning ges av <math>k= y'(1)</math>, dvs.
+
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 gradient of <math>k= y'(1)</math>, i.e.
-
{{Fristående formel||<math>y' = 2x,\qquad y'(1) = 2\cdot 1 = 2</math>.}}
+
{{Displayed math||<math>y' = 2x,\qquad y'(1) = 2\times 1 = 2</math>.}}
-
Tangentlinjen ska också passerar genom punkten <math>(1,2)</math> och därför måste <math>(1,2)</math> uppfylla tangentens ekvation
+
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
+
{{Displayed math||<math>2 = 2 \times 1 + m \quad \Leftrightarrow \quad
m = 0</math>.}}
m = 0</math>.}}
-
Tangentens ekvation är alltså <math>y=2x</math>.
+
The tangents equation is thus <math>y=2x</math>.
-
Riktningskoefficienten för normalen är <math>k_N = -\frac{1}{k_T} = -\frac{1}{2}</math> .
+
The gradient of the normal is <math>k_N = -\frac{1}{k_T} = -\frac{1}{2}</math> .
-
Vidare går normalen också genom punkten <math>(1, 2)</math> , dvs.
+
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
+
{{Displayed math||<math>2= -\frac{1}{2}\times 1 + m
\quad \Leftrightarrow \quad m = \frac{5}{2}</math>.}}
\quad \Leftrightarrow \quad m = \frac{5}{2}</math>.}}
-
Normalen har ekvationen <math>y= -\frac{x}{2} + \frac{5}{2} = \frac{5-x}{2}</math>.
+
The normal has the equation <math>y= -\frac{x}{2} + \frac{5}{2} = \frac{5-x}{2}</math>.
 +
 
 +
 
<center>
<center>
{| align="center"
{| align="center"
-
| align="center" |{{:1.1 - Figur - Tangentlinjen y = 2x}}
+
| align="center" |{{:1.1 - Figure - The tangent y = 2x}}
| width="30px" |
| width="30px" |
-
| align="center" |{{:1.1 - Figur - Normallinjen y = (5 - x)/2}}
+
| align="center" |{{:1.1 - Figure - The normal y = (5 - x)/2}}
|-
|-
-
| align="center" |<small>Tangentlinjen <math>y=2x</math></small>
+
| align="center" |<small>Tangent <math>y=2x</math></small>
| width="30px" |
| width="30px" |
-
| align="center" |<small>Normallinjen <math>y=(5-x)/2</math></small>
+
| align="center" |<small>Normal <math>y=(5-x)/2</math></small>
|}
|}
</center>
</center>
 +
 +
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 14'''
+
''' Example 14'''
-
Kurvan <math>y = 2 \, e^x - 3x</math> har en tangent vars riktningskoefficient är <math>–1</math>. Bestäm tangeringspunkten.
+
The curve <math>y = 2 \, e^x - 3x</math> has a tangent with a gradient of <math>–1</math>. Determine the point of tangency (where the tangent touches the curve).
<br>
<br>
<br>
<br>
{| width="100%"
{| width="100%"
| width="90%" |
| width="90%" |
-
Derivatan av högerledet är <math>y' = 2 \, e^x -3</math> och i tangeringspunkten ska derivatan vara lika med <math>-1</math>, dvs.
+
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>, och detta ger oss ekvationen
+
<math>y' = -1</math>, and this gives us the equation
-
{{Fristående formel||<math>2 \, e^x - 3=-1</math>}}
+
{{Displayed math||<math>2 \, e^x - 3=-1</math>}}
-
som har lösningen <math>x=0</math>. I punkten <math>x=0</math> har kurvan <math>y</math>-värdet <math>y(0) = 2 \, e^0 - 3 \cdot 0 = 2</math> och därmed är tangeringspunkten <math>(0,2)</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 \times 0 = 2</math> and therefore the point of tangency is <math>(0,2)</math>.
| width="10%" |
| width="10%" |
-
||{{:1.1 - Figur - Kurvan y = 2e^x - 3x och tangentlinjen genom (0,2)}}
+
||{{:1.1 - Figure - The curve y = 2e^x - 3x and its tangent through (0,2)}}
|}
|}
</div>
</div>

Current revision

       Theory          Exercises      

Contents:

  • Definition of the derivative (overview).
  • Derivative of \displaystyle x^\alpha, \displaystyle \ln x, \displaystyle e^x, \displaystyle \cos x, \displaystyle \sin x and \displaystyle \tan x.
  • Derivative of sums and differences.
  • Tangents and normals to curves.

