1.1 Einführung zur Differentialrechnung
Aus Online Mathematik Brückenkurs 2
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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. | * 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>. |
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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)= at
xx=0 ). - To differentiate
x ,lnx ,ex ,cosx ,sinx 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.
| 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 \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.
- Mean rate of change (mean gradient) from \displaystyle x = 1
to \displaystyle x = 2 is
and the function increases in this interval.\displaystyle \frac{\Delta y}{\Delta x}= \frac{f(2)-f(1)}{2-1} = \frac{4-3}{1}=1\,\mbox{,} - Mean rate of change from \displaystyle x = 2 to \displaystyle x = 4 is
and the function decreases in this interval.\displaystyle \frac{\Delta y}{\Delta x}= \frac{f(4)-f(2)}{4-2} = \frac{0-4}{2}=-2\,\mbox{,} - Between \displaystyle x = 1 and \displaystyle x = 4 the mean rate of change is
On average, the function decreases in this interval, although the function both increases and decreases within this interval.t.\displaystyle \frac{\Delta y}{\Delta x} = \frac{f(4)-f(1)}{4-1} = \frac{0-3}{3} = -1\,\mbox{.}
| 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
\displaystyle \frac{\Delta y}{\Delta x}
= \frac{f(x+h)-f(x)}{(x+h)-x}= \frac{f(x+h)-f(x)}{h}\,\mbox{.}
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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{.}
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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
- \displaystyle f(2)=3\ means that the value of the function is \displaystyle 3 at \displaystyle x=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
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\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*} |
1.1 - Figure - The graph of y = f(x) with points x = a, b, c, d, e and g |
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:
- After 10 minutes the temperature is 80°.
\displaystyle T(10)=80 - 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".
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{.}
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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{.}
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Additionally, if k is a constant, then
\displaystyle D(k \, f(x))
= k \, f^{\,\prime}(x)\,\mbox{.}
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Example 8
- \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 - \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\,.
- \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\,.
- \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
- \displaystyle f(x) = \frac{1}{x} = x^{-1} \quad gives that \displaystyle \quad f^{\,\prime}(x) = -1 \times x^{-2} = -\frac{1}{x^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}\,.
- \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}\,.
- \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{.}
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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{.}
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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{.}
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Calculate and explain the meaning of the following expressions.
- \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. - \displaystyle T'(120)
The derivative is given by \displaystyle T^{\,\prime}(x)= 370 - 0\textrm{.}18x and therefore, is
Marginal costs ("the cost to produce an additional 1 object") of 120 manufactured objects is approximately 348 dollars.\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}.
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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).
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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) |
