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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Standard Derivatives
1.
y = xn ;dy
dx= nxn−1
2.
y = ex ;dy
dx= ex
3.
y = ekx ;dy
dx= kekx
4.
y = ln x ;dy
dx=
1
x
5.
y = sin x ;dy
dx= cos x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Standard Derivatives
6.
y = cos x ;dy
dx= − sin x
7.
y = tan x ;dy
dx= sec2 x
8.
y = sec x ;dy
dx= sec x tan x
9.
y = cot x ;dy
dx= −cosec2x
10.
y = cosec2x ;dy
dx= −cosecx cot x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Symbols
dy
dx,
d2y
dx2,
d3y
dx3, · · ·
fx fxx , fxxx , · · ·
f ′(x), f ′′(x), f ′′′(x), · · ·
y ′, y ′′, y ′′′, · · ·
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
The General Rule
Giveny = xn
thendy
dx= nxn−1
d
dx(x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example
If y = x8, finddy
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= 8x7
d
dx(x) = 8x7(1)
= 8x7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
If n is a positive integer, then
d
dx(xn) = nxn−1 · d
dx(x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Examples
1.d
dx(x5) = 5x4
2.d
dx(x12) = 12x11
3.d
dx(x) = 1x0 = 1
4.d
dx(c) = 0;
d
dx(12) = 0
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
5.d
dx(4x8) = 4
d
dx(x8) = 4(8x7) = 32x7
6.
d
dx(x4 − 6x11) =
d
dx(x4)− d
dx(x11)
= 4x3 − 6 · 11x10
= 4x3 − 66x10
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Using The Power Rule to find derivvatives of functions
d
dx(X n) = nX n−1 · d
dx(X ), where X is a function.
Given f (u) = Un .Then
d
dx(Un) =
d
dx(Un)
du
dx
= nUn−1 du
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 1
Findd
dx(x2 − 1)50
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
d
dx(x2 − 1)50 = 50(x2 − 1)49
d
dx(x2 − 1)
= 50(x2 − 1)49(2x)
= 100x(x2 − 1)49
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 2
Findd
dx(x2 − 1)4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
d
dx(x2 − 1)4 = 4(x2 − 1)3
d
dx(x2 − 1)
= 4(x2 − 1)3(2x)
= 8x(x2 − 1)3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
EXERCISE 1
y = x13
Solution
dy
dx= 13x12
EXERCISE 2
y = (5x)8
Solution
dy
dx= 8(5x)7(5) = 40(5x)7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 3
Findy = (x2 + 4)5
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
d
dx(x2 + 4)5 = 5(x2 + 4)4
d
dx(x2 + 4)
= 5(x2 + 4)4(2x)
= 10x(x2 + 4)4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 4
Findy = 3
√(1 + x2)4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y =(1 + x2
) 43
d
dx=
4
3
(1 + x2
) 13d
dx(1 + x2)
=4
3(1 + x2)
13 (2x)
=8
3(1 + x2)
13
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 5
Findy =
√(2x + 7)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = (2x + 7)12
d
dx=
1
2(2x + 7)−
12d
dx(2x)
=1
2(2x + 7)−
12 (2)
= (2x + 7)−12
=1√
2x + 7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 6
y = (7 + 3x)5,dy
