Tangent Line,
Velocity, Derivative
and Differentiability
instant rate of change
and approximation of the function
Motivation:
what do tangent line and
velocity have in common?
Euclid: The tangent is a line that touches a
curve at just one point.
Works for circles:
Euclid: The tangent is a line that touches a
curve at just one point.
Works for circles:
x
Does NOT work for sin(x)!
x
Modern: The tangent is the limiting position of
the secant line (if exists).
Tangent = lim Secant
f(t)
t→x
Slope of
Slope of
= lim
Tangent t→x Secant
f(t)-f(x)
slope of ______
= t-x
secant
f(t)-f(x)
t-x
f(x)
x
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
t
EXAMPLES
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
EXAMPLE 1. Find slope of the tangent line to
the graph of f (x) = x2 − 4 at point (3, 5).
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
EXAMPLE 1. Find slope of the tangent line to
the graph of f (x) = x2 − 4 at point (3, 5).
2
2
(t
−
4)
−
(x
− 4)
Slope of
= lim
Tangent t→x
t−x
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
EXAMPLE 1. Find slope of the tangent line to
the graph of f (x) = x2 − 4 at point (3, 5).
2
2
(t
−
4)
−
(x
− 4)
Slope of
= lim
Tangent t→x
t−x
t2 − x2
(t − x)(t + x)
= lim
= lim
= 2x
t→x t − x
t→x
t−x
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
EXAMPLE 1. Find slope of the tangent line to
the graph of f (x) = x2 − 4 at point (3, 5).
2
2
(t
−
4)
−
(x
− 4)
Slope of
= lim
Tangent t→x
t−x
t2 − x2
(t − x)(t + x)
= lim
= lim
= 2x
t→x t − x
t→x
t−x
At x = 3, (y = 5 is irrelevant)
Slope of Tangent = 6
Velocity: How fast the position s(x) is changing.
x=3
h
x=0
4
0
h4
h
h x=4 4
h 4 h x=2
x=1 h
4
1
s
h
x=5
Average
=
Velocity
s(x + h) − s(x)
=
h
s(x + h) − s(x)
Instantaneous
= lim
Velocity
h→0
h
EXAMPLES
s(x + h) − s(x)
Instant.
= lim
Velocity h→0
h
EXAMPLE 2. Motion of the particle along a
line is described by s(x) = 1 + cos(x). Find
instantaneous velocity at moment x = 2.
s(x + h) − s(x)
Instant.
= lim
Velocity h→0
h
EXAMPLE 2. Motion of the particle along a
line is described by s(x) = 1 + cos(x). Find
instantaneous velocity at moment x = 2.
(1 + cos(x + h)) − (1 + cos(x))
Instant.
= lim
Velocity h→0
h
s(x + h) − s(x)
Instant.
= lim
Velocity h→0
h
EXAMPLE 2. Motion of the particle along a
line is described by s(x) = 1 + cos(x). Find
instantaneous velocity at moment x = 2.
(1 + cos(x + h)) − (1 + cos(x))
Instant.
= lim
Velocity h→0
h
cos(x + h) − cos(x)
= lim
h→0
h
cos(x + h) − cos(x)
= lim
h→0
h
(using cos(x + h) = cos(x) cos(h) − sin(x) sin(h))
cos(x + h) − cos(x)
= lim
h→0
h
(using cos(x + h) = cos(x) cos(h) − sin(x) sin(h))
cos(x) cos(h) − sin(x) sin(h) − cos(x)
= lim
h→0
h
cos(x + h) − cos(x)
= lim
h→0
h
(using cos(x + h) = cos(x) cos(h) − sin(x) sin(h))
cos(x) cos(h) − sin(x) sin(h) − cos(x)
= lim
h→0
h
cos(h) − 1
sin(h)
= cos(x) lim
− sin(x) lim
x→0
x→0
h
h
= − sin(x)
cos(x + h) − cos(x)
= lim
h→0
h
(using cos(x + h) = cos(x) cos(h) − sin(x) sin(h))
cos(x) cos(h) − sin(x) sin(h) − cos(x)
= lim
h→0
h
cos(h) − 1
sin(h)
= cos(x) lim
− sin(x) lim
x→0
x→0
h
h
= − sin(x)
At x = 2, Instantaneous Velocity = − sin(2)
Compare
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
s(x + h) − s(x)
Instant.
