GUC
MATH103
For Engineering
Winter 2023
Lecture # 10
Chapter 3
App’s of limits: Differentiation of functions
1
Remark
dy
Find
if :
dx
(1) y = x 2 / 3
Solution :
dy 2 −1 / 3 2 1
(1)
= x
= .3
x
3
( 2) y = x .e
dx 3
3 x
dy 3
x
x 1 −2 / 3
( 2)
= x .(e ) + e .( x )
dx
3
Conclusion: For any rational number n, the derivative of x(n) is n. x(n -1).
2
Rule
The proof follows
from the definition
Example : Find the points at which the tangent lines
to the curve : y = x 4 − 2 x 2 + 2, are horizontal.
Solution : y = x 4 − 2 x 2 + 2
Horizontal lines have
zero slope !
 y = 4 x 3 − 4 x = 4 x ( x 2 − 1)
For horizontal tangent y = 0.
 4 x ( x 2 − 1) = 0  x = 0, x = −1, x = 1,
for x = 0  y = 2.
for x = −1  y = 1.
for x = 1  y = 1
 the three points are (0,2), ( −1,1), (1,1).
3
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Derivative of Functions
By Definition.
Power function rule
Square root function rule.
Absolute value function rule.
Product “of 2 functions” rule.
Quotient “of 2 functions” rule.
Exponential function rule.
Logarithmic function rule.
Trigonometric functions rules and their inverses.
Hyperbolic functions rules and their inverses.
4
The proof follows from
the definition. How?
Ex :
d
[ x ( x 2 + 3 x − 1)] =
dx
=
x ( x 2 + 3 x − 1)'+ ( x 2 + 3 x − 1)( x )'
x ( 2 x + 3) + ( x + 3 x − 1)
2
1
2 x
The proof follows from
the definition. How?
d 2 x2 − 1
( x 2 + 1)( 2 x 2 − 1)'−( 2 x 2 − 1)( x 2 + 1)'
Ex : [ 2
]=
dx x + 1
( x 2 + 1)2
( x 2 + 1)(4 x ) − ( 2 x 2 − 1)( 2 x )
6x
=
=
( x 2 + 1)2
( x 2 + 1)2
5
7- Derivative of the exponential function
d x
x
(a ) = a . ln(a ) ??
dx
ah − 1
We know that : lim
= ln(a ) ?
h→0
h
Proof :
x+h
x
d
a
−
a
( a x ) = lim
dx
h
h→ 0
= lim
a x .a h − a x
h
= lim
a x .[a h − 1]
h
h→ 0
h→ 0
= a x . lim
h→ 0
ah − 1
= a x . ln(a )
h
See problem
13 in ws4.
6
8- Derivative of the logarithmic function
d
1
(ln x ) = , x  0
dx
x
1
y
We know that : lim(1 + y ) = e ?
y →0
Proof :
ln(
x+h
)
x
h
d
ln( x + h) − ln x
(ln x ) = lim
= lim
dx
h
h→ 0
h→ 0
h
ln(1 + )
ln(1 + y )
x =
By the continuity =
,
lim
lim
h
x. y
of “ ln(.) ” function
h→ 0
y →0
(put h = y )
x
1
1
1
1
= lim . ln(1 + y ) = lim ln(1 + y ) y
x y →0 y
x y →0
1
y
1
1
1
= ln lim (1 + y ) = ln e = .
x
x
x
y→o
Prove that
d
1
(ln x ) =
dx
x
7
Derivative of the general logarithmic function
HW. Prove it
by Definition?
d
1 1
(loga x ) = .
dx
x ln a
Proof :
ln x
Since loga x =
,
ln a
then,
d
d ln x
1 d
1 1
(loga x ) =
(
)=
(ln x ) = .
dx
dx ln a
ln a dx
x ln a
8
The chain rule
An easier form
of the theorem
Example
dy
Find
if : y = x 3 + 5 x
dx
1
1
Solution : y' =
( x 3 + 5 x )' =
( 3 x 2 + 5)
2 x3 + 5 x
2 x3 + 5 x
9
Summary: special case for “x” and general case for a function “u” “chain rule”
10
Summary: special case for “x” and general case for a function “u” “chain rule”
11
7- Derivative of the exponential function
HW. Prove it
by Definition?
d x
(e ) = e x
dx
Can you find another
function whose derivative
gives the same function?
Proof : Let y = e  x = ln y
x
d
d