Learning outcomes:

After this section, you will have learned:

  • That the first derivative \displaystyle f^{\,\prime}(a) is the gradient of the curve \displaystyle y=f(x) at the point \displaystyle x=a.
  • 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 \displaystyle f(x)=\vert x\vert at \displaystyle x=0).
  • To differentiate \displaystyle x^\alpha, \displaystyle \ln x, \displaystyle e^x, \displaystyle \cos x, \displaystyle \sin x and \displaystyle \tan x as well as the sums / differences of such terms.
  • To determine the tangent and normal to the curve \displaystyle y=f(x).
  • That the derivative can be denoted by \displaystyle f^{\,\prime}(x) or \displaystyle df/dx(x).

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 (\displaystyle y) is changing per unit increase in the variable (\displaystyle x). 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

\displaystyle \frac{\Delta y}{\Delta x}= \frac{\text{increment in @(i)y@(/i)}}{\text{increment in@(i)x@(/i)}}

Example 1

The linear functions \displaystyle f(x)=x and \displaystyle g(x)=-2x change by the same amount everywhere. Their rates of change are \displaystyle 1 and. \displaystyle −2, which are the gradients of these straight lines.



[Image]

[Image]

Graph of f(x) = x has gradient 1. Graph of g(x) = - 2x has gradient - 2.


Thus for a linear function the rate of change is the same as the gradient.

If a car is moving at a speed of 80 km/h then the distance traveled, s km, after t hours is given by the function \displaystyle s(t)=80 t. 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 km/h.

For non-linear functions however, the gradient of the graph of the function (that is, the function's rate of change) varies from point to point. We can either give the function's average (mean) rate of change between two points on the curve of the function, or the instantaneous rate of change at one point on the curve. The mean rate of change is fairly easy to calculate; how to calculate the instantaneous rate of change forms the main focus of this section.

Example 2

For the function \displaystyle f(x)=4x-x^2 one has \displaystyle f(1)=3, \displaystyle f(2)=4 and \displaystyle f(4)=0.

  1. Mean rate of change (mean gradient) from \displaystyle x = 1 to \displaystyle x = 2 is
    \displaystyle \frac{\Delta y}{\Delta x}= \frac{f(2)-f(1)}{2-1} 
     = \frac{4-3}{1}=1\,\mbox{,}
    and the function increases in this interval.
  2. Mean rate of change from \displaystyle x = 2 to \displaystyle x = 4 is
    \displaystyle \frac{\Delta y}{\Delta x}= \frac{f(4)-f(2)}{4-2} 
     = \frac{0-4}{2}=-2\,\mbox{,}
    and the function decreases in this interval.
  3. Between \displaystyle x = 1 and \displaystyle x = 4 the mean rate of change is
    \displaystyle \frac{\Delta y}{\Delta x} = \frac{f(4)-f(1)}{4-1} 
     = \frac{0-3}{3} = -1\,\mbox{.}
    On average, the function decreases in this interval, although the function both increases and decreases within this interval.t.

[Image]

[Image]

Between x = 1 and x = 2 the function has the mean rate of change 1/1 = 1. Between x = 1 and x = 4 the function has the mean rate of change (-3)/3 = -1.



Definition of the derivative

To calculate the instantaneous rate of change of a function, that is, the gradient 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:

[Image]

Increment ratio

\displaystyle \frac{\Delta y}{\Delta x} 
 = \frac{f(x+h)-f(x)}{(x+h)-x}= \frac{f(x+h)-f(x)}{h}\,\mbox{.}


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

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

The derivative of a function \displaystyle f(x), is defined as

\displaystyle f^{\,\prime}(x) 
 = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \,\mbox{.}

If \displaystyle f^{\,\prime}(x_0) exists, one says that \displaystyle f(x) is differentiable at the point \displaystyle x=x_0.

