Derivate of a function


Derivates
Lets f a continuous real variabled funcion, ie
$$\begin{matrix} f: \mathbb{R} \rightarrow \mathbb{R}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x \in \mathbb{R} \rightarrow y=f(x) \in\mathbb{R}\end{matrix}$$

We are interested in the instantaneous rate of change of function f in a given point \(x_{0}\). First, we will take an «small increment» in a neighborhood right of the point point \(x_{0}\).
With «small change» of x we means, that we take some small interval \((x_{0}, x_{0}+h)\) and we will denote it as \( \triangle x\). Continuity of f guarantees that any «small change» at x-axis produces a «small change» at y-axis \((f(x_{0}), f(x_{0}+h))\) called \( \triangle y\).

Then we want to take
$$\frac{\triangle y}{\triangle x}=\frac{f(x_{0}+h) – f(x_{0})}{(x_{0}+h)-(x_{0})}=\frac{f(x_{0}+h) – f(x_{0})}{h}$$

And after take the limit when x maps to 0.

In following interactive graphic you can see in fact what are we doing. This time you can drag the point \(x_{0}\) to the point you preffer and slice h to the size what you want too.

Derivate definition
Lets f a real variabled funcion, ie
$$\begin{matrix} f: \mathbb{R} \rightarrow \mathbb{R}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;x \in \mathbb{R} \rightarrow y=f(x) \in\mathbb{R}\end{matrix}$$
It is said that f is differentiable at point \(x_{0}\) if the limit

$$\lim_{h \mapsto 0} \frac{f(x_{0}+h) – f(x_{0})}{h}$$

exists and it is finite. In this case, the value of this limit is called derivate of f at \(x_{0}\) and it is denoted by

$$f'(x_{0})=\left.\begin{matrix}\frac{df(x)}{dx}\end{matrix}\right|_{x_{0}}$$

If f has derivate in all points into any open set \(U \subset \mathbb{R}\), then the function results by calculating \(\frac{df(x)}{dx}\) in all \(x \in U\) is denoted by \(f'(x)\).

Derivate value is, as said, the measure of instantaneous rate of change of function f at any \(x{0}\). But there is an important geometric interpretation for this

If we take straigth rect passing througth the points \((x_{0}, f(x_{0}))\) and \((x_{0}, f(x_{0}+h))\). When h tends to zero, the secant rects becomes, at the limit in the slope of tangent straight line at point \((x_{0}, f(x_{0}))\) (see the interactive graphic up).

Example
Lets
$$f(x)=x^{2}$$
Then
$$f(x+h)=(x+h)^{2} = (x^{2}+h^{2}+2xh)$$
So, we have
$$\frac{f(x+h)-f(x)}{h}=\frac{x^{2}+h^{2}+2xh-x^{2}}{h}=\frac{h^2+2xh}{h}=h+2x$$
We take the limit when h tends to 0
$$\lim_{h \mapsto 0} \frac{f(x+h) – f(x)}{h} = \lim_{h \mapsto h} h+2x = 2x $$
So
$$\frac{d(x^2)}{dx}=2x$$

Derivate is a central concept in math analysis. The establishment of this important concept is owned to Leibnitz and Newton in the seventeenth century. It allowed progress in many branches in different sciences and therefore in engineering, economics, and so on… This is one of those concepts that made our idea of current progress owes all its success. Integral, Fourier Series, differential equations are just some few examples that start from initial concept of derivative.

Example
Lets
$$f(x)=|x|$$
Note that f can be defined as
$$f(x)=\left\{\begin{matrix}x \;\;\;\; x>0\\-x \;\;\;\; x\leq0 \end{matrix}\right.$$
Then
$$f'(x)=\left\{\begin{matrix}1 \;\;\;\; x>0\\-1 \;\;\;\; x\leq0 \end{matrix}\right.$$

Note that f'(x) is not continuous at x=0. Then it can be said that f has not derivate or it is not differentiable at x=0.

0 comentarios

Dejar un comentario

¿Quieres unirte a la conversación?
Siéntete libre de contribuir!

Deja una respuesta