Calculus
1. Area Problem
A1
A2
A4
A3
A4
A5
A10
…
…
A3
A = lim An = πr2
A = A1 + A2 + A3 + A4
y
n -> ∞
y
y=x2
y
A
0
Volume Problem
x
0
x
0
x
Integral Calculus
Calculus
2. Tangent Problem
Tangent to a curve at a point P. Slope?
y=x2
Q
y
P
k = lim kPQ
P
kPQ
0
y=x2
y
k
x
0
QP
x
Differential Calculus
Integral calculus ↔ Differential calculus
inverse problems
Calculus: deals with limits
Main objects in Calculus:
FUNCTIONS
A function f is a rule that assigns to each element x in a set A (domain of f)
exactly one element f(x) in a set B (range of f).
r
x – independent variable, f(x) – dependent variable.
x1
f(x1)
f(x2)
A(r)=πr2
rule, machine
x
x2
Example:
input
black box
output
f(x)
function
x3
f(x3)
domain A
range B =
all possible values
Graph of a function: { (x,f(x)) | xA}
y
y=x2 1
0
1
x
Some properties:
y
1. Piecewise defined
y=x2
1  x, x  1
f ( x)   2
 x , x  1
y=1-x 1
2. Symmetric
0
x
1
even: f(-x) = f(x), e.g. f(x) = x2, symmetric w.r.t. y-axis
odd: f(-x) = - f(x), e.g. f(x) = x3, symmetric about the origin
y=x2 y
3. Increasing/Decreasing
decreasing on I: if f(x1)>f(x2) for any x1<x2 in I
f(x2)
increasing on I: if f(x1)<f(x2) for any x1<x2 in I
f(x1)
4. Periodic
increasing on [0,∞)
decreasing on (∞,0]
with period T, if f(x+T) = f(x), e.g. cos(x+2π) = cos(x)
0
x1
x2
x
Some basic functions:
Linear: f(x) = kx+b, graph is a line with slope k and y-intersect b, grow at constant rate
Polynomial: f(x) = anxn+an-1xn-1+…+a1x+a0 , n ≥ 0 integer, coefficients ai – constants, i=0..n
if an ≠ 0 then n is called degree of polynomial, domain (−∞,∞)
y
Power: f(x) = xa, a – constant. n > 0 integer
n
if a = n  polynomial (i.e. x );
if a = 1/n  root (i.e. n x );
1
if a = -n  reciprocal (i.e. n )
x
Rational: f(x) = P(x)/Q(x) – ratio of two polynomials,
domain: such x that Q(x) ≠ 0
3
x  3x  1
(e.g. f ( x) 
, domain x ≠  2)
2
x 4
Algebraic: algebraic operations on polynomials (i.e. + , − ,  ,  , )
(e.g. f ( x)  x2  1 )
y=1/x
hyperbola
0
x
Trigonometric: f(x) = {sin(x), cos(x), tan(x), cot(x), sec(x), csc(x)}, x in radians
y
1
y=sin(x)
-π/2
0
domain: (−∞,∞)
range: [-1,1]
period: 2π (waves)
zeros:
πn for sin(x)
x
π
π/2
π/2+πn for cos(x)
sin ( x)
cos( x)
domain: cos(x) ≠ 0
range: (−∞,∞)
period: π
zeros: πn for sin(x)
y
y=cos(x) 1
y
1
tan( x) 
cos 2 x sin 2 x  1
0
-π/2
-π/2
0
x
π/2 π
y=tan(x)
π/2 π
x
Exponential: f(x) = ax, constant a > 0 – base, x - exponent
 p
1
f (0)  a0  1, f (n)  a  n  n , f    q a p
a
q
x
x
a  x y
x y
x y
x y
a, b  0, x, y  R : a  a a , a  y , a  a xy , ab   a x b x
a
Special base e = 2.71828…
Logarithmic: f(x) = logax. Inverse of exponential: logax = y  ay = x. lnx:= loge`x, ln e=1
lnx
log a ( xy)  log a x  log a y, log a ( x r )  r log a x, log a x 
lna
Hyperbolic: certain combination of ex and e-x:
e x  e x
sinh x 
2
e x  ex
cosh x 
2
sinh x
tanh x 
cosh x
1
sinh x
1
sech x 
cosh x
cosh x
coth x 
sinh x
csch x 
y
1/2
y=sinhx
y
domain: (−∞,∞)
range:
(−∞,∞) for sinh(x)
[1,∞) for cosh(x)
y=coshx
1
1/2
0
x
0
1/2
x
Application of cosh: shape of hanging wire = catenary (catena=chain in Latin) y=c+acosh(x/a)
y
1
asymptote y=1
0
x
y=tanhx
sinh ( x)  sinh x
cosh ( x)  cosh x
cosh 2 x sinh 2 x  1
1  tanh2 x  sech 2 x
sinh ( x  y ) sinh x  cosh y  cosh x  sinh y
cosh ( x  y ) cosh x  cosh y  sinh x  sinh y
-1
asymptote
y=-1
1
Trigonometric vs. Hyperbolic
cos 2 sin 2  1
circle
cosh 2 sinh 2  1
(cos,sin)
(cosh,sinh)
hyperbola
0
0
 = twice the area of this region
Inverse hyperbolic:
 = twice the area of this region
y  sinh 1 x  sinh y  x
y  cosh 1 x  cosh y  x and y  0
y=sinh-1x
0
y  tanh1 x  tanh y  x
sinh 1 x  ln ( x  x 2  1), x  R
cosh 1 x  ln ( x  x 2  1), x  1
tanh1 x  12 ln ( 11 xx ), 1  x  1
0 1
y=cosh-1x
y=tanh-1x
-1
0
1
1
Show that sinh x  ln ( x 
Proof.
x 2  1)
e y  e  y z  z 1 z 2  1
y  sinh x  x  sinh y 


