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DIFFERENTIATION Differentiation is about rates of change.
Differentiation is all about finding rates of change of one quantity compared to another. We need differentiation when the rate of change is not constant.
Constant Rate of Change Let's take an example of a car travelling at a constant 60 km/h. The
distance-time graph would look like this:
We notice that the distance from the starting point increases at a constant rate of 60 km each hour, so after 5 hours we have travelled 300 km.
We notice that the slope (gradient) is always 300/5 = 60 for the whole graph. There is a constant rate of change of the distance compared to the time. The slope is positive all the way (the graph goes up as you go left to right along the graph.)
Rate of Change that is Not Constant Now let's throw a ball straight up in the air. Because gravity acts on the
ball it slows down, then it reverses direction and starts to fall. All the time during this motion the velocity is changing. It goes from positive (when the ball is going up), slows down to zero, then becomes negative (as the ball is coming down).
Cont’d Notice this time that the slope of the graph is changing throughout the
motion. At the beginning, it has a steep positive slope (indicating the large velocity we give it when we throw it). Then, as it slows, the slope get less and less until it become 0 (when the ball is at the highest point and the velocity is zero). Then the ball starts to fall and the slope becomes negative (corresponding to the negative velocity) and the slope becomes steeper (as the velocity increases).
Important Concept – Approximations (limit) of the Slope Notice that if we zoom in close enough to a curve, it begins to look like a
straight line. We can find a very good approximation to the slope of the curve at the point t = 1 (it will be the slope of the tangent to the curve, marked in dark red) by observing the points that the curve passes through near t = 1. (A tangent is a line that touches the curve at one point only.)
Cont’d Observing the graph, we see that it passes through (0.9, 36.2) and (1.1,
42). So the slope of the tangent at t = 1 is about:
The units are m/s, as this is a velocity. We have found the rate of change by looking at the slope.
Tangent line Clearly, if we were to zoom in closer, our curve would look even
more straight and we could get an even better approximation for the slope of the curve.
This idea of "zooming in" on the graph and getting closer and closer to get a better approximation for the slope of the curve (thus giving us the rate of change) was the breakthrough that led to the development of differentiation.
The slope of the straight line through is given byax
afxfm
ax
)()(lim
))(,( afa
Example 1: Find the equation of the tangent line to the curve , at the point .)4,1(
Solution
1
)1()(lim
)()(lim
1
x
fxf
ax
afxfm
xax
1
)1)1(41()14(lim
1
)1()(lim
22
11
x
xx
x
fxfm
xx
2
1 1
1
4 3 ( 1)( 3)lim lim
1 1lim( 3) 2
x x
x
x x x x
x xx
The derivative (or differentiation) of a function at a point using tangent line, is defined as:
provided the limit exists and is a small increment.
The process of getting the tangent line is called
( ) ( )'( ) lim
x a
f x f af a
x a
0
( ) ( )limh
f a h f a
h
differentiation from the first principles
differentiation using definition
Example 2 Find the derivate of at using the First Principle Method.xxxf 4)( 2 7x
Solution
ax
afxfaf
ax
)()(lim)('
7
))7(47(4lim
7
)7()(lim)7('
22
77
x
xx
x
fxff
xx
7
)3)(7(lim
7
214lim
7
2
7
x
xx
x
xxxx
10)3(lim7
xx
Example 3 Find the derivative of the function , using the First Principle.
3)( xxf
Solution
h
xhx
h
xfhxfxf
hh
33
00
)(lim
)()(lim)('
h
xhxhhxxh
33223
0
33lim
h
hxhhxh
322
0
33lim
22
033lim hxhx
h
23x
Find the derivatives of the following functions using definitions.
