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G. M. Fichtenholz – Integral and differential calculus, vol. 1, PWN Warsaw 1999 (originally in Russian)
Let's consider a function f(x) defined in a given range γ. Next move from the x = x_{0} point with the step Δx > 0 or Δx < 0 such that the new value of x = x_{0} + Δx is still within the range γ. Then the value of the f function changes from f(x_{0}) to f(x_{0} + Δx). If the the limit of the expression
when
exists it is called the derivative of f function over variable x in the x_{0} point.
by Leibniz
by Lagrange
by Cauchy.
Let the function f(x) have in the point x_{0} its finite derivative. If this derivative is
then for x values that are close enough to x_{0} on its right the function f(x) is
and for x values that are close enough to x_{0} on its left the function f(x) is
In other words, one can say that function f(x) in the point x_{0} grows (or decays).
Let's consider the case when f'(x) > 0. From definition of derivative of f(x)
we can find such a surrounding of x_{0}
where
and x ≠ x_{0}. Then we have
such that
Then
thus
on the other hand when
and
we get
thus
what was to be demonstrated (Q. E. D. quod erat demonstrandum).
Let a function f(x) that is defined in a range γ have its minimum or maximum in an inner point p of γ. If in that point both right and left derivative of f over x exist and are finite then
Let the function f have maximum in the point p. Assume that
Then if f'(p) > 0 and if x > p then f(x) > f(p) or if f'(p) < 0 and if x < p then f(x) > f(p). In both cases f(p) cannot be the greatest value of f function in the γ range.
If a function f(x) have in the range [a, b] the finite derivative then this derivative i. e. the function f'(x) takes at least once each value between f'(a) and f'(b).
Let's take a number P from the range (f'(a), f'(b)) such that
We can define a new continuous function g(x)
having, in the range [a, b] the derivative
Because
and g(x) function is continuous, the second Weierstrass theorem implies that must be such a point p where g(x) function takes its maximum value and, in this case, this point is within the [a, b] range i. e. a < p < b. So from the Fermat's theorem we have
and it gives
what was to be demonstrated (Q. E. D. quod erat demonstrandum).
Let's assume that a function f(x):
that
The function f(x) is continuous in the range [a, b] and from the second theorem of Weierstrass has in the range [a, b] maximum M and minimum m. Let's consider two cases:
what was to be demonstrated (Q. E. D. quod erat demonstrandum).
Let's assume that:
then between a and b is a point p such that
for which the following formula is fulfilled
Let's define in the range (a, b) a new function F(x)
The function fulfills all assumptions of the Rolle's theorem. Its derivative
is finite in the range (a, b). And at both ends of [a, b]
The Rolle's theorem implies that within the range (a, b) is a point p such that
Thus we have
and finally
what was to be demonstrated (Q. E. D. quod erat demonstrandum).
Let's assume that a function f(x) is continuous in the range [x_{0}, x_{0} + H] (where H > 0) and has the finite derivative f'(x) for x > x_{0}. If a finite or infinite limit of
exists and equals K then the right derivative in the point x_{0} (because the limit is in x_{0}^{+0}) has the same value.
Let's consider the function
with the derivative
the limit of f'(x) when
equals
and the limit of f'(x) when
equals
It means that the function f(x) has only one-sided derivatives i. e. the right derivative +∞ and the left derivative -∞.
Let's assume that:
then between a and b is a point p such that
for which the following formula is fulfilled
From the Rolle's theorem g(b) ≠ g(a). Let's consider in the range (a, b) a new function F(x) such that
The function fulfills all assumptions of the Rolle's theorem. The function F(x) is continuous in the range [a, b] because in this range are continuous both functions f(x) and g(x). In (a, b) exists the derivative F'(x) and equals
And at both ends of [a, b]
The Rolle's theorem implies that within the range (a, b) is a point p such that
Thus we have
and finally
what was to be demonstrated (Q. E. D. quod erat demonstrandum). This theorem is called also the generalized theorem of average value.
Constant functions.
If y = const then always Δy = 0. It does not depend on the Δx value.
Polynomial functions.
If y = x^{n} where n is a natural number. Let's define Δx then Δy is
thus
and
when Δx → 0
1/x Function
If y = 1/x then y + Δy is
which gives
and finally when Δx → 0
x^{1/2} Function
If y = x^{1/2} (for x > 0) then
then
and finally
when Δx → 0
Power functions.
If y = x^{μ} where μ is a real number. Let's assume x ≠ 0
when Δx → 0
because when Δx/x → 0
Exponential functions.
If y = a^{x} where a > 0, -∞ < x < +∞
when Δx → 0
when a = e
Logarithmic functions.
If y = log_{a} x where 0 < a ≠ 1, 0 < x < +∞
when Δx → 0
In the case of natural logarithm
Trigonometric functions.
Let's consider y = sin(x) then
when Δx → 0
because when Δx → 0
Inverse functions.
Let's consider y = f(x) that has its inverse function and has in the point x_{0} finite and different from 0 derivative f'(x_{0}). Then also the derivative of its inverse function g(y) such that x = g(y) equals 1/f(x_{0}).
when Δx → 0
Ciclometric functions.
Let's consider y = arcsin(x) (-1 < x < 1 and -π/2 < y < π/2). y is the inverse function of x = sin(y). The derivative of x equals
it implies that also the derivative of y exists
the sign is set as plus because cos(y) > 0.
Theorem.
Let's assume that:
or shorter
y = const | y' = 0 |
y = x; | y' = 1 |
y = x^{μ}; | y' = μ x^{μ - 1} |
y = a^{x}; | y' = a^{x}ln a |
y = log^{a}(x); | y' = log^{a}(e)/x |
y = sin(x); | y' = cos(x) |
y = cos(x); | y' = -sin(x) |
y = tg(x); | y' = 1/cos^{2}(x) |
y = ctg(x); | y' = -1/sin^{2}(x) |
y = arcsin(x); | y' = 1/[1 - x^{2}]^{1/2} |
y = arccos(x); | y' = -1/[1 - x^{2}]^{1/2} |
y = arctg(x); | y' = 1/[1 + x^{2}] |
Let a function u = f(x) has in the point x the derivative u'.
then the function z = const u has also its derivative in the point x
Let a function v = g(x) has in the point x the derivative v'. Then the function y = u ± v has also the derivative in that point that equals
and
finally in the limit when Δx → 0
Let a function v = g(x) has in the point x the derivative v'. Then the function y = uv has also the derivative in that point that equals
and
finally in the limit when Δx → 0 also Δv → 0
Let a function v = g(x) differ from 0 and has in the point x the derivative v'. Then the function y = u/v has also the derivative in that point that equals
and
finally in the limit when Δx → 0 also Δv → 0
Presented-below examples only show the way of finding and presenting solutions.
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