differentiation
difference quotient
(Differenzenquotient)
The difference quotient of f where y = f(x) $${ \frac{f(x) - f(x_0)}{x-x_0}}$$
the difference quotient is the gradient of a line passing through the points P(x0,f(x0)) and Q(x, f(x)).
Decreasing the distance of x0 and x to an infinitissimal amount leads to the differential quotient.
the rate of change
The difference quotient is the rate of change between to points. To get the rate of a change at a point, one point approaches the other point which leads to the differential quotient.
differential quotient
definition - differentiability (Differenzierbarkeit)
A function f is differentiable at $${x_0 \in D}$$ if the limit $${\lim\limits_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0} = f’(x_0)}$$ exits. This limit is denoted by $${f’(x_0)}$$ and is called the first derivative of f at x0. f’(x0) is also the slope (gradient, Steigung) of the tangent to the graph of f at x0.
Differentialquotient (1. Ableitung) von f(x) an der Stelle x0
method I
$${\lim\limits_{x \to x_0} \frac{f(x) - f(x_0)}{x-x_0} = f’(x_0)}$$
Another way to write this is using the h-method.
method II
$${\lim\limits_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} = f’(x_0)}$$
where h becomes infinitissimaly small as it approaches 0
example - using the h-method
Calculate the gradient of the function f at x0. Berechnen Sie die Steigung der Funktion f and der Stelle x0. $${f(x) = x^2 + 4x \space , x_0 = 1}$$ Solution: $${f’(1) = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}}$$ $${ = \lim\limits_{h \to 0} \frac{(1+h)^2 + 4(1 + h) - 5}{h}}$$ $${ = \lim\limits_{h \to 0} \frac{1+2h+h^2 + 4 + 4h - 5}{h}}$$ $${ = \lim\limits_{h \to 0} \frac{6h+h^2}{h}}$$ $${ = \lim\limits_{h \to 0} (6+h)}$$ $${ = 6}$$
additional exercise
Find the gradient (Steigung) of the above function $${f(x) = x^2 + 4x \space , x_0 = -1}$$
Also
Find the gradient (Steigung) of the function $${f(x) = 2x^2 + 3x \space , x_0 = 3}$$ see Cornelsen MA-1, page 51, Übung 1e)
the derivative function - 1. Ableitungsfunktion
If every point of the function f has a gradient, we can map each x to a unique gradient and we get a new function that is called the derivative function of f. We write $${f’(x)}$$ It is defined as follows.
$${y’ = f’(x_0) = \lim\limits_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} }$$
or
$${\frac{dy}{dx} = \lim\limits_{\Delta x \to 0} \frac{\Delta y}{\Delta x}}$$
higher derivatives - höhere Ableitungen
A derivative function can also be derived which leads to higher derivatives.
second derivative - 2. Ableitung
$${y’,’ = [f’(x)]’ = f’\thinspace’(x) = \frac{\Delta^{2} y}{\Delta x^{2}} }$$
is often used to determine whether there are changes of curvature in the function (Die zweite Ableitung wird bei der Kurvediskussion oft benützt um Wendepunkte zu bestimmen.)
third derivative - 3. Ableitung
To be sure about whether a point where f’'(x) = 0 is in fact a point of inflexion (Wendepunkt), we need to check that the 3rd derivative is not equal to zero $${f^{\prime\prime\prime}(x) \neq 0}$$
nth derivative - n-te Ableitung
$${y^{n} = [f^{n-1}(x)]’ = f^{n}(x) = \frac{\Delta^{n} y}{\Delta x^{n}} }$$
table of common derivatives
Kurvendiskussion
Applications
- gradient
- tangent to a function f(x) for an x-value
- angle of the tangent
- to determine where two functions meet in one point
- minima and maxima
- point of inflexion Wendepunkt
chain rule - Kettenregel
$${\left[f(g(x))\right]’ = f’(g(x)) \cdot g’(x)}$$
product rule - Produktregel
$${\left[f(x) \cdot g(x)\right]’ = f’(x) \cdot g(x) + f(x) \cdot g’(x)}$$
quotient rule - Quotientenregel
$${\left[\frac{f(x)}{g(x)}\right]’ = \frac{f’(x) \cdot g(x) - f(x) \cdot g’(x)}{g(x)^2}}$$