Derivativity is key notion in analysis, because it underlies all physics and science in general.
The ideo of derivativity of a function \( f \) is expressed as follows:
$$ f'(x) = lim_{h \to 0} \enspace \frac{f(x+h) - f(x)}{h} $$
The is the limit of the rate variation when \( h \to 0 \).
It can also be found in this form:
$$ f'(x) = lim_{x \to a} \enspace \frac{f(x) - f(a)}{x - a} $$
At this point, it will be the limit of rate variation when \( x \to a \).
We then say that \( f \) is derivable at a point \( (x = a) \) if and only if \( f'(a) \), the derived number from \( f \) at point \( a \) is a real number, so:
$$ f \ derivable \ at \ a \Longleftrightarrow lim_{h \to 0} \enspace \frac{f(x+h) - f(x)}{h}= f'(a) \in \mathbb{R} $$
Derivativity implies continuity
$$ f \ derivable \ at \ a \Longrightarrow f \ continuous \ at \ a $$
The sign of derivative indicates the variation
$$ \forall x \in [a,b], \ f'(x) \geqslant 0 \ \Longleftrightarrow f \ increasing \ on \ [a,b] $$
$$ \forall x \in [a,b], \ f'(x) \leqslant 0 \ \Longleftrightarrow f \ decreasing \ on \ [a,b] $$
Let \( f :x \longmapsto f(x) \) be a function, continuous on an interval \( [x_0, \ x_0 +h] \) such as the following figure:
Let us mark two points on the abscissa axis, \( x_0 \) et \( x_0 +h \) (\( h \) being a relatively short distance).
Their respective image being \( f(x_0) \) and \( f(x_0 + h) \), we obtain two points: \( A(x_0; f(x_0)) \) and \( B(x_0 + h; f(x_0 + h)) \).
Drawing the straight line joining them, we can calculate a mean variation of this function between \( A \) and \( B \).
We can calculate this slope by the following formula:
$$ m = \frac{ \Delta _y}{\Delta _x}$$
$$ m = \frac{ y_B- y_A}{x_B- x_A}$$
In our case, this gives:
$$ m = \frac{f(x_0+h) - f(x_0)}{ x_0 + h -x_0} $$
So:
$$ m = \frac{f(x_0+h) - f(x_0)}{h} \qquad (1) $$
Let us gradually reduce the distance \( h \) which separates our two points on the abscissa axis, such as the following figure.
We see that the values of \( x_0 \) and \( x_0 + h \) started to get closer, and the line which connects \( A \) and \( B \) begin to draw a tangent to the curve.
In the same way, we will further reduce the distance \( h \), the latter begins to reduce it to \( 0 \).
We now see that our two points \( A \) and \( B \) are almost coincident, and that we obtain an almost perfect tangent to the curve at the point of abscissa \( x_0 \).
By imaginating that \( h \) becomes smaller and smaller approching to \( 0 \), our formula \( (1) \) can be expressed as a limit:
$$ m = lim_{h \to 0} \enspace \frac{f(x_0+h) - f(x_0)}{h} $$
This number \( m\) obtained, for a \( x_0 \) arbitrarily chosen, will be called the derived number of the function \( f \) at point \( x_0 \). It will be noted \( f'(x_0) \).
If this number cannot be calculated, the derivative is not defined at this point \( x_0 \).
We will then say that \( f \) is derivable \( (x = a) \) if and only if \( f'(a) \), the derived numer of the function \( f \) at the point \( a \) is a real number, so:
$$ f \ derivable \ at \ a \Longleftrightarrow lim_{h \to 0} \enspace \frac{f(x+h) - f(x)}{h}= f'(a) \in \mathbb{R} $$
By generalizing it, that is to say for all \( x \), we call \( f' \) the derivative function of the function \( f \).
$$ f'(x) = lim_{h \to 0} \enspace \frac{f(x+h) - f(x)}{h} $$
The definition set of \( f' \) will then depend of its expression, and will be restricted to the definition set of the function \( \textcolor{#A65757}f \).
For example, the \( arccosh(x) \) function is only defined on \( I= [1, \hspace{0.1em} +\infty[\).
Then, its derivative function:
$$ arccosh(x)' = \frac{1}{\sqrt{x^2 -1}} $$
is also restricted to this interval, whereas the function \( f: x \longmapsto \frac{1}{\sqrt{x^2 -1}} \) is usually defined on the larger interval: \( J= ]-\infty , \hspace{0.1em} -1] \cup [1, \hspace{0.1em} +\infty[\).
