| Course ID | Course | Professor | Time | Location |
| Introduction to differential calculus | - |
Introduction to differential calculus
General information about functions
The most common abbreviation of the mathematical funktion is f (x). Instead of that, any other letter may stand for f: q (x), s (x), p (x), etc. In physics, Greek letters are often used. The same applies to the variable x. Often it is also written: y, z, t or Greek letters are used. The x in f (x) is any real number from which an associated value of the function is assigned.
Real numbers are all numbers including those with infinite and aperiodic decimal fractions, but without alphabetic numbers (without letters). These include the whole numbers (1, 2, 3 …) and the natural numbers (… -2, -1, 0, 1, 2 …) the rational numbers, with abrupt and periodic decimal fractions (eg ¼, = 0.25, 1/3 = 0.33333333 …) and the irrational numbers with aperiodic and infinite decimal fractions (eg the number Phi 3.14 … ..). The pure definition of real numbers is abstract and is treated in higher mathematics.
From this x, of the f(x) the value of the function is calculated or assigned. This value of the function f (x) may be of any size. For example the value of a funkction may figure in temperature, in velocity, in an amount (e.g., in dollars, $, Euros or pounds, £) in population density or in any other measurable quantity.
The function does not have to include a mathematically derived formula, but usually one is searched for. A function can be displayed in tabular and graphical form.
The function for the currency pair EURO / USD is the function that represents the value of this currency pair in Euro at a certain time. The value is based on the respective market conditions.
A mathematical formula can not be derived here, otherwise there would be no markets.
Here are some mathematical function formulas briefly listed up below:
The polynomial functions
f (x) = ax² + bx + c
We already know this as a quadratic equation and is a 2nd order equation, which was discussed above.
f (x) = ax³ + bx² + cx + d
This is a 3rd degree function. The solutions were presented above as a theorem of Vieta.
Rational functions
f (x) = f1(x) ⁄ f2(x). It should be noted that these functions have definition gaps, the denominator may not be zero, because division by zero is “forbidden”.
f(x) = [f1 (x)] / [f2 (x)] oder
f(x) =1 / f(x) z.B. f(x) = 1/x
Ohter samples of rational functions:
- (x – a)/(b + x)
- 1/[1 + √(1 + ax)]
- 1/x, see picture below: This is a hyperbole
F(x) = 1/x, see picture below
potency functions
f (x) = xª The exponent can be a real number (positive or negative)
f (x) = x¹⁄ª is the inverse function of xª, it is a root function.
The exponential function
f (x) = a exp x: Here, in contrast to the potency function, the variable x is an exponent.
f (x) = e exp x: exponential function with base e, (repeat chapter about Euler)
see picture below
Function of the logarithm naturalis
f(x) = lnx
see picture below
The trigonometric functions
f (x) = sinx,
f (x) = cosx,
f (x) = tanx
f (x) = cotx
(see chapter trigonometry)
Derivation
Please repeat
Before we look at derivation rules, here are the most important terms such as
- Limit
- Differentiability
- Continuity
limit
We take some mathematical function with a variable, x,
and put some fixed number, xo. Infinitely close to xo we choose any number x. (x <0 or x> 0)
Now we put xo and x in the function and divide
f (x) -f (xo) / (x-xo)
xo is fixed, x is variable.
Here is a practical example:
f (x) = x²
This function is a parabola graph
xo is 3 and is fixed.
We choose the number x infinitely close to 3.
We are looking for the limit value: If x = 3, numerator and denominator = 0, this is useless.
What if x is approximately = 3, e.g. x = 2.99?
f (xo) = 3² = 9, f'(x) = 2.99² = 8.9401
f (x) – f ‘(xo) = 9 – 8.9401 = -0.0599
x – xo = 2.99 – 3 = – 0.01
As a limit, we get: f (x) – f (xo) / (x – xo) = -.0599 / 0.01 = 5.99
The same result is obtained even if x = 3.01 With infinite approximation of x to xo one gets closer and closer to f ‘(xo) = 5,999999999 = 6.
Generally formulated:
f (x) has a limit, L, at xo if f (x) is arbitrarily close to L,
for x> xo and x <xo, but x does not equal xo!
One writes: Lim f (x) = L x -> xo (Lim, speak Limes)
or simply:
f (x) -> L for x -> xo
This limit, L, is nothing more than the tangent slope at xo. See the
It is the first derivative of f (x) and is defined as follows:
f ‘(xo) = lim [f (x) – f (xo)] / (x – xo)
x -> xo
Common symbols
for the
1st derivative:
f ‘(x), df / dx, z’, v’
And for higher derivatives, e.g.
2nd Derivative:
f ” (x), df² / dx², z”, v″
Differentiability and continuity
Here only the most essential can be explained.
A function f (x) is differentiable at any point x (0) if the limit f (x) -f (xo) / (x-x0) exists.
Not differentiable: Is a function
• Where she has a definitional gap, eg 1 / x at the point xo = 0
• Where the graph is vertical. Because here xo = xo and this would give division by zero, which is “forbidden”. For the root function f (x) = 2 √x this is the case for xo = 0
• Where the graph has a corner, for example f (x) = ΙxΙ + 2, ΙxΙ between absolute lines.