xxf2
1)(
xxf 3)(
2
2( ) ,
3f x
x 2x
Differentiation Formulas or where c is
constant function.0)( c
dx
d 0)( cf f(x) = 180
1)( nn nxxdx
d 1)( nn nxxf or where n is any number either integer or rational
10)( xxf
)()( xfdx
dcxcf
dx
d or where c is constant function. )()( xfcxcf ( ) 2sinf x x
)()()()( xgdx
dxf
dx
dxgxf
dx
d
)()()()( xgdx
dxf
dx
dxgxf
dx
d
)()()()()()( xfdx
dxgxg
dx
dxfxgxf
dx
d 2( ) (1 )f x x x
2)(
)()()()(
)(
)(
xg
xgxfxfxg
xg
xf
( )x
f xx
18 23( ) 3 2f x x x x 4
52
1 2( ) 3 1
3f x x
xx
Power Rule
6)3( xy
325 )1234( xxxy
12 2 xy
)(')]([)]([ 1 xfxfnxfdx
d nn
5 5' 6( 3) ( 3) ' 6( 3)y x x x
5 2 2 5 2
5 2 2 4 1
' 3(4 3 2 1) (4 3 2 1) '
3(4 3 2 1) (20 6 2)
y x x x x x x
x x x x x
2 1/2
2 1/2 2
2
' [(2 1) ]'
1(2 1) (2 1) '
22
(2 1)
y x
x x
x
x
Exercise4 3 12(8 5) ( 3)y x x
5
8
(3 1)
(2 )
xy
x
5
12 )13(
2
x
y
)(')]([)]([ 1 xfxfnxfdx
d nn
Derivatives of Trigonometric Functions
(sin ) cosd
x xdx
(Derivative of the sine function)
xxdx
dsin)(cos (Derivative of the cosine function)
xxdx
d 2sec)(tan
)tan()sec()(sec xxxdx
d
)(csc)(cot 2 xxdx
d
)cot()csc()(csc xxxdx
d
(Derivative of the tangent function)
(Derivative of the secant function)
(Derivative of the cotangent function)
(Derivative of the cosecant function)
Example
2( ) sin(3 )f x x x 2
2 2
2
'( ) [sin(3 )]'
cos(3 )(3 ) '
(6 1)cos(3 )
f x x x
x x x x
x x x
Example 12Differentiate of following functions
)csc(4)cot()sin(5)( xxxxy
)csc(4))(sin()cot()cot()sin(5)( xdx
dx
dx
dxx
dx
dxxy
SOLUTION
5 sin( )( csc( )) cot( )(cos( )) 4 csc( )cot( )x x x x x x
5cos( )cot( ) 5csc( ) 4 csc( )cot( )x x x x x
)()()()()()( xfdx
dxgxg
dx
dxfxgxf
dx
d
)cos(23
)sin()(
t
tty
Differentiate of following functions
SOLUTION
2)cos(23
)cos(23)sin()sin()cos(23)(
t
tdx
dtt
dt
dt
ty
2
3 2cos( ) cos( ) sin( ) 2sin( )
3 2cos( )
t t t t
t
2
3cos( ) 2
3 2cos( )
t
t
2)(
)()()()(
)(
)(
xg
xgxfxfxg
xg
xf
Logarithmic Differentiation
xx
dx
d 1)(ln
dx
du
uu
dx
d 1)(ln
)(log1
ln
1)(log e
xaxx
dx
daa
)(xfu where
Example 13Differentiate the following functions with
respect to x.27ln xy
Solution
)7(7
1 22
xdx
d
xdx
dy
2
14 2
7
x
x x
dx
du
uu
dx
d 1)(ln )(xfu where
)23ln(2)( 23 xxxg
Differentiate the following functions with respect to x.
3 2 2 3'( ) 2 ln(3 2) ln(3 2) (2 )d d
g x x x x xdx dx
3 2 22
62 ( ) 6 ln(3 2)
3 2
xx x x
x
4
2 22
12( ) 6 ln(3 2)3 2
xx x
x
(using the product rule)
Solution
)()()()()()( xfdx
dxgxg
dx
dxfxgxf
dx
d
dx
du
uu
dx
d 1)(ln
6)23ln( xy
66
)23()23(
1
x
dx
d
xdx
dy
56
1(6)(3 2) (3)
(3 2)x
x
)23(
18
x
Differentiate the following functions with respect to x.
Solutiondx
du
uu
dx
d 1)(ln
)(')]([)]([ 1 xfxfnxfdx
d nn
Differentiations of the exponential function
xx eedx
d)(
dx
duee
dx
d uu )(
)1(,ln)( aaaadx
d xx
)1()(,)ln()( axfuwheredx
duaaa
dx
d uu
)(xfu where
Example 14
Find for each of the following exponential function23xey
Solution
2
2
3
23
6
)3(
x
x
xe
xdx
de
dx
dy
dx
duee
dx
d uu )(
Find for each of the following exponential function
Solution
1123 xx eey
2 1 1(3) (2) . ( 1)x xdy dye e x
dx dx
dx
duee
dx
d uu )(
2 1 1 1 1(6) .
2 1x xe e
x
12 16
2 1
xx e
ex
Find for each of the following exponential function
Solution
127 xy
2 1(7 ) ln 7 (2 1)xdy dux
dx dx
( ) ( ln )u ud dua a a
dx dx
2 1(2)(7 ) ln 7x
Chain RuleSuppose that we have two functions f(x) and g(x) and they are both differentiable.
