sin(x)
lim ------- = 1
x->0 x
|
tan(x) sin(x) sin(x) 1
lim ------- = lim --------- = lim ------ .------- = 1.1 = 1
0 x 0 x.cos(x) x cos(x)
tan(x)
lim ------- = 1
x->0 x
|
lim ( cot(x) - 1/x) = lim ( 1/tan(x) - 1/x)
0 0
1 - tan(x)/x 1 - 1
= lim (--------------) = ------- = 0
0 tan(x)/x 1
(f(a+h)-f(a))
The slope of PQ is ---------------
h
This slope can be viewed as the mean slope of G in the
curve segment [PQ].
Now, let h become smaller and smaller. Then the line PQ
approaches to the tangent line to the curve at point P.
If the following limit exists and if it is real, then
(f(a+h)-f(a))
the slope of that tangent line = lim --------------
h->0 h
The last limit is called the derivative of f(x) at point P.
This value is noted f'(a).
the slope of the tangent line in point P(x,f(x))
(f(x+h)-f(x))
= lim ---------------
h->0 h
= the derivative of f(x)
= f'(x) (notation)
df
= --- (notation)
dx
d
= --- f(x) (notation)
dx
If, for x = b, this limit is not a real number, then we say
that the derivative does not exist for x = b,
or that f(x) is not differentiable for x = b.
Proof:
Since f(x) is differentiable for x = b, we have
(f(b+h)-f(b))
lim --------------- = a real number = f'(b)
h->0 h
Now,
f(b+h) - f(b)
lim f(b+h) = lim ( ---------------- .h + f(b) )
h->0 h->0 h
f(b+h) - f(b)
lim ( ---------------- .h ) + f(b)
= h->0 h
= f'(b) .0 + f(b)
= f(b)
Let x = b+h ; if h ->0 then x -> b
lim f(x) = f(b)
x->b
And f(x) is continuous in b.
d
-- c = 0
dx
|
d
-- x = 1
dx
|
d f(x+h)+g(x+h) -(f(x)+g(x))
-- (f(x)+g(x)) = lim -----------------------------
dx h->0 h
(f(x+h)-f(x)) + (g(x+h)-g(x))
= lim -----------------------------
h->0 h
(f(x+h)-f(x)) (g(x+h)-g(x))
= lim --------------- + lim ---------------
h->0 h h->0 h
d d
= -- f(x) + -- g(x)
dx dx
d d d -- (f(x)+g(x)) = -- f(x) + -- g(x) = f'(x) + g'(x) dx dx dx |
d f(x+h).g(x+h) - f(x).g(x)
-- (f(x).g(x)) = lim ---------------------------
dx h->0 h
f(x+h).g(x+h) -f(x+h)g(x) + f(x+h)g(x) - f(x).g(x)
= lim --------------------------------------------------
h->0 h
(g(x+h)-g(x)) (f(x+h)-f(x))
= lim (f(x+h). --------------- + g(x). --------------)
h->0 h h
= f(x).g'(x) + f'(x).g(x)
d -- (f(x).g(x)) = f(x).g'(x) + f'(x).g(x) dx |
d --(u.v.w) = u'.v.w + u.v'.w + u.v.w' dx |
d --(c.u) = c.u' dx |
d n d n-1 --( u ) = --(u.u.u. ... .u) = n.(u ).u' dx dx
d --( un ) = n.(un-1).u' dx |
d f(x+h)/g(x+h) -(f(x)/g(x))
-- (f(x)/g(x)) = lim -----------------------------
dx h->0 h
f(x+h).g(x) - f(x).g(x+h)
= lim -----------------------------
h->0 h.g(x).g(x+h)
f(x+h).g(x) -f(x)g(x) +f(x)g(x) - f(x).g(x+h)
= lim ---------------------------------------------
h->0 h.g(x).g(x+h)
g(x) . (f(x+h)-f(x))/h - f(x) . (g(x+h)-g(x))/h
= lim ------------------------------------------------
h->0 g(x).g(x+h)
g(x).f'(x) - f(x).g'(x)
= -------------------------
g(x).g(x)
(u / v)' = ( v u' - u v') / v2
|
n (n-1)
d -n d 1 u . 0 - n.u .u' (-n-1)
--( u ) = -- (---) = --------------------- = -n.u .u'
dx dx n 2n
u u
From this it follows that the formula
d n n-1 --( u ) = n.(u ).u' dxholds for all integer values of n.