• which is not steady.
Now back to the limit And to the function f (x) = x²
Now we are already near the derivation rules.
If we use the above formula in order to determine the Limit:
We put in f (x) = x² and f (xo) = xo², so we obtain
x² – xo² / (x – xo)
the binomial Theorem reads:
a² -b² = (a – b) (a + b)
and the limit of f (x) = x² is thus
f ‘(xo) = (x – xo) (x + xo) / (x – xo)
Lim x → xo
we shorten by (x – xo) and get
f ‘(x) = lim (x + xo)
If x approaches xo more and more, one can say that
x + xo = xo + xo = 2xo
this corresponds to the general derivation rule
nx high n-1 or n*xexp (n – 1) = f ‘(x)
Another example:
f (x) = x³
f ‘(x) = (x³ – xo³) / (x – xo)
x3 -xo3 can be broken down into the following factors:
(x – xo) (x² + x*xo + xo²)
We shorten f (x) by (x – xo) and get
x² + x*xo + xo² and can say repeatedly
that as x gets closer and closer to xo, we get closer and closer to the result
xo² + xo² + xo² = 3x².
This also corresponds to the derivative formula:
nx exp n-1
Not all functions can be derived algebraically, because then again and again the division by zero stands in the way.
In the derivation then help mostly numerical methods, such as at
f (x) = e high x and lnx
f (x) = e high x
We write the limit formula
f ‘(x) = (e high x – e high xo) / (x – xo)
and put
x0 = 2 and x = 1.99999,
we now insert these values into the limit equation and get
f ‘(x) = 0.00738 / .00001 = 7.389
e high 2 = 7,389
We perform the same calculation with
xo = 3 and choose x = 2.9999
f ‘(3) = 2.00854 / .00001 = 20,085
e high 3 = 20,085
The derivative
f ‘(x) e high x = e high x
e high x does not obey the derivation rule n * x high n-1
An analogue exercise will now be carried out with
f (x) = lnx
Again we choose xo = 2 and x = 1.99999
And get by using the limit formula:
F ‘(2) = 0.000005 / .00001 = 0.5 = 1/2
The same with xo = 3 and x = 2.99999
F ‘(3) = .000003333 / .00001 = 0.333 = 1/3
As you just noted, the derivative
f ‘(x) of f (x) = ln x = 1 / x
f ‘(x) lnx = 1 / x
Of course, more than 2 examples would have to be calculated numerically.
Special derivations (see list below)
(This list is incomplete)
Applications and examples of special derivation rules
Special derivation rules also apply to the sum rule and the difference rule.
The summands and those that are negative are calculated according to the general derivation rules. Please refer Basics differential calculus
f(x) = 3x² + 4x – 1
f ‘(x) = 6x + 4 – 0 = 6x + 4
Product rule
f(x) = e high x * cosx
f ‘(x) = e high x * cosx + e hoch x * (-sinx)
f ‘(x) = e high x *(cosx – sinx)
Explanation: Derivation of e high x = e high x
Derivation of cosx = -sinx
Quotient rule
f(x) / g(x)
If only the denominator (below the fraction bar) has a variable, here is another general formula:
If f (x) = 1 and g (x) = g (x) that means: f (x) = 1 / g (x) Derivation according to the formulation of the quotient rule (see list above):
f ‘(x) = [0 * g (x) – 1 * g’ (x)] / (g (x))² = – g ‘(x) / (g (x))²
Here is a simple example: f (x) = 1 / x or = x¯¹, f ‘(x) = -1 / x² according to the sum rule
f (x) = 1 / x, g (x) = x and thus g ‘(x) = 1 and g (x)² = x²
-g ‘(x) / (g (x))² = -1 / x² according to the general quotient rule with variables in the denominator.
another example:
f(u) = (u³ – 1)/(u² + 1)
The counter, u³ – 1, is derived as in the difference rule and is 3u². Derivation of the denominator = 2u
f ‘(u) = 3u²(u² + 1) – (u³ – 1)2u / (u² + 1)² = (u high 4 + 3u² + 2u) / (u² + 1)²
Chain rule
f[g(x)]
The chain rule divides the function, f (x), into
- a external function and
- a inner function
external and internal functions are derived.
The rule is then just a
multiplication of Outer derivative with the internal derivative = f ‘[g(x)] * g ‘(x)
the rest is doing exercise.
Let’s show 2 simple examples, which can also be derived by means of sum and difference rule:
f(x) = (x + 2)²
Derivation with sum rule: (derive each summand individually)
f(x) = (x + 2)² = x² + 4x + 4
f ‘(x) = 2x + 4 = 2(x + 2)
Derivation with chain rule:
Outer function f(u) = u² , u = (x + 2)
Outer derivation, f ‘(u) = 2u
Inner funktion, I(x) = x + 2
Inner derivation, I ‘(x) = 1
Outer derivation multiplied with inner derivation, f ‘ (u) * I ‘(x) = 2u*1 = 2u, = 2(x + 2)