1.1 Einführung zur Differentialrechnung
Aus Online Mathematik Brückenkurs 2
(Edited for UK idiom and terminology.) |
K |
||
Zeile 25: | Zeile 25: | ||
* 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 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=f(x)</math>. | * To determine the tangent and normal to the curve <math>y=f(x)</math>. | ||
- | * That the derivative can be denoted by <math>f^{\,\prime}(x)</math> | + | * That the derivative can be denoted by <math>f^{\,\prime}(x)</math> or <math>df/dx(x)</math>. |
}} | }} | ||
Version vom 17:48, 6. Jan. 2009
Theory | Exercises |
Contents:
- Definition of the derivative (overview).
- Derivative of
x ,lnx ,ex ,cosx ,sinx andtanx . - Derivative of sums and differences.
- Tangents and normals to curves.
Learning outcomes:
After this section, you will have learned:
- That the first derivative
f is the gradient of the curve(a)
y=f(x) at the pointx=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
f(x)= atx
x=0 ). - To differentiate
x ,lnx ,ex ,cosx ,sinx andtanx as well as the sums / differences of such terms. - To determine the tangent and normal to the curve
y=f(x) . - That the derivative can be denoted by
f or(x)
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 (
![]() ![]() |
Example 1
The linear functions
1.1 - Figure - The graph of f(x) = x | 1.1 - Figure - The graph of f(x) = -2x | |
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
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
- Mean rate of change (mean gradient) from
x=1 tox=2 isx
y=2−1f(2)−f(1)=14−3=1,
- Mean rate of change from
x=2 tox=4 isx
y=4−2f(4)−f(2)=20−4=−2,
- Between
x=1 andx=4 the mean rate of change isx
y=4−1f(4)−f(1)=30−3=−1.
1.1 - Figure - Mean change of f(x) = x(4 - x) between x = 1 and x = 2 | 1.1 - Figure - Mean change of f(x) = x(4 - x) between x = 1 and x = 4 | |
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:
Increment ratio
![]() ![]() |
If we allow Q to approach P (that is allow 0
The derivative of a function (x)
The derivative of a function
![]() ![]() |
If (x0)
Different notations for the derivative are used, for example,
Function | derivative |
---|---|
| ![]() |
| ![]() |
| |
| |
| ![]() |
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:
-
f (positive gradient) means that(x)
0
f(x) is increasing. -
f (negative gradient) means that(x)
0
f(x) is decreasing. -
f (gradient zero) means that(x)=0
f(x) is stationary (horizontal).
Example 3
f(2)=3 means that the value of the function is3 atx=2 .f means that the value of the derivative is(2)=3
3 whenx=2 , which in turn means that the function's graph has a gradient3 atx=2 .
Example 4
From the figure one can obtain that
|
1.1 - Figure - The graph of y = f(x) with points x = a, b, c, d, e and g |
Note the different meanings of (x)
Example 5
The temperature in a thermos is given by a function, where
- After 10 minutes the temperature is 80°.
T(10)=80 - After 2 minutes the temperature is dropping in the thermos by 3° per minute
T (the temperature is decreasing, which is why the derivative is negative)(2)=−3
Example 6
The function x
0)
One can express this, for example, in one of the following ways:"(0)
(0)
Differentiation rules
Using the definition of differentiation one can determine the derivatives for the standard types of functions.
Example 7
If
If we then let
In a similar way, we can deduce general differentiation rules:
Function | Derivative |
---|---|
![]() | |
![]() |
In addition, for sums and differences of expressions of functions one has
![]() ![]() |
Additionally, if k is a constant, then
![]() |
Example 8
D(2x3−4x+10−sinx)=2Dx3−4Dx+D10−Dsinx
=2 3x2−4
1+0−cosx
y=3lnx+2ex gives thaty .=3
x1+2ex=x3+2ex
ddx .53x2−2x3
=ddx
53x2−21x3
=53
2x−21
3x2=56x−23x2
s(t)=v0t+2at2 gives thats .(t)=v0+22at=v0+at
Example 9
f(x)=x1=x−1 gives thatf .(x)=−1
x−2=−1x2
f(x)=13x2=31x−2 gives thatf .(x)=31
(−2)x−3=−32
x−3=−23x3
g(t)=tt2−2t+1=t−2+t1 gives thatg .(t)=1−1t2
y= x2+x1
2=(x2)2+2x2
x1+
x1
2=x4+2x+x−2
gives that y .=4x3+2−2x−3=4x3+2−2x3
Example 10
The function
![]() |
This means, for example, that (2)=2
2−2
23=4−41=415
(−1)=2
(−1)−2
(−1)3=−2+2=0
(0)
Example 11
An object moves according to (3)
Differentiating with respect to the time
![]() ![]() ![]() ![]() |
This might suggest that after 3 hours the object's speed is 8 km/h.
Example 12
The total cost
![]() ![]() ![]() ![]() |
Calculate and explain the meaning of the following expressions.
T(120)
T(120)=40000+370 .120−0
09
1202=83104
The total cost to manufacture 120 objects is 83104 dollars.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{.}
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}.
1.1 - Figure - The tangent y = 2x | 1.1 - Figure - The normal y = (5 - x)/2 | |
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
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). | 1.1 - Figure - The curve y = 2e^x - 3x and its tangent through (0,2) |