dx=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= 5(7 + 3x)4
d
dx(7 + 3x)
= 5(7 + 3x)4(3)
= 15(7 + 3x)4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 7
y = (2x − 3)−2,dy
dx=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= −2(2x − 3)−3
d
dx(2x − 3)
= −2(2x − 3)4(2)
= −4(2x − 3)−3
= − 4√(2x − 3)3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 8
y = (3x2 + 5)−3,dy
dx=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= −3(3x2 + 5)−4
d
dx(3x2 + 5)
= −3(3x2 + 5)−4(6x)
= − 18x√(3x2 + 5)4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 9
y = (x3 − 3x2 + 7x − 3)4,dy
dx=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= 4(x3 − 3x2 + 7x − 3)−4
d
dx(x3 − 3x2 + 7x − 3)
= 4(x3 − 3x2 + 7x − 3)3(3x2 − 6x + 7)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 10
f (x) = 7x5 − 3x4 + 6x2 + 3x + 4, f ′(x) =?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f ′(x) = 7(5)x4 − 3(4)x3 + 6(2)x + 3
= 35x4 − 12x3 + 12x + 3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 10
f (x) = 4x6 + 2x5 − 7x2 + 2x + 5
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
fx = 24x5 + 10x5 − 14x + 2
fxx = 120x4 + 40x3 − 13
fxxx = 480x3 + 120x2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercises
Finddy
dx
1.y = −8x5 +
√3x3 − 7x
2.y = −7x6 +
√3x2 + 2πx
3.2x50 + 3x12 − 14x2 +
3√
7x +√
5
4.y = x
13
5.(3x2 + 5)−4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
LEVEL 2Exercise 1
If
f (x) = x6 +1
3√x2
, find f ′(x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f (x) = x6 + x−23
f ′(x) = 6x5 +
(−2
3
)x−
53
= 6x5 −(
2
3
)x
53
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 2
If
f (x) = 7x3 +10√x
, find f ′(9)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f (x) = 7x3 + 10x−12
f ′(x) = 21x2 + 10
(−1
2x−
32
)= 21x2 +
(−5x−
32
)= 21x2 − 5x−
32
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
f ′(9) = 21(92)− 5(9)−32
= 21(81)− 5
27
= 1708− 5
27= 1700.815
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 3
Find the derivative of y =1
(4x2 − 3)5
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y =1
(4x2 − 3)5= (4x2 − 3)−5
dy
dx= −5(4x2 − 3)−6
d
dx(4x2 − 3)
= −5(4x2 − 3)−6(8x)
= −40x(4x2 − 3)−6
= − 40x
(4x2 − 3)6
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 4
Find the derivative of y = 3√
3x − 2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = 3√
3x − 2 = (3x − 2)13
dy
dx=
1
3(3x − 2)−
23d
dx(3x − 2)
=1
3(3x − 2)−
23 (3)
= (3x − 2)−23
=1
3√
(3x − 2)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 5
Finddy
dxif y =
3√x− 2√x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y =3√x− 2√x = 3x
12 − 2x
12
dy
dx= 3
(−1
2
)x−
32 − 2
(1
2
)x−
12
=
(−3
2
)x−
32 − x−
12
= − 3
2x( 32 )− 1
x( 12 )
= − 3
2√x3− 1√
x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 6
Find the derivative of y = (2x5 − 4x3 − x)3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = (2x5 − 4x3 − x)3