Different notations for the derivative are used, for example,

Function derivative
\displaystyle f(x) \displaystyle f^{\,\prime}(x)
\displaystyle y \displaystyle y^{\,\prime}
\displaystyle y \displaystyle Dy
\displaystyle y \displaystyle \dfrac{dy}{dx}
\displaystyle s(t) \displaystyle \dot s(t)


The sign of the derivative

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

  • \displaystyle f^{\,\prime}(x) > 0 (positive gradient) means that \displaystyle f(x) is increasing.
  • \displaystyle f^{\,\prime}(x) < 0 (negative gradient) means that \displaystyle f(x) is decreasing.
  • \displaystyle f^{\,\prime}(x) = 0 (gradient zero) means that \displaystyle f(x) is stationary (horizontal).


Example 3

  1. \displaystyle f(2)=3\ means that the value of the function is \displaystyle 3 at \displaystyle x=2.
  2. \displaystyle f^{\,\prime}(2)=3\ means that the value of the derivative is \displaystyle 3 when \displaystyle x=2, which in turn means that the function's graph has a gradient \displaystyle 3 at \displaystyle x=2.

Example 4

From the figure one can obtain that

\displaystyle \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*}

[Image]

Note the different meanings of \displaystyle f(x) and \displaystyle f^{\,\prime}(x).

Example 5

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

  1. After 10 minutes the temperature is 80°.

    \displaystyle T(10)=80

  2. After 2 minutes the temperature is dropping in the thermos by 3° per minute

    \displaystyle T'(2)=-3 (the temperature is decreasing, which is why the derivative is negative)

Example 6

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

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

[Image]

graph of the function f(x) = |x|


Differentiation rules

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


Example 7

If \displaystyle f(x)=x^2 then, according to the definition of the increment ratio

\displaystyle \frac{(x+h)^2-x^2}{h}=\frac{x^2+2hx+h^2-x^2}{h} 
 = \frac{h(2x+h)}{h} = 2x + h\,\mbox{.}

If we then let \displaystyle h go to zero, we see that the gradient at the point becomes \displaystyle 2x. We have thus shown that the gradient of an arbitrary point on the curve \displaystyle y=x^2 is \displaystyle 2x. That is, the derivative of \displaystyle x^2 is \displaystyle 2x.

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

Function Derivative
\displaystyle x^n \displaystyle nx^{n-1}
\displaystyle \ln x \displaystyle 1/x
\displaystyle e^x \displaystyle e^x
\displaystyle \sin x \displaystyle \cos x
\displaystyle \cos x \displaystyle -\sin x
\displaystyle \tan x \displaystyle 1/\cos^2 x


In addition, for sums and differences of expressions of functions one has

\displaystyle D(f(x) +g(x)) 
 = f^{\,\prime}(x) + g'(x)\,\mbox{.}

Additionally, if k is a constant, then

\displaystyle D(k \, f(x)) 
 = k \, f^{\,\prime}(x)\,\mbox{.}


Example 8

  1. \displaystyle D(2x^3 - 4x + 10 - \sin x) = 2\,D x^3 - 4\,D x + D 10 - D \sin x
    \displaystyle \phantom{D(2x^3 - 4x + 10 - \sin x)}{} = 2\times 3x^2 - 4\times 1 + 0 - \cos x
  2. \displaystyle y= 3 \ln x + 2e^x \quad gives that \displaystyle \quad y'= 3 \times\frac{1}{x} + 2 e^x = \frac{3}{x} + 2 e^x\,.
  3. \displaystyle \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}\times 2x - \tfrac{1}{2}\times 3x^2 = \tfrac{6}{5}x - \tfrac{3}{2}x^2\,.
  4. \displaystyle s(t)= v_0t + \frac{at^2}{2} \quad gives that \displaystyle \quad s'(t)=v_0 + \frac{2at}{2} = v_0 + at\,.

Example 9

  1. \displaystyle f(x) = \frac{1}{x} = x^{-1} \quad gives that \displaystyle \quad f^{\,\prime}(x) = -1 \times x^{-2} = -\frac{1}{x^2}\,.
  2. \displaystyle f(x)= \frac{1}{3x^2} = \tfrac{1}{3}\,x^{-2} \quad gives that \displaystyle \quad f^{\,\prime}(x) = \tfrac{1}{3}\times(-2)x^{-3} = -\tfrac{2}{3} \times x^{-3} = -\frac{2}{3x^3}\,.
  3. \displaystyle g(t) = \frac{t^2 - 2t + 1}{t} = t -2 + \frac{1}{t} \quad gives that\displaystyle \quad g'(t) = 1 - \frac{1}{t^2}\,.
  4. \displaystyle y = \Bigl( x^2 + \frac{1}{x} \Bigr)^2 = (x^2)^2 + 2 \, x^2 \times \frac{1}{x} + \Bigl(\frac{1}{x} \Bigr)^2 = x^4 + 2x + x^{-2}
    \displaystyle \qquad\quad gives that\displaystyle \quad y' =4x^3 + 2 -2x^{-3} = 4x^3 + 2 - \frac{2}{x^3}\,.