= lim
Velocity τ →0
h
Compare
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
s(x + h) − s(x)
Instant.
= lim
Velocity τ →0
h
Equivalent up to the substitution
f = s,
t−x=h
Compare
f (t) − f (x)
Slope of
= lim
Tangent t→x
t−x
s(x + h) − s(x)
Instant.
= lim
Velocity τ →0
h
Equivalent up to the substitution
f = s,
Is it the same thing?
t−x=h
Definition of the Derivative
putting things together
Slope of
Tangent
Instant.
Velocity
s(x + h) − s(x)
f (t) − f (x)
lim
lim
t→x
h→0
h
t−x
Rate of
Change
s(t + ∆t) − s(t)
lim
∆t→0
∆t
Slope of
Tangent
Rate of
Change
Instant.
Velocity
s(x + h) − s(x)
f (t) − f (x)
lim
lim
t→x
h→0
h
t−x
s(t + ∆t) − s(t)
lim
∆t→0
∆t
Definition: Derivative of f (x) at point x is
f (t) − f (x)
f (x) = lim
t→x
t−x
0
f (x + h) − f (x)
or, f (x) = lim
h→0
h
0
Illustration
Slope of
Tangent
y
slope=f '(x)
Instant.
Velocity
Rate of
Change
y=f(x)
x=3
h
x
x=0
4
0
x=4
h4
4
x=2
4
4 h c s'(x)
1
s
x=1
Velocity = s'(x)
h
Rate of = Q'(t)
Change
x=5
Q(t)
EXAMPLES
f (t) − f (x)
f (x) = lim
t→x
t−x
0
EXAMPLE 3. Evaluate f 0(x) using definition of
derivative, if f (x) = x3 + 2x.
f (t) − f (x)
f (x) = lim
t→x
t−x
0
EXAMPLE 3. Evaluate f 0(x) using definition of
derivative, if f (x) = x3 + 2x.
3
3
(t
+
2t)
−
(x
− 2x)
0
f (x) = lim
t→x
t−x
f (t) − f (x)
f (x) = lim
t→x
t−x
0
EXAMPLE 3. Evaluate f 0(x) using definition of
derivative, if f (x) = x3 + 2x.
3
3
(t
+
2t)
−
(x
− 2x)
0
f (x) = lim
t→x
t−x
(t3 − x3) + (2t − 2x)
= lim
t→x
t−x
f (t) − f (x)
f (x) = lim
t→x
t−x
0
EXAMPLE 3. Evaluate f 0(x) using definition of
derivative, if f (x) = x3 + 2x.
3
3
(t
+
2t)
−
(x
− 2x)
0
f (x) = lim
t→x
t−x
(t3 − x3) + (2t − 2x)
= lim
t→x
t−x
t3 − x3
2t − 2x
= lim
+ lim
t→x t − x
t→x t − x
Consider
t3 − x3
2t − 2x
= lim
+ lim
t→x t − x
t→x t − x
Consider
t3 − x3
2t − 2x
= lim
+ lim
t→x t − x
t→x t − x
(t − x)(t2 + tx + x2)
2(t − x)
= lim
+ lim
t→x
t→x (t − x)
(t − x)
= 1 · (x2 + x2 + x2) + 2 = 3x2 + 2
f (x + h) − f (x)
f (x) = lim
h→0
h
0
0
EXAMPLE 4. Evaluate f (x) using definition of
√
derivative, if f (x) = x.
f (x + h) − f (x)
f (x) = lim
h→0
h
0
0
EXAMPLE 4. Evaluate f (x) using definition of
√
derivative, if f (x) = x.
√
√
x+h− x
0
f (x) = lim
h→0
h
f (x + h) − f (x)
f (x) = lim
h→0
h
0
0
EXAMPLE 4. Evaluate f (x) using definition of
√
derivative, if f (x) = x.
√
√
x+h− x
0
f (x) = lim
h→0
h
√
√ √
√
( x + h − x) ( x + h + x)
√
= lim
√
h→0
h
( x + h + x)
f (x + h) − f (x)
f (x) = lim
h→0
h
0
0
EXAMPLE 4. Evaluate f (x) using definition of
√
derivative, if f (x) = x.