dy
dx= 3(2x5 − 4x3 − x)2
d
dy(2x5 − 4x3 − x)
= 3(2x5 − 4x3 − x)2(10x4 − 12x2 − 1)
= 3(10x4 − 12x2 − 1)(2x5 − 4x3 − x)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 7
If y = 1000 + 8x − 150
x2, find
dy
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = 1000 + 8x − 150x−2
dy
dx= 0 + 8− 150(−2x−3)
= 8 +300
x3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 8
Differentiate f (x) =5
x2− 6
xand find f ′
(1
2
)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f (x) = 5x−2 − 6x−1
f ′(x) = −10x−3 + 6x−2
=10
x3+
6
x2
f ′(1
2) = − 10(
1
2
)3 +6(1
2
)2
= −80 + 24
= −56
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 9
If y =4
3x2 − x + 5,find y ′
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y ′ = 4d
dx(3x2 − x + 5)−1
= 4(−1)(3x2 − x + 5)−2d
dx(3x2 − x + 5)
= −4(3x2 − x + 5)−2(6x − 1)
=−4(6x − 1)
(3x2 − x + 5)2=
4(1− 6x)
(3x2 − x + 5)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 10
Given y = 3√
(4x2 + 3)2,, finddy
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
d
dx(4x2 + 3)
23
=2
3(4x2 + 3)−
13 · d
dx(4x2 + 3)
=2
3
(1
4x2 + 3
) 13
(8x)
=16x
3(4x2 + 3)13
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 11
Find the derivative of y = (6x5 − 4x3 − 5)7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = (6x5 − 4x3 − 5)7
dy
dx= 7(6x5 − 4x3 − 5)6
d
dx(6x5 − 4x3 − 5)
= 7(6x5 − 4x3 − 5)6(30x4 − 12x2)
= 7(30x4 − 12x2)(6x5 − 4x3 − 5)6
= 7(6)(5x4 − 2x2)(6x5 − 4x3 − 5)6
= 42x2(5x2 − 2)(6x5 − 4x3 − 5)6
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 12
Find the derivative of y =2
4√x3 − x2 − x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y =2
4√x3 − x2 − x
= 2(x3 − x2 − x)−14
dy
dx= 2
(−1
4
)(x3 − x2 − x
)− 54d
dy(x3 − x2 − x)
= −1
2
(x3 − x2 − x
)− 54 (3x2 − 2x − 1)
= − (3x2 − 2x − 1)
2(x3 − x2 − x)54
=1 + 2x − 3x2
2( 4√x3 − x2 − x)5
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Exercise 13
d
dx
(x +
1
x
)−3=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
d
dx
(x +
1
x
)−3= −3(x +
1
x)−4
d
dx(x +
1
x)
= −3(x +1
x)−4(1− 1
x2)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
If u and v are differentiable functions of x , thend
dx(uv) = u
dv
dx+ v
du
dx
⇒ If we are asked to differentiate the product of a function:
(a) We may have to keep the first function constant and differentiate thesecond function plus
(b) Keep the second function constant and differentiate the first.
Note: If u, v and w are differentiable functions of x , thend
dy(uvw) = uv
dw
dx+ uw
dv
dx+ vw
du
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Proof:
d
dy(uvw) =
d
dx[(uv)w ] = uv
dw
dx+ w
d
dx(uv)
= uvdw
dx+ w [u
dv
dx+ v
du
dx]
= uvdw
dx+ uw
dv
dx+ vw
du
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 1
Differentiate y = (x2 + 1)(x4 − 2x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = (x2 + 1)(x4 − 2x)