Example 10

The function \displaystyle f(x)=x^2 + x^{-2} has the derivative

\displaystyle f^{\,\prime}(x) = 2x^1 -2x^{-3} 
 = 2x - \frac{2}{x^3}\,\mbox{.}

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

Example 11

An object moves according to \displaystyle s(t) = t^3 -4t^2 +5t, where \displaystyle s(t) km is the distance from the starting point after \displaystyle t hours. Calculate \displaystyle s'(3) and explain what the value stands for.

Differentiating with respect to the time

\displaystyle s'(t) = 3t^2 - 8t +5\qquad 
 \text{which gives that}\qquad s'(3) = 3 \times 3^2 - 8 \times 3 + 5
 = 8\,\mbox{.}

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

Example 12

The total cost \displaystyle T dollars for the manufacture of \displaystyle x objects is given by the function

\displaystyle T(x) = 40000 + 370x -0{,}09x^2, \quad 
 \text{for} \ 0 \le x \le 200\,\mbox{.}

Calculate and explain the meaning of the following expressions.

  1. \displaystyle T(120)

    \displaystyle T(120)=40000 + 370 \times 120 - 0{,}09 \times 120^2 = 83104\,.
    The total cost to manufacture 120 objects is 83104 dollars.
  2. \displaystyle T'(120)

    The derivative is given by \displaystyle T^{\,\prime}(x)= 370 - 0\textrm{.}18x and therefore, is
    \displaystyle T^{\,\prime}(120) = 370 - 0\textrm{.}18 \times 120 
           \approx 348\textrm{.}
    Marginal costs ("the cost to produce an additional 1 object") of 120 manufactured objects is approximately 348 dollars.


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 gradients is \displaystyle –1, i.e. if the tangents gradient is \displaystyle k_T and the normals is \displaystyle k_N then \displaystyle k_T \, k_N = -1. Since we can determine the gradient 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.


Example 13

Determine the equation for the tangent and the normal to the curve \displaystyle y=x^2 + 1 at the point \displaystyle (1,2).

We write the tangents equation as \displaystyle y = kx + m. Since it is to tangent (touch) the curve at \displaystyle x=1 it must have a gradient of \displaystyle k= y'(1), i.e.

\displaystyle y' = 2x,\qquad y'(1) = 2\times 1 = 2.

The tangent also passes through the point \displaystyle (1,2) and therefore \displaystyle (1,2) must satisfy the tangents equation

\displaystyle 2 = 2 \times 1 + m \quad \Leftrightarrow \quad 
 m = 0.

The tangents equation is thus \displaystyle y=2x.


The gradient of the normal is \displaystyle k_N = -\frac{1}{k_T} = -\frac{1}{2} .

In addition, the normal also passes through the point \displaystyle (1, 2) , i.e.

\displaystyle 2= -\frac{1}{2}\times 1 + m 
 \quad \Leftrightarrow \quad m = \frac{5}{2}.

The normal has the equation \displaystyle y= -\frac{x}{2} + \frac{5}{2} = \frac{5-x}{2}.


[Image]

[Image]

Tangent \displaystyle y=2x Normal \displaystyle y=(5-x)/2


Example 14

The curve \displaystyle y = 2 \, e^x - 3x has a tangent with a gradient of \displaystyle –1. Determine the point of tangency (where the tangent touches the curve).

The derivative of the right-hand side is \displaystyle y' = 2 \, e^x -3 and at the point of tangency the derivative must be equal to \displaystyle -1, that is, \displaystyle y' = -1, and this gives us the equation

\displaystyle 2 \, e^x - 3=-1

which has a solution \displaystyle x=0. At the point \displaystyle x=0 the curve has \displaystyle y-value \displaystyle y(0) = 2 \, e^0 - 3 \times 0 = 2 and therefore the point of tangency is \displaystyle (0,2).

[Image]

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