Ch.3 The Derivative
Differentiation
Given a curve y = f(x)
Want to compute the slope of tangent at
some value x=a.
Let A=(a, f(a)) and
B=(a + x , f(a + x )) = (a + x , f(a) + y)
where the change in y-value (y) is given by
y = f(a + x ) – f(a)
and B is a point on the curve y = f(x) close to A.
Therefore slope of chord
y f (a x ) f (a )
AB
x x
As x 0, the chord AB tangent
Definition The derivative of f is defined by
f (a x ) f (a )
f ' (a ) lim
x 0 x
This gives the slope of tangent at x = a
f ( x ) f (a )
f ' (a ) lim
x a xa
after setting x =a +x. We say that f is
differentiable at a if this limit exists.
Thus the derivative of f at x is
dy y f ( x h) f ( x )
lim lim f ' ( x)
dx x 0 x h0 h
Just set h = x and a=x.
We say that f is differentiable at x if
the above limit exists.
This defines a new function f’(x) called
the derivative of f(x) .
Other notations for f’(x)
dy df
or or Df ( x )
dx dx
Examples
dy
1.
4 x 3 )x ( f y
, show that 3 f ' ( x)
dx
dy y
lim
dx x 0 x
f ( x h) f ( x )
lim
h0 h
[3( x h) 4] [3 x 4]
lim
h0 h
[3 x 3h 4] [3 x 4]
lim
h0 h
3h
lim lim 3 3, as required.
h0 h h0
Note : y f ( x ) 3 x 4 is the equation of a straight line so the slope
f ( x h) f ( x )
of the chord AB, , is identically 3 as expected.
h
dy
2. f ( x ) x show that
2
2 x f ' ( x ).
dx
dy y
Now lim
dx x 0 x
f ( x h) f ( x )
lim
h0 h
( x h) 2 x 2
lim
h0 h
( x 2 2 xh h 2 ) x 2
lim
h 0 h
2 xh h 2
lim
h 0 h
lim (2 x h) 2 x 0 2 x , as required.
h 0
3. If y f ( x ) x n
dy
nx n 1 f ' ( x )
dx
since
f ( x h) f ( x ) ( x h) n x n
f ' ( x ) lim lim
h 0 h h0 h
[ x n nx n 1 h 1 n(n 1) x n 2 h 2 h n ] x n
lim 2
h 0 h
lim [nx n 1 1 n(n 1) x n 2 h terms in h 2 , h 3 , etc
2
h 0
nx n 1 1 n(n 1) x n 2 0 0 0 0
2
nx n 1 .
Trigonometry identity:
sin(a b) sin a cosb cosa sin b
Example dy
y f ( x ) sin x cos x f ' ( x )
dx
f ( x h) f ( x ) sin( x h) sin x
Proof : f ' ( x ) lim lim
h 0 h h 0 h
sin x cos h cos x sin h sin x
lim
h 0 h
cos h 1 sin h
lim sin x cos x
h 0
h h
cos h 1 sin h
sin x lim cos x lim
h 0
h h 0 h
sin x 0 cos x 1 cos x.
cos h 1 (cosh 1)(cosh 1)
Note : lim lim
h0 h h0 h(cosh 1)
cos 2 h 12 sin h sin h
lim lim
h0 h(cosh 1) h0 h (cosh 1)
sin 0
1 1 0 0.
(cos0 1)
Example: Is f(x)=|x| differentiable at x=0?
Must consider
f ( x h) f ( x )
f ' ( x ) lim at x 0
h0 h
|0h||0| |h|
f ' (0) lim lim .
h 0 h h0 h
| h | 1 if h 0
But
h 1 if h 0
So the limit does not exists that is y=|x| is
not differentiable at 0
Thus continuity does not imply
differentiability.
Rate of Change
Recall that the slope of the tangent at a
point measures the (instantaneous) rate
of change of y with respect to x at that
point.
Example. Let s(t) be the distance of a
car that has traveled at time t.
Speed v= rate of change of distance
ds
v s' (t )
dt
Similarly
acceleration a = rate of change of speed
dv d ds d 2 s
a 2 s' ' (t ) s ( 2) (t )
dt dt dt dt
other notation
Example
A cylindrical tank holds 50 litres of water and can be drained
from the bottom of the tank in 100 seconds. Find the rate of
change of volume after 30 seconds given volume V of water in
the tank after t seconds can be shown to be
t 2 t2
V (t ) 50 (1 ) 50 t for 0 t 100
100 200
Rate of change of volume
dV 2t
0 1 1 10 2 t liter/sec.
dt 200
At t 20,
dV 2 1
1 10 20 1 0.8 litres/sec
dt 5
that is, volume is decreasing at rate 0.8 litres/second.
Theorem
If f(x) is differentiable at a then f is continuous
at x=a.
Proof. Assume f(x) differentiable at x=a.
Must show lim f ( x) f (a )
x a
lim ( f ( x ) f (a )) 0
x a
f ( x ) f (a )
But f ( x ) f (a ) ( x a)
xa
f ( x ) f (a )
lim ( f ( x ) f (a )) lim lim ( x a )
x a x a xa x a
f ' (a ) 0 0.
as required.
Basic Differentiation Rules
Read Section 3.2 (or the same topic
from other textbooks)
You should be able to use the
differentiation rules/theorems to find the
derivatives of functions