( x) =
(ln y )
dx
dx
1 dy
Proof by chain rule

1=
y dx

dy
d x
x
= y
(e ) = e
dx
dx
12
Derivative of the general exponential function
d x
(a ) = a x ln a ??
dx
We know that : loga x =
ln y
Proof : Let y = a  x = loga y =
ln a
d
d ln y

( x) =
(
)
dx
dx ln a
1 dy
Proof by chain rule

1=
y ln a dx
x

dy
d x
= y ln a 
(a ) = a x ln a
13
dx
dx
ln x
ln a
Examples
dy
Find y' =
:
dx
(1) y = e . ln( 2 x + 1) ( 2) y = e
3x
x
+2
( x2 +7 x )
( 3) y = (1 + x + 1 )
10
Solutions :
(1) y = e .
3x
1
2 x +1
( 2) y = e . 2
x
1
x
( 2) + ln( 2 x + 1).e ( 3)
3x
+2
( x2 +7 x )
( 3) y = 10(1 + x + 1 )
9
. ln 2.( 2 x + 7 )
1
2 x +1
14
Examples
dy
Find y' =
for the following :
dx
(1) y = ln( x 2 − 6)
( 2) y = 1 + ln 2 x
ln x
( 3) y =
1 + ln x
(4) y = ( x 2 + 1)[ln( x 2 + 1)]2
(5) y = log 2 x + 2
x
(6) y = log
x +1
HomeWork
Find y' for the following :
y = ln( x 2 )
y = ln2 ( x )
y = ln(ln x )
15
1
(1) y' = 2
.[2 x ] Solutions
x −6
1
1
( 2 ) y' =
.[0 + 2 ln x . ]
2
x
2 1 + ln x
1
1
(1 + ln x ).[ ] − ln x .[ ]
1
x
x
( 3 ) y' =
=
2
(1 + ln x )
x .(1 + ln x )2
(4) y' = ( x 2 + 1).2[ln( x 2 + 1)].[ x 21+1 ][2 x ] + [ln( x 2 + 1)]2 .[2 x ]
( 5 ) y' =
1
1
1
.
.
x + 2 ln 2 2 x + 2
1 1
1
1
(6) y = log10 x − log10 ( x + 1)  y' = .
−
.
x ln 10 x + 1 ln 10
16
The rate of change of a
function as the Derivative
Given a function f (x):
If x moves from “a” to “a + h”, the change in x: ∆x = (a + h) – (a)
Then, the corresponding change in f (x): ∆y = f (a + h) – f (a).
I - The Average Rate of Change of f (x); on [a, a + h] or for 2 points “a”
& “a + h”; is defined as:
 f
f (a + h) − f (a )
(a ) =
x
h
II - The instantaneous rate of change of f(x) at a point “a”; is obtained
when “the point Q moves to P” i.e., “as ∆x = h → 0”; is defined by:
df
f (a + h) − f (a )
f ( a ) =
(a ) = lim
dx
h
h→0
17
Example: If the distance( in meters) traveled by a moving
object as a function of time ( in seconds) is given by:
s( t ) = 2t + 3
2
Find the average velocity of the object during the first three
seconds, and its instantaneous velocity in third second.
Solution: s(t) = 2t2 + 3, then, x0 = 0, x0+h = 3, h = 3,
hence
s(x0) = s(0) = 3,
s(x0+h) = s(3) = 21,
s( 3) − s(0) 21 − 3 18
hence, the average velocity = 3 − 0 = 3 = 3 = 6 m / sec
The instantaneous velocity as a function of time is s( t ) = 4t
Therefore, the instantaneous velocity in the third second is
s( 3) = 12 m / sec
18
Thank you
20