d sin(x+h) - sin(x)
--sin(x) = lim ------------------
dx h->0 h
2.cos(x + h/2).sin(h/2)
= lim ------------------------
h->0 h
cos(x + h/2).sin(h/2)
= lim ------------------------
h->0 h/2
sin(h/2)
= lim cos(x + h/2).--------
h->0 h/2
= cos(x)
d --cos(x) = - sin(x) dx
d d sin(x) cos2(x) - sin(x).(-sin(x))
--tan(x) = --(------) = -----------------------------
dx dx cos(x) cos2(x)
1
= ---------
cos2(x)
d -1 --cot(x) = --------- dx sin2(x)
d f(g(x+h)) - f(g(x))
--(f(g(x)) = lim ------------------
dx h->0 h
f(g(x+h)) - f(g(x)) g(x+h) - g(x)
= lim --------------------. ----------------
h->0 g(x+h) - g(x) h
let g(x+h) = g(x) + k = u + k
f(u + k) - f(u) g(x+h) - g(x)
= lim -------------------. ----------------
h->0 k h
if h -> 0 then k -> 0
f(u + k) - f(u) g(x+h) - g(x)
= lim -------------------. lim ----------------
k->0 k h->0 h
d d
= -- f(u) . -- g(x)
du dx
We can also write
d d -- f(u) = -- f(u) . u' dx duThis formula is known as the chain rule. |
d d --sin(u) = -- sin(u) . u' = cos(u) . u' dx du d d --cos(u) = -- cos(u) . u' = - sin(u) . u' dx du d d 1 --tan(u) = -- tan(u) . u' = ---------- . u' dx du cos2(u) d d 1 --cot(u) = -- cot(u) . u' = - -------- . u' dx du sin2(u) d --(un ) = n . (un-1).u' dx ...
d --sin(u) = cos(u) . u' dx d --cos(u) = - sin(u) . u' dx d 1 --tan(u) = ---------- . u' dx cos2(u) d 1 --cot(u) = - -------- . u' dx s in2(u) d --(un ) = n . (un-1).u' dx |
d
cos( arcsin(x) ). --( arcsin(x) ) = 1 (2)
dx
But let b = arcsin(x). Then b is in [-pi/2,+pi/2] .
_____________ _______
| 2 | 2
cos(b) = \| 1 - sin (b) = \| 1 - x
________
| 2
Thus, cos( arcsin(x) ) = \| 1 - x
From (2) we can write now :
________
| 2 d
\| 1 - x . --( arcsin(x) ) = 1
dx
d 1
<=> --( arcsin(x) ) = -------------
dx _______
| 2
\| 1 - x
and with the chain rule :
d 1
--( arcsin(u) ) = -------------.u'
dx _______
| 2
\| 1 - u
|
d - 1
--( arccos(u) ) = -------------.u'
dx _______
| 2
\| 1 - u
d 1
--( arctan(u) ) = -------------.u'
dx 1 + x2
d - 1
--( arccot(u) ) = -------------.u'
dx 1 + x2
|
(x(1/n))n= x
So, these two functions must have the same derivative.
(1/n) n-1 d (1/n)
n.(x ) . -- (x ) = 1
dx
<=> ...
d (1/n) 1 1/n - 1
<=> -- (x ) = ---. (x )
dx n
d (1/n) 1 1/n - 1
-- (u ) = ---. (u ).u'
dx n
d (m/n) d (1/n) m (1/n) m-1 1 1/n - 1
-- (x ) = -- (x ) = m.(x ) . --. (x )
dx dx n
<=> ...
m
<=> = --- . xm/n - 1
n
d (m/n) m
-- (u ) = --- . um/n - 1.u'
dx n
We write f(x,y,z).
Consider, for a moment, in such function f(x,y,z) y and z as constant,
then f(x,y,z) only depends on the variable x. Then, we can
calculate the derivative with respect to x.