√
√
x+h− x
0
f (x) = lim
h→0
h
√
√ √
√
( x + h − x) ( x + h + x)
√
= lim
√
h→0
h
( x + h + x)
1
1
(x + h) − x
√ = √
= lim √
√ =√
h→0 h( x + h +
x+ x 2 x
x)
The Differential
problem of linear approximation
Approximating function with a line
f (x+h) = f (x)+m·h+o(h)
where
o(h)
lim
=0
h→0 h
Motion along a line:
y = y0 + m · h
slope = m
A
f(x)
y
mh
h
x0
o(h)
mh
h
x
y0
A
f(x+h)
slope=m
x+h
Approximating function with a line
f (x+h) = f (x)+m·h+o(h)
where
o(h)
lim
=0
h→0 h
Motion along a line:
y = y0 + m · h
What is m ?
slope = m
A
f(x)
y
mh
h
x0
o(h)
mh
h
x
y0
A
f(x+h)
slope=m
x+h
Definition. Function f (x) is differentiable at x if
there exists a number m such that
f (x + h) = f (x) + m · h + o(h)
o(h)
where lim
=0
h→0 h
(o(h) “o”-little is a quantity of the order of
magnitude smaller then h).
Differential:
df (x)hhi = m · h
Compute m. Assume function f (x) is differenti
able: f (x + h) = f (x) + m · h + o(h). Solving for
m and rearranging,
f (x + h) − f (x)
o(h)
=m+
h
h
Compute m. Assume function f (x) is differenti
able: f (x + h) = f (x) + m · h + o(h). Solving for
m and rearranging,
f (x + h) − f (x)
o(h)
=m+
h
h
differentiability requires
o(h)
limh→0 h
= 0 therefore,
f (x + h) − f (x)
lim
=m
h→0
h
Compute m. Assume function f (x) is differenti
able: f (x + h) = f (x) + m · h + o(h). Solving for
m and rearranging,
f (x + h) − f (x)
o(h)
=m+
h
h
differentiability requires
o(h)
limh→0 h
= 0 therefore,
f (x + h) − f (x)
lim
=m
h→0
h
or f 0(x) exists and f 0(x) = m.
Differentiability implies Derivative!
f (x + h) = f (x) + m · h + o(h)
EXAMPLE 5. Differentiate f (x) = sin(x).
f (x + h) = f (x) + m · h + o(h)
EXAMPLE 5. Differentiate f (x) = sin(x).
sin(x + h) = sin(x) + cos(x)h
+[− cos(x)h + sin(x + h) − sin(x)]
f (x + h) = f (x) + m · h + o(h)
EXAMPLE 5. Differentiate f (x) = sin(x).
sin(x + h) = sin(x) + cos(x)h
+[− cos(x)h + sin(x + h) − sin(x)]
Now, m = f 0(x) = cos(x) and o(h) = [. . .].
f (x + h) = f (x) + m · h + o(h)
EXAMPLE 5. Differentiate f (x) = sin(x).
sin(x + h) = sin(x) + cos(x)h
+[− cos(x)h + sin(x + h) − sin(x)]
Now, m = f 0(x) = cos(x) and o(h) = [. . .]. Notice
o(h) = cos(x)[sin(h) − h] + sin(x)[cos(h) − 1]
f (x + h) = f (x) + m · h + o(h)
EXAMPLE 5. Differentiate f (x) = sin(x).
sin(x + h) = sin(x) + cos(x)h
+[− cos(x)h + sin(x + h) − sin(x)]
Now, m = f 0(x) = cos(x) and o(h) = [. . .]. Notice
o(h) = cos(x)[sin(h) − h] + sin(x)[cos(h) − 1]
Therefore,
o(h)
lim
=0
h→0 h
Evaluating differentials. Common notations:
0
df (x) = f (x)dx
(Compare to df (x)hhi = m · h, substituting m =
f 0(x), h = dx)
Evaluating differentials. Common notations:
0
df (x) = f (x)dx
(Compare to df (x)hhi = m · h, substituting m =
f 0(x), h = dx)
EXAMPLE 6.
d(sin x) = cos xdx,
√
1
d( x) = √ dx,
2 x
1
d(arctan(x)) =
dx
1 + x2