The Tangent Line Problem
Lesson 4.1
The Rate of Change of a Curved Line
slope – the rate of change of a function
How do we find the slope of a linear function?
y 2 – y1
x 2 – x1
=
y
x
What is the rate of change of a curved line?
Can we estimate it?
2 – 1
f (x) = 2x
1 – 0
= 1 …Or is it 2? …or ½?
Which secant line is correct?
…All of them? …or none?
…Oh! It’s not constant?
The Rate of Change of a Curved Line
^
How can we find the instantaneous rate of change of
a curved line at a given point?
If we start with a secant…
(1, 2) & (2, 4) 
4 – 2
2 – 1
= 2
…Then get the points closer…
(1, 2) & (15, 2828)  = 1657
…And closer yet…
f (x) = 2x
(1, 2) & (11, 2144)  = 1435
…Are we getting somewhere?
This Looks Like a Calculus Problem!
How can we find the instantaneous rate of change of a
curved line at a given point?
If we start with a point (x, f (x)) on the curve…
And make a secant w/ point (x + x, f (x + x)) on the curve…
…Then moved x really close to x…
(x+x, f (x+x))
…Then calculated the slope…
(x, f(x)) & (x + x, f (x + x))
f (x) = 2x (x, f (x))
=
f(x +  x) – f(x)
x +x – x
=
f(x +  x) – f(x)
…And applied a LIMIT to x as it
approaches 0…
x
f
(x + x) – f (x)
lim
x→0
x
BEHOLD!
derivative – a measurement of the instantaneous rate of
change (i.e. sensitivity of change) of a given variable as
determined by another variable.
f ( x  x)  f ( x)
f ' ( x)  lim
x 0
, provided the limit exists. For all
x
x which the limit exists, f ’(x) is a function of x.
Clarifications & Terminology:
f‘(x)
• f ’(x) is read as “f-prime of x”
• f ’(x) is a tangent line to f (x) at (c, f (c))
(c, f (c))
f (x)
• f ’(x) is also called “the slope of the graph
of f at x = c”
• f ’(x) also notated as y’, dy/dx , or d/dx[f (x)]
So, um...How Does It Work?
Find f’(x) for the graph of f(x) = x2 + 1 at (1, 2).
[(x+x)2 + 1] – [x2 + 1]
f ( x  x)  f ( x)
lim
f ' ( x)  lim
=
x 0
x
x
x →0
f’(x)
= lim
[x2 + 2xx + (x)2 + 1] – x2 – 1
x
x →0
= lim
f (x)
x2 + 2xx + (x)2 + 1 – x2 – 1
x
x →0
(1, 2)
= lim
x →0
2xx + (x)2
x
= lim 2x + x = 2x
x →0
So, when x = 1, f ’(x) = 2(1) = 2
Another Example
Find f’(x) for the graph of f(x) = √x at (4, 2).
√(x+x) + √x
√(x+x) – √x
f ( x  x)  f ( x)
lim
f ' ( x)  lim
=
x 0
x
x
x →0
= lim
x →0
= lim
x →0
f’(x)
f (x)
(4, 2)
= lim
x →0
√(x+x) + √x
[√(x+x)]2 – [√x]2
x [√(x+x) + √x]
x + x - x
x [√(x+x) + √x]
1
√(x+x) +√x
=
1
√x + √x
=
1
2√x
So, when x = 4, f ’(x) = 1/(2•√4) = 1/4
Differentiation
differentiation – finding the derivative of a function f(x)
dt
f (t  t )  f (t )
 lim
dy t 0
t
Differentiate y = 2/t .
= lim
t →0
= lim
2
2
–
t
t+t
t →0
t
2t – 2t – 2t
t2 + tt
t →0
= lim
= lim
t →0 t2
t
-2
+ tt
2t – 2(t+t)
t(t+t)
t
= lim
t →0
y’=
-2
t2
-2t
t(t2 + tt)
at t = 1, y’ = -2/(1)2 = -2
at t = 2,
y’ = -2/(2)2 = -1/2
Footnotes
• The derivative of a function is also a function.
• If f is differentiable at x = c, then f is continuous at c.
• If f is not continuous at x = c, then f is not
differentiable at c.
• The instantaneous rate of change
of a function at c can be 0 or ∞.
f ’(0) doesn’t exist
• If one-sided limits are not
equal at c, then f is not
differentiable at c.
(Ex: |x| at x = 0.)
f ’(0) = ∞
f ’(0) = 0
Lesson Practice
1) Find the instantaneous rate of change of
f(x) = 2x at x = -3.
Classwork
Pg. 259 (1, 4 (w/ expl.), 7, 9, 19, 21, 23, 39, 40, 45,
46, 48, 72 - 73)