dy
dx= (x2 + 1)
d
dx(x4 − 2x) + (x4 − 2x)
d
dx(x2 + 1)
= (x2 + 1)(4x3 − 2) + (x4 − 2x)(2x)
= 4x5 − 2x2 + 4x3 − 2 + 2x5 − 4x2
= 6x5 + 4x3 − 6x2 − 2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 2
d
dx
(x +
1
x
)−3=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
d
dx
(x +
1
x
)−3= −3(
(x +
1
x
)−4d
dx
(x +
1
x
))
= −3(
(x +
1
x
)−4(1− 1
x2
)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 3
Finddy
dxif y = x5(3x3 − 2x + 1)3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = x5(3x3 − 2x + 1)3
= x5d
dx(3x3 − 2x + 1)3 + (3x3 − 2x + 1)3
d
dx(x5)
= x5 · 3(3x3 − 2x + 1)2d
dx(3x3 − 2x + 1) + (3x3 − 2x + 1)3(5x4)
= 3x5(3x3 − 2x + 1)2(9x2 − 2) + 5x4(3x3 − 2x + 1)3
= x4(3x3 − 2x + 1)2[3x(9x2 − 2) + 5(3x3 − 2x + 1)
]= x4(3x3 − 2x + 1)2(27x3 − 6x + 15x3 − 10x + 5)
= x4(3x3 − 2x + 1)2(42x3 − 16x + 5)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 4
Calculatedy
dxif y = (2x4 − 1)4(x8 − 3x2)7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
y = (2x4 − 1)4(x8 − 3x2)7
dy
dx= (2x4 − 1)4
d
dx(x8 − 3x2)7 + (x8 − 3x2)7
d
dx(2x4 − 1)4
= (2x4 − 1)4 · 7(x8 − 3x2)6dy
dx(x8 − 3x2)
+ (x8 − 3x2)7 · 4(2x4 − 1)3dy
dx(2x4 − 1)
= 7(2x4 − 1)4(x8 − 3x2)6(8x7 − 6x) + 4(2x4 − 1)3(x8 − 3x2)7(8x3)
= (2x4 − 1)3(x8 − 3x2)6[7(2x4)(8x7 − 6x) + 32x3(x8 − 3x2)
]= (2x4 − 1)3(x8 − 3x2)6(112x11 − 84x5 − 56x7 + 42x + 23x11− 96x5)
= (2x4 − 1)3(x8 − 3x2)6(144x11 − 56x7 − 180x5 + 42x)
= (2x4 − 1)3(x8 − 3x2)6 · x(144x10 − 56x6 − 180x4 + 42)
= x(2x4 − 1)3(x8 − 3x2)6(144x10 − 56x6 − 180x4 + 42)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 5
Finddy
dxif y = (2x3 − 1)5
√x − 1
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= (2x3 − 1)5
d
dx(x − 1)
12 + (X − 1)
12d
dx(2x3 − 1)5
= (2x3 − 1)51
2((x − 1)−
12 )
d
dx(x − 1) + (x − 1)
12 · 5(2x3 − 1)4
d
dx(2x3 − 1)
= (2x3 − 1)5 · 1
2((x − 1)−
12 )(1) + (x − 1)
12 · 5(2x3 − 1)4(6x2)
=1
2(2x3 − 1)5 · ((x − 1)−
12 ) + (x − 1)
12 (2x3 − 1)4(30x2)
=
1
2(2x3 − 1)5
(x − 1)12
+ (30x2)(x − 1)12 (2x3 − 1)4
=(2x3 − 1)5
2√x − 1
+ (30x2)(2x3 − 1)4√x − 1
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 6
Finddy
dxif y = x2(2x − 1)(6x + 5)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= x2(2x − 1)
d
dx(6x + 5) + x2(6x + 5)
d
dx(2x − 1)
+ (2x − 1)(6x + 5)d
dx(x2)
= x2(2x − 1)(6) + x2(6x + 5)(2) + (2x − 1)(6x + 5)(2x)
= 6x2(2x − 1) + 2x2(6x + 5) + (2x − 1)(12x2 + 10x)
= 48x3 + 12x2 − 10x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 7
Given that y = (3x − 1)2(2x2 − 3)(x3 + 4)5 ;dy
dx=?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= (3x − 1)2(2x2 − 3)
d
dx(x3 + 4)5 + (3x − 1)2(x3 + 4)5
d
dx(2x2 − 3)
+ (2x2 − 3)(x3 + 4)5d
dx(3x − 1)2
= (3x − 1)2(2x2 − 3) · 5(x3 + 4)4d
dx(x3 + 4) + (3x − 1)2(x3 + 4)5(4x)
+ (2x2 − 3)(x3 + 4)5(2)(3x − 1)d
dx(3x − 1)
= (3x − 1)2(2x2 − 3) · 5(x3 + 4)4(3x2) + 4x(3x − 1)2(x3 + 4)5
