Transcript Document

2.1
The Derivative and the
Tangent Line Problem
I’m going nuts
over derivatives!!!
Calculus grew out of 4 major problems that
European mathematicians were working on
in the seventeenth century.
1.
The tangent line problem
2.
The velocity and acceleration problem
3.
The minimum and maximum problem
4.
The area problem
The tangent line problem
c  x, f (c  x)
secant line
f (c  x)  f (c)
(c, f(c))
x
(c, f(c)) is the point of tangency and f(c+ x ) – f(c)
is a second point on the graph of f.
The slope between these two points is
msec
f (c  x)  f (c)

c  x  c
f (c  x)  f (c)

x
Definition of Tangent Line with Slope m
f (c  x)  f (c)
lim
m
x 0
x
Find the slope of the graph of f(x) = x2 +1 at
the point (-1,2). Then, find the equation of the
tangent line.
(-1,2)
f ( x  x)  f ( x)
lim
x 0
x

f(x) = x2 + 1

( x  x)  1  x  1
 lim
x 0
x
2
2
Therefore, the slope
at any point (x, f(x))
is given by m = 2x
x  2 xx  x   1  x  1
 lim
x 0
x
2
2
x(2 x  x)
 lim

2
x
x 0
x
2
What is the slope
at the point (-1,2)?
The equation of the tangent line is
m = -2
y – 2 = -2(x + 1)
The limit used to define the slope of a tangent
line is also used to define one of the two fundamental operations of calculus --- differentiation
Definition of the Derivative of a Function
f ( x  x)  f ( x)
f ' ( x)  lim
x 0
x
f’(x) is read “f prime of x”
Other notations besides f’(x) include:
dy
d
, y ' , [ f ( x)], Dx [ y ]
dx
dx
Find f’(x) for f(x) = x, and use the result to find
the slope of the graph of f at the points (1,1) &
(4,2). What happens at the point (0,0)?
f ( x  x)  f ( x)
f ' ( x)  lim
x 0
x
f ' ( x)  lim
x 0
x  x  x  x  x  x 

 

x
x


x

x


x  x  x 1
( x  x)  x
 lim
 lim
x 0 x
x 0 x
x  x  x
x  x  x

 lim
x 0

1
x  x  x




1
2 x

f ' ( x)  m 
1
2 x
Therefore, at the point (1,1), the
slope is ½, and at the point (4,2),
the slope is ¼.
What happens at the point (0,0)?
The slope is undefined, since it produces division
by zero.
1
m
4
1
m
2
1
2
3
4
Find the derivative with respect to t for the
2
function y  .
t
dy
f (t  t )  f (t )
 lim
dx t 0
t
2t  2(t  t )
2
2

t (t  t )
t


t
t
 lim
 lim
t 0
t 0
t
t
1
2t  2t  2t 1
2
 lim
  2
t 0
t
t (t  t )
t
Theorem 3.1 Alternate Form of the Derivative
The derivative of f at x = c is given by
f ( x )  f (c )
f ' (c)  lim
x c
xc
(x, f(x))
f ( x)  f (c)
(c, f(c))
x  x  c
c
x
Derivative from the left and from the right.
f ( x )  f (c )
lim
x c
xc
f ( x )  f (c )
lim
x c
xc
Example of a point that is not differentiable.
f ( x)  x  2 is continuous at x = 2 but let’s
look at it’s one sided limits.
x2 0
f ( x)  f (2)
lim
 lim
 -1
x2
x 2
x2
x2
x2 0
f ( x)  f (2)
lim
 lim
 1
x2
x 2
x2
x2
The 1-sided limits are not equal.
 , x is not differentiable at x = 2. Also, the
graph of f does not have a tangent line at the
point (2, 0).
A function is not differentiable at a point at
which its graph has a sharp turn or a vertical
tangent line(y = x1/3 or y = absolute value of x).
Differentiability can also be destroyed by
a discontinuity ( y = the greatest integer of x).