We note this derivative as
fx'(x,y,z)
Similarly we can calculate
fy'(x,y,z) and fz'(x,y,z)
f(x,y,z) = 3 x y - x z + 6 y -4
fx'(x,y,z) = 3 y - z
fy'(x,y,z) = 3 x + 6
fz'(x,y,z) = - x
Suppose that x, y and z are differentiable functions of t. We denote the three derivatives for short as x',y' and z'. We can calculate the derivative of f with respect to t with the formula
d
--- f(x,y,z) = fx'(x,y,z) . x' + fy'(x,y,z) . y' + fz'(x,y,z) . z'
dt
|
f(x,y,z) = x2 + x.y + z2 + y.z
and x = 3t ; y = 5 t2 ; z = t3
Then
d
--- f(x,y,z) = (2x + y).3 + (x + z). 10t + (2z + y). 3t2
dt
= (6t + 5t2).3 + (3t + t3) .10t + (2t3 + 5t2).3t2
= ...
= 6 t5 + 25 t4 + 45 t2 + 18 t
x + 6
y = -----
x - 5
If x is not 5, the same function is defined by one the implicit forms
y.(x - 5) = x + 6 or x y = x + 5 y + 6 or x y - x - 5 y - 6 = 0
g(x,y) = 0
or g(x,y) = h(x,y)
can define y implicitly as a function of x.
g(x,y) = h(x,y)
If we think y as a function of x, the previous form expresses
an identity for all x-values.d g(x, y) d h(x, y) --------- = --------- dx dxThis defines implicitly y'.
x y = x + 5 y + 6
defines y implicitly as a function of x.
1.y + x.y' = 1 + 5 y' + 0
<=>
(x-5) y' = 1 - y
<=>
1 - y
y' = -----
x - 5
Previous procedure to calculate y' is called implicit differentiation
g(x,y) = h(x,y)
defines two or more different y values as a function of x.2 y = 4x defines two different y values as a function of x.Thinking of y as a function of x, we obtain
2 y y' = 4
<=>
y' = 2/y
This result holds for the two different functions!!
|
If f'(x) exists in an open interval that contains the value x = t,
and suppose that f reaches a maximum for x = t.
Then, f'(t) = 0. |
f'(x) exists for x = t
(f(t+h)-f(t))
=> lim --------------- = a number g
h->0 h
We consider the right and left limit separately
(f(t+h)-f(t)) (f(t+h)-f(t))
=> lim --------------- = g and lim --------------- = g
> 0 h < 0 h
Since f(x) reaches a maximum for x = t , we have
=> ( g > or = 0 ) and ( g < or = 0 )
=> g = 0
=> f'(t) = 0
In the same way we can prove :
|
If f'(x) exists in an open interval that contains the value x = t,
and suppose that f reaches a minimum for x = t.
Then, f'(t) = 0. |
If
|
Now, suppose f is not constant in [a,b].
Then there is a number d in ]a,b[ such that f(d) is different from f(a) and f(b).
Suppose first that f(d) > f(a). Since f is continuous in [a,b], f attains, according to Weierstrass, a greatest image.
Thus, there is a c-value in ]a,b[ such that f(c) is maximum and since f is differentiable in ]a,b[, f'(c) = 0.
In the same way, you can prove the theorem for f(d) < f(a).
If
f(b) - f(a)
f'(c) = ---------------
b - a
|
C is the graph of y = f(x).
The slope of PQ is ( f(b) - f(a) ) / (b - a).
f'(c) is the slope of the tangent line in point R(c,f(c)).
The theorem says that there is a suitable point R on the curve C
such that the tangent line in R is parallel to PQ.
Proof:
The equation of PQ is
f(b) - f(a)
y - f(a) = -------------- (x - a)
b - a
<=>
f(b) - f(a)
y = f(a) + -------------- (x - a)
b - a
Now, we consider three functions
f : x --> y = f(x)
f(b) - f(a)
g : x --> y = f(a) + -------------- (x - a)
b - a
h : x --> y = f(x) - g(x)
It is not difficult to verify that the h(x) satisfies the three
conditions of Rolle's theorem.