+ 2(2x2 − 3)(x3 + 4)5(3x − 1) · 3= 15x2(3x − 1)2(2x2 − 3)(x3 + 4)4 + 4x(3x − 1)2(x3 + 4)5
+ 6(2x2 − 3)(x3 + 4)5(3x − 1)
= (3x − 1)(x3 + 4)4[15x2(2x2 − 3)(2x + 1) + 4x(3x − 1)(x3 + 4)
+ 6(2x2 − 3)(x3 + 4)]
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 8
Finddy
dxif y = (8x3 − x2)5(7x2 − 3x)2(x5 − 2)10
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= (8x3 − x2)5(7x2 − 3x)2
d
dx(x5 − 2)10
+ (8x3 − x2)5(x5 − 2)10d
dx(7x2 − 3x)2
+ (7x2 − 3x)2(x5 − 2)10d
dx(8x3 − x2)5
dy
dx= (8x3 − x2)5(7x2 − 3x)2 · 10(x5 − 2)9
d
dx(x5 − 2)
+ (8x3 − x2)5(x5 − 2)10 · 2(7x2 − 3x)d
dx(7x2 − 3x)
+ (7x2 − 3x)2(x5 − 2)10 · 5(8x3 − x2)4d
dx(8x3 − x2)
dy
dx= (8x3 − x2)5(7x2 − 3x)2 · 10(x5 − 2)9 · 5x4
+ (8x3 − x2)5(x5 − 2)10 · 2(7x2 − 3x)(14x − 3)
+ (7x2 − 3x)2(x5 − 2)10 · 5(8x3 − x2)4(24x2 − 2x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= 50x4(7x2 − 3x)2(8x3 − x2)5(x5 − 2)9
+ 2(8x3 − x2)5(x5 − 2)10(7x2 − 3x)(14x − 3)
+ 5(7x2 − 3x)2(x5 − 2)10(8x3 − x2)4(24x2 − 2x)
dy
dx= (7x2 − 3x)(8x3 − x2)4(x5 − 2)9[50x4(8x3 − x2)(7x2 − 3x)
+ 2(x5 − 2)(8x3 − x2)(14x − 3) + 5(x5 − 2)(7x2 − 3x)2(24x2 − 2x)]
dy
dx= (7x2 − 3x)(8x3 − x2)4(x5 − 2)9[(400x7 − 50x6)(7x2 − 3x)
+ (2x5 − 4)(8x3 − x2)(14x − 3) + (5x5 − 10)(7x2 − 3x)2(24x2 − 2x)]
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 9
Finddy
dxif y = (2x5 − 3x3)4(5x − 4x2)3(x6 − 6x2)7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= (2x5 − 3x3)4(5x − 4x2)3
d
dx(x6 − 6x2)7
+ (2x5 − 3x3)4(x6 − 6x2)7d
dx(5x − 4x2)3
+ (5x − 4x2)3(x6 − 6x2)7d
dx(2x5 − 3x3)4
dy
dx= (2x5 − 3x3)4(5x − 4x2)3 · 7(x6 − 6x2)6
d
dx(x6 − 6x2)
+ (2x5 − 3x3)4(x6 − 6x2)7 · 3(5x − 4x2)2d
dx(5x − 4x2)
+ (5x − 4x2)3(x6 − 6x2)7 · 4(2x5 − 3x3)3d
dx(2x5 − 3x3)
dy
dx= 7(2x5 − 3x3)4(5x − 4x2)3(x6 − 6x2)6(6x5 − 12x)
+ 3(2x5 − 3x3)4(x6 − 6x2)7(5x − 4x2)2(5− 8x)
+ 4(5x − 4x2)3(x6 − 6x2)7(2x5 − 3x3)3(10x4 − 9x2)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= 7(6x5 − 12x)(2x5 − 3x3)4(5x − 4x2)3(x6 − 6x2)6
+ 3(5− 8x)(2x5 − 3x3)4(x6 − 6x2)7(5x − 4x2)2
+ 4(2x5 − 3x3)3(10x4 − 9x2)(5x − 4x2)3(x6 − 6x2)7
dy
dx= (2x5 − 3x3)3(5x − 4x2)2(x6 − 6x2)6[7(6x5 − 12x)(2x5 − 3x3)(5x − 4x2)
+ 3(5− 8x)(2x5 − 3x3)4(x6 − 6x2) + 4(10x4 − 9x2)(5x − 4x2)(x6 − 6x2)]
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Note:If f and g are differentiable at x , thend
dx[f (x)g(x)] = f (x)
d
dx[g(x)] + g(x)
d
dx[f (x)]
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 10
Finddy
dxif y = (4x2 − 1)(7x3 + x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
d
dx[(4x2 − 1)(7x3 + x)]
= (4x2 − 1)d
dx(7x3 + x) + (7x3 + x)
d
dx(4x2 − 1)
= (4x2 − 1)(21x2 + 1) + (7x3 + x)(8x)
= 140x4 − 9x2 − 1
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 11
If f (x) = (x2 + x)(2x3 − 3) find f ′(x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f ′(x) = (x2 + x)d
dx(2x3 − 3) + (2x3 − 3)
d
dx(x2 + x)
= (x2 + x)(6x2) + (2x3 − 3)(2x + 1)
= 10x4 + 8x2 − 6x