1) If we define then the derivative of F(x) is,
2) If we have y=f(u) and u=g(x) then the derivative of y is,
))(()( xgfxF
( ) ( ( )) ( )F x f g x g x
dx
du
du
dy
dx
dy
EXAMPLE 9
Find if 1)( 2 xxf)(xf
Solution
Let )(12 xux
So, uufuy )()( then we get u
uufdu
dy
2
1
2
1)( 2
1
Also xxudx
du2)(
By using Chain Rule we get,
112
22
2
122
x
x
x
xx
udx
du
du
dy
dx
dy
Chain RuleSuppose that we have two functions f(x) and g(x) and they are both differentiable.
1) If we define then the derivative of F(x) is,
2) If we have y=f(u) and u=g(x) then the derivative of y is,
))(()( xgfxF
( ) ( ( )) ( )F x f g x g x
dx
du
du
dy
dx
dy
EXAMPLE 9
Find if 1)( 2 xxf)(xf
Solution
Let )(12 xux
So, uufuy )()( then we get u
uufdu
dy
2
1
2
1)( 2
1
Also xxudx
du2)(
By using Chain Rule we get,
112
22
2
122
x
x
x
xx
udx
du
du
dy
dx
dy
2( ) sin(3 )f x x x cos( ) 6 1
dy dy duu x
dx du dx
6( 3)y x 5 56 1 6( 3)dy dy du
u xdx du dx
23u x x
siny u
3u x
6y u
2(6 1)(cos(3 ))x x x
22 1y x 1/21
42
dy dy duu x
dx du dx
5 2 3(4 3 2 1)y x x x 2 43 (20 6 2)
dy dy duu x x
dx du dx
5 24 3 2 1u x x x
3y u
22 1u x
1/2y u u
5 2 2 4
4 5 2 2
3(4 3 2 1) (20 6 2)
(60 18 6)(4 3 2 1)
x x x x x
x x x x x
2 1/21(2 1) 4
2x x
2
2
14
2 (2 1)
4
2 (2 1)
xx
x
x
4 2( 1) sin( 1)y x x 2( 1)u x
4( 1) siny x u
2 4( 1)u x
2( ) sinu u
2(2 sin cos ) 2( 1)dy dy du
u u u u xdx du dx
2 2( ) ' (sin ) (sin ) 'dy
u u u udu
2 siny u u
2( 1)u x
2( 1)du
xdx
22 sin cosdy
u u u udu
2 2 4 2(2( 1) sin( 1) ( 1) cos( 1) ) 2( 1)x x x x x
3 2 5 24( 1) sin( 1) 2( 1) cos( 1)x x x x
3 2 2 22( 1) 2sin( 1) ( 1) cos( 1)x x x x
Revisions : Find the derivation
3 5 2( ) ( 4 7)(3 6 )p x x x x x x
3 5 2 3 5 2'( ) ( 4 7) '(3 6 ) ( 4 7)(3 6 ) 'p x x x x x x x x x x x
)()()()()()( xfdx
dxgxg
dx
dxfxgxf
dx
d
2 5 2 3 4(3 4)(3 6 ) ( 4 7)(15 2 6)x x x x x x x x
2
2
3 5( )
2 3
xf x
x x
2 2 2 2
2 2
(2 3)(3 5) ' (3 5)(2 3) ''( )
(2 3)
x x x x x xf x
x x
2)(
)()()()(
)(
)(
xg
xgxfxfxg
xg
xf
2 2
2 2
(2 3)(6 ) (3 5)(4 1)
(2 3)
x x x x x
x x
( )cos
th t
t
1/2
'( )cos
d th t
dt t
1/2 1/2
2
cos cos
cos
d dt t t t
dt dtt
2)(
)()()()(
)(
)(
xg
xgxfxfxg
xg
xf
1/2 1/2
2
1cos sin
2cos
t t t t
t
1/2 1/2
2
1cos sin
2cos
t t t t
t
1/2
2
1(cos 2 sin )
2cos
t t t t
t
2
(cos 2 sin )
2 cos
t t t
t t
( ) 3 secf
'( ) (3 sec )d
fd
3 sec sec 3d d
d d
3 sec tan 3sec
)()()()()()( xfdx
dxgxg
dx
dxfxgxf
dx
d
sec sec tand
d
ln( )
sin
xf
x
2
(sin ) (ln ) (ln ) (sin )'( )
sin
d dx x x x
dx dxf xx
2
1(sin )( ) (ln )(cos )
sin
x x xx
x
2
(sin ) (ln )(cos )
sin
x x x x
x x
2)(
)()()()(
)(
)(
xg
xgxfxfxg
xg
xf
Implicit differentiationsThe equation
Is define explicitly as for
Is define implicitly as for andwhen y is differentiable function of x.
2( ) 1y x x
21y x 1 1x
2 2 1x y 0 1y 1 1x