We apply Rolle's theorem on the function h(x). There is a c-value in ]a,b[ such that h'(c) = 0;
h'(x) = f'(x) - g'(x)
f(b) - f(a)
= f'(x) - 0 - ------------- .1
b - a
There is a c-value in ]a,b[ such that
f(b) - f(a)
f'(c) - ------------- = 0
b - a
<=>
f(b) - f(a)
f'(c) = ------------
b - a
Proof:
Take two values c and d in [a,b]. We'll show that f(c) = f(d).
Since f'(x) exists for each x in [a,b], f'(x) exists in [c,d]
and f(x) is continuous in [c,d].
We apply lagrange's-theorem in [c,d].
There is a value e in ]c,d[ such that
f(c) - f(d)
f'(e) = -------------
c - d
but f'(x) = 0 for all x in [c,d]. Thus,
f(c) - f(d)
------------- = 0
c - d
and f(c) = f(d)
Proof:
Take two values c < d in [a,b]. We'll show that f(c) < f(d).
Since f'(x) exists for each x in [a,b], f'(x) exists in [c,d]
and f(x) is continuous in [c,d].
We apply lagrange's-theorem in [c,d].
There is a value e in ]c,d[ such that
f(c) - f(d)
f'(e) = -------------
c - d
but f'(x) > 0 for all x in [c,d]. Thus,
f(c) - f(d)
------------- > 0
c - d
Since c < d , we have f(c) < f(d).
(proof similar to previous theorem)
If f'(x) > 0 in a suitable interval ]t-e,t[ , the f(x) is increasing at
the left side of t.
If f'(x) < 0 in a suitable interval ]t,t+e[ , the f(x) is decreasing at
the right side of t.
In that case f(x) has a relative maximum for x = t.
If f'(x) < 0 in a suitable interval ]t-e,t[ , the f(x) is decreasing at the left side of t.
If f'(x) > 0 in a suitable interval ]t,t+e[ , the f(x) is increasing at the right side of t.
In that case f(x) has a relative minimum for x = t.
|
f(x) has a maximum = f(t) if f'(x) changes from + to - for x=t.
f(x) has a minimum = f(t) if f'(x) changes from - to + for x=t. |
f(x) = 3x5- 5x3 then f'(x) = 15x4-15x2= 15 x2.(x-1)(x+1)
Investigation of the sign gives
x | -1 0 1
---------------------------------------
f'(x) | + - - +
So there is a maximum for x = -1 and a minimum for x = 1
| If, in an interval, f"(x) > 0 then f(x) is concave upward in that interval. |
| If, in an interval, f"(x) < 0 then f(x) is concave downward in that interval. |
f(x) = 3x5- 5x3 then f"(x) = 15(x4 - x2) = 30x(2x2- 1)
Investigation of the sign gives
x | -sqrt(1/2) 0 sqrt(1/2)
----------------------------------------------------
f"(x) | - + - +
So,
for x < -sqrt(1/2) the graph is concave downward.
for -sqrt(1/2) < x < 0 the graph is concave upward.
for 0 < x < sqrt(1/2) the graph is concave downward.
for x > sqrt(1/2) the graph is concave upward.
The points where the concavity changes sign are the points with
x = -sqrt(1/2) ; x = 0 ; x = sqrt(1/2).
These points are called points of inflection.
| The points where the concavity changes sign are the points of inflection. |
If f(x) and g(x) are continious in [a,b] and if f'(x) and g'(x) exist
in ]a,b[ and have no common roots in ]a,b[ , then there is a value c
in ]a,b[ such that
f(b) - f(a) f'(c)
------------- = -------
g(b) - g(a) g'(c)
|
Denote
f(b) - f(a)
------------- = K (1)
g(b) - g(a)
Then f(b) - f(a) - K (g(b) - g(a)) = 0
We create the function f(x) - K g(x).There is a value c in ]a,b[ such that
f(b) - K g(b) - ( f(a) - K g(a) ) = (b-a).(f'(c) - K g'(c))
=> 0 = (f'(c) - K g'(c))
If g'(c) = 0 then f'(c) = 0 and this is impossible because f'(x) and g'(x)
have no common roots in ]a,b[.
