Limit Definition of the Derivative:
Limit Definition of the Derivative:
WHY DOES THE POWER RULE WORK?
To Introduce the Limit Definition of the
Derivative, let’s pose this question:
If f(x)=x2 and you needed to find the
dy
f ‘(1) without power rule or dx on your
calculator, how could you get a
reasonable and accurate answer?
HOW ABOUT THIS………….
How about approximating f’(1), the slope of
The tangent line, by finding the slope of a
secant line?
f(x)=x2
Recall: a secant line is a line that intersects f(x) in
two points.
Let’s take a graphical look…….
Recall f’(1), graphically, is the slope of
the tangent line.
Let’s find a secant line whose slope
is close to f’(1).
By picking a point at x=1 and x=3 on f(x),
The slope of the blue line (secant) would give
an approximation of f’(1), but we could do
better if we picked a second point closer
to x=1.
Now the slope of the secant line (blue) is
closer to f ‘ (1). Do you agree??
As point B gets closer to x=1, the slope
of the secant line gets closer to the slope
of the tangent line.
Does this
make sense?
Recall: slope=
y2 − y1
x2 − x1
f (x + h) − f (x)
f '(x) = lim
h→0
h
f (x + h) − f (x)
f '(x) = lim
h→0
h
THIS IS THE LIMIT DEFINITION OF THE
DERIVATIVE.
THIS LIMIT DEFINITION OF THE
DERIVATIVE CAN BE WRITTEN
WITH SOME SLIGHT VARIATIONS
BUT THE SAME IDEA IS CAPTURED.
THIS LIMIT DEFINITION OF THE
DERIVATIVE CAN BE WRITTEN
WITH SOME SLIGHT VARIATIONS
BUT THE SAME IDEA IS CAPTURED
How about this……….
f (x +x) − f (x)
f (x) = lim
 x→0
x
AND ONE MORE:
?
f '(x) = lim
?→? ?
Can you fill in for the question marks?
Answer:
f (g) − f (x)
f '(x) = lim
g→x
g− x
How does this all work algebraically????
Let’s use the Limit Definition of the Derivative to
to find f ‘ (x) for f(x)=4x-2.
Let’s use the Limit Definition of the Derivative to
to find f ‘ (x) for f(x)=4x-2.
RECALL THE LIMIT DEFINITION OF THE DERIVATIVE:
f (x + h) − f (x)
f '(x) = lim
h→0
h
Let’s use the Limit Definition of the Derivative to
to find f ‘ (x) for f(x)=4x-2.
RECALL THE LIMIT DEFINITION OF THE DERIVATIVE:
f (x + h) − f (x)
f '(x) = lim
h→0
h
APPLY THE DEFINITION TO f(x)=4x-2:
Let’s use the Limit Definition of the Derivative to
to find f ‘ (x) for f(x)=4x-2.
RECALL THE LIMIT DEFINITION OF THE DERIVATIVE:
f (x + h) − f (x)
f '(x) = lim
h→0
h
APPLY THE DEFINITION TO f(x)=4x-2:
4(x + h) − 2 − (4 x − 2)
f '(x) = lim
h→0
h
4(x + h) − 2 − (4 x − 2)
f '(x) = lim
h→0
h
4(x + h) − 2 − (4 x − 2)
f '(x) = lim
h→0
h
4 x + 4h − 2 − 4 x + 2
f '(x) = lim
h→0
h
4(x + h) − 2 − (4 x − 2)
f '(x) = lim
h→0
h
4 x + 4h − 2 − 4 x + 2
f '(x) = lim
h→0
h
4h
f '(x) = lim
h→0 h
4(x + h) − 2 − (4 x − 2)
f '(x) = lim
h→0
h
4 x + 4h − 2 − 4 x + 2
f '(x) = lim
h→0
h
4h
f '(x) = lim
h→0 h
f '(x) = lim 4
h→0
4(x + h) − 2 − (4 x − 2)
f '(x) = lim
h→0
h
4 x + 4h − 2 − 4 x + 2
f '(x) = lim
h→0
h
4h
f '(x) = lim
h→0 h
f '(x) = lim 4
h→0
f '(x) = 4