,
2
2
2z
y
where z  e  0.
1
Then
z 2 1
x
 z 2  2 x  z  1  0.
2z
Solving this quadratic equation with respect to z:
But x 
Therefore
z  x  x 2  1.
x 2  1  x  x 2  1  0, while z should be positive.
z  e y  x  x 2  1  y  ln ( x  x 2  1)  sinh 1 x.
Limit of a function:
lim f ( x)  L, x  a!
xa
A function f(x) is continuous
•
f(a) defined
•
lim f ( x ) exists
•
and lim f ( x)  f (a).
f ( x)  f (a),
at a point a if lim
xa
i.e.
x a
xa
A function f(x) is continuous
on an interval if it is continuous at every point
of this interval.
Itermediate Value Theorem: f – continuous on [a,b].
N between f (a) and f (b), where f (a)  f (b), there exists c  (a, b), s.t. f (c)  N ,
i.e. continuous function takes on every intermediate value between the function values f(a) and f(b).
f(b)
N
f(a)
a
c
b
P(a,f(a)) q
q
tanq 
f(a+h)-f(a)
y=f(x)
Q(a+h,f(a+h))
Tangent line has the same direction as the curve at the point of contact.
(tangent = touching in Latin)
h
f ( a  h)  f ( a )
h
Tangent line to the curve y=f(x) at the point P(a,f(a)) is the line through P with slope
k  lim
h 0
if this limit exists.
f ( a  h)  f ( a )
,
h
The tangent line to the curve y=f(x) at the point (a,f(a)) is the line through (a,f(a)) whose slope
is equal to f(a) = the derivative of f at a, i.e.
f (a)  lim
h 0
f ( a  h)  f ( a )
.
h
If this limit exists then the function f is differentiable
at a point a.
( a  h) 2  a 2
a 2  2ah  h 2  a 2
 lim
 lim (2a  h)  2a.
Example. f ( x)  x . f (a)  lim
h 0
h 0
h 0
h
h
2
Derivative as a function (let the point a vary):
f ( x)  lim
h 0
f ( x  h)  f ( x )
.
h
Given any x for which this limit exists  assign to x the number f(x).
( x  h) 2  x 2
 2 x  f ( x)  2 x.
h 0
h
Example. f ( x)  x . f ( x)  lim
2
Other notations: f ( x),
df
, Df ( x), Dx f ( x), ...
dx
Th. If f is differentiable at a, then it is continuous at a. (Proof: see section 2.9)
a
a
Example: f(x)=|x|, a=0
not differentiable at a,
but continuous
Example: discontinuity
not continuous at a 
not differentiable at a
a
Example: vertical tangent
f is continuous at a and
f (x)  
not differentiable: lim
x a
Derivatives of some basic functions. Rules.
d
(c )  0
dx
d n
Power:
( x )  nx n 1 , n  any real number
dx
Polynomial: derivative of polynomial = sum of all corresponding derivatives:
Constant c:
Constant Multiple Rule:
Sum Rule:
Difference Rule:
(cf ( x))  cf ( x)
( f ( x)  g ( x))  f ( x)  g ( x)
( f ( x)  g ( x))  f ( x)  g ( x)
a xh  a x
ah 1
x
Exponential: f ( x)  a . f ( x)  lim
 a  lim
 a x  f (0)
h 0
h 0
h
h
x
 rate of change of any exponential function is proportional to the function itself.
eh 1
 1  (e x )  e x .
e is defined s.t. lim
h 0
h
( f ( x)  g ( x))  f ( x)  g ( x)  f ( x)  g ( x)
Product Rule:

 f ( x) 
f ( x)  g ( x)  f ( x)  g ( x)

 
Quotient Rule:
( g ( x)) 2
 g ( x) 
Trigonometric:
( sinx )  cosx; (cosx )   sinx ; (tanx)  sec 2 x 
1
cos 2 x
Hyperbolic:


 e x  e x 
  e x  ex 
  cosh x
  sinh x
sinh x  
cosh x  
2
2





  sinh x 
1
 
tanh x  
 sech 2 x
2
 cosh x  cosh x






Chain Rule:

if F ( x)  ( f  g )( x)  f ( g ( x))
 F ( x)  f ( g ( x))  g ( x)
Implicit differentiation…
for inverse functions
(a x )  (e xln a )  e xln a  ln a  a x  ln a
Logarithmic:
log a x  y  a y  x  (a y )  a y  ln a  y  x  ln a  y  1
(log a x) 
1
,
x  ln a
(ln x) 
1
x
Inverse hyperbolic: y  sinh 1 x  sinh y  x  cosh y  y  1; cosh y  1  sinh 2 y  1  x 2
sinh x  
1
1
1 x2

1
tanh1 x 
1 x2


cosh x  
1
1
x2 1
sinh x  
1
Show that
Proof.
1
x 1
2
.
y  sinh 1 x  x  sinh y.
Let us differentiate both sides with respect to x and find y’:
1  cosh y  y  y 
But
1
.
cosh y
cosh 2 y  sinh 2 y  1  cosh y  1  sinh 2 y  1  x 2 ,
since x  sinh y.
Therefore,
y 
1
1

cosh y
1 x2