So, g'(c) is not 0 and then
f'(c)
------ = K (2)
g'(c)
From (1) and (2), we have that
there is a value c
in ]a,b[ such that
f(b) - f(a) f'(c)
------------- = -------
g(b) - g(a) g'(c)
If ( lim f(x) = lim g(x) = 0 )
a a
f'(x)
and lim ------- = can be found
a g'(x)
Then
f(x) f'(x)
lim ------ = lim -------
a g(x) a g'(x)
|
f'(x)
First, suppose that lim ------- = finite = A
a g'(x)
Since f'(x) exists in the environment of a, f(x) is continuous in the
environment of a and
lim f(x) = f(a)
a
But we have also that
lim f(x) = 0
a
Therefore f(a) = 0.
Similarly g(a) = 0.
f(x) f(x) - f(a) f'(c)
----- = ------------- = ------- with c between x and a.
g(x) g(x) - g(a) g'(c)
If x --> a , c --> a.
f(x) f'(c)
lim ------ = lim ------- = A
a g(x) a g'(c)
And, in this first case, the theorem is proved.
f'(x)
Now, suppose that lim ------- = + infinity
a g'(x)
Then
g'(x)
lim ------- = +0
a f'(x)
and appealing on the first case
g(x) g'(c)
lim ------ = lim ------- = +0
a f(x) a f'(c)
So,
f(x) f'(c)
lim ------ = lim ------- = + infinity
a g(x) a g'(c)
(similar for - infinity)
If ( lim f(x) = lim g(x) = infinite )
a a
f'(x)
and lim ------- = can be found
a g'(x)
Then
f(x) f'(x)
lim ------ = lim -------
a g(x) a g'(x)
|
f'(x)
First, suppose that lim ------- = finite = A
a g'(x)
Consider the identity for all x
f(x) f(x) - f(b) 1 - g(b)/g(x)
----- = -------------. ---------------
g(x) g(x) - g(b) 1 - f(b)/f(x)
Since f'(x) exists in the environment of a, f(x) is continuous in the
environment of a and similarly for g(x).
With Cauchy's theorem, previous identity becomes, with b and x in the same environment of a :
f(x) f'(c) 1 - g(b)/g(x)
----- = ------- . --------------- with c between x and a.
g(x) g'(c) 1 - f(b)/f(x)
If x --> a , c --> a.
f(x) f'(c) 1 - 0
lim ------ = lim -------. ------ = A
a g(x) a g'(c) 1 - 0
And, in this first case, the theorem is proved.
f'(x)
Now, suppose that lim ------- = + infinity
a g'(x)
Then
g'(x)
lim ------- = +0
a f'(x)
and appealing on the first case
g(x) g'(c)
lim ------ = lim ------- = +0
a f(x) a f'(c)
So,
f(x) f'(c)
lim ------ = lim ------- = + infinity
a g(x) a g'(c)
(similar for - infinity)
cos(2x) - 1 -2 sin(2x)
lim ------------ = lim ------------ = 0
0 sin(3x) 0 3 cos(3x)
cos(x)-cos(2x) - sin(x) + 2 sin(2x)
lim -------------- = lim ---------------------
0 cos(x)-cos(3x) 0 - sin(x) + 3 sin(3x)
- cos(x) + 4cos(2x) 3
= lim -------------------- = ---
0 - cos(x) + 9cos(3x) 8
ln(x) 1/x
lim ------- = lim ------- = 0
infty x infty 1
xn n.xn-1
lim ------ = lim ---------
infty ex ex
n.(n-1)xn-2
= lim --------------
ex
= ...
n!
= lim ------ = 0
ex
ln(x) 1/x
lim x. ln(x) = lim ------ = lim ------ = - lim x = 0
0 1/x -1/x2
lim xx = lim ex.ln(x) = elim x.ln(x)
0 0
and from previous example
= e0 = 1
lim (cos(x))1/x
0
= lim e(1/x).ln(cos(x))
0
= elim (1/x).ln(cos(x))
but lim (1/x).ln(cos(x))
0
ln(cos(x))
= lim ---------------
0 x
-sin(x)/cos(x)
= lim ---------------- = 0
0 1
So lim (cos(x))1/x = e0 = 1
0
You can find solved problems about differentiation and
limits
here and about maximum, minimum
and inflection points
here