We say that the derivative is the limit of the variation rate when \( h \) goes to \( 0 \).
We will also find it through this form:
$$ f'(x) = lim_{x \to a} \enspace \frac{f(x) - f(a)}{x - a} $$
At this stage, it will be the limit of the variation rate when \( x \to a \).
In physics, we can also use Leibniz's differential notation \( \frac{df}{dx} \), or that of Newton \( \overset{.}{f} \).
Especially for integral calculus, it is convenient to use Leibniz's.
We saw above that if a function can be derivated at point \( a\), tehen:
$$ lim_{h \to 0} \enspace \frac{f(a+h) - f(a)}{h}= f'(a) \in \mathbb{R} $$
And as a result,
$$ lim_{h \to 0} \ f(a+h) = hf'(a) + f(a) $$
$$ lim_{h \to 0} \ f(a+h) = f(a)$$
Which implies a continuity of the function \( f \) at point \( x = a\).
$$ f \ derivable \ at \ a \Longrightarrow f \ continuous \ at \ a $$
Let \(f\) be a positive continuous function on \([a,b]\), and derivable on \( \hspace{0.1em} ]a,b[\).
Now let \( (x_1, x_2) \in \hspace{0.1em} ]a,b[ \), be two inner points of \( \hspace{0.1em} ]a,b[\) in this order.
According to the mean value theorem:
$$ f \ continuous \ on \ [a,b] \ and \ derivable \ on \ ]a,b[ \ \Longrightarrow \ \exists c \in \hspace{0.05em} ]a, b[, \ f'(c) = \frac{ f(b) - f(a)}{b-a}$$
In our case,
$$ \forall (x_1, x_2) \in \hspace{0.1em} ]a,b[ ,$$
$$ \exists x_3 \in \hspace{0.05em} ]x_1, x_2[, \ f'(x_3) = \frac{ f(x_2) - f(x_1)}{x_2-x_1}$$
The interval \((x_2-x_1)\) being always positive, if the function \(f\) is increasing on \([a,b]\), it is also the case on \(]x_1, x_2[\), and in this case:
$$ f(x_2) - f(x_1) \geqslant 0 \Longleftrightarrow f'(x_3) \geqslant 0 $$
The derivative function \(f'\) will be therefore positive for all \( x \in [a,b]\).
The same reasoning can be applied for a decreasing function.
$$ \forall x \in [a,b], \ f'(x) \geqslant 0 \ \Longleftrightarrow f \ croissante \ sur \ [a,b] $$
$$ \forall x \in [a,b], \ f'(x) \leqslant 0 \ \Longleftrightarrow f \ décroissante \ sur \ [a,b] $$
Let us study the variations of a function \(f\) such as:
$$f(x) = \frac{1}{x} - 2\sqrt{x} $$
This function is only defined on: \( D_f = \ ] 0, +\infty[\).
Calculating its derivative \(f'\), we do have:
$$f(x) = \hspace{0.1em} \underbrace{-\frac{1}{x^2}} _\text{ \( < \hspace{0.2em} 0\)} - \hspace{0.1em} \underbrace{\frac{1}{\sqrt{x}} } _\text{ \( < \hspace{0.2em} 0\)} $$
\(f'(x)\) is always negative on \(D_f\).
Thus, \(f(x)\) will be decreasing on this interval.
$$ x $$ |
$$ 0 $$ |
$$ \dots $$ |
$$ +\infty $$ |
---|---|---|---|
$$ sign \ of \ f' $$ |
$$ \bigl ]-\infty \bigr] $$ |
$$- $$ |
$$ \bigl [ 0^- \bigr] $$ |
$$ variations \ of \ f $$ |
$$ \bigl [+\infty\bigr] $$ |
$$ \searrow $$ |
$$ \bigl ]-\infty \bigr] $$ |
Furthermore:
$$ \Biggl \{ \begin{align*} lim_{x \to 0} \ f(x) = +\infty \\ lim_{x \to +\infty} \ f(x) = -\infty \end{align*} $$
$$ \Biggl \{ \begin{align*} lim_{x \to 0} \ f'(x) = -\infty \\ lim_{x \to +\infty} \ f'(x) = 0^- \end{align*} $$