Differentiation

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

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

$${\left[\frac{f(x)}{g(x)}\right]’ = \frac{f’(x) \cdot g(x) - f(x) \cdot g’(x)}{g(x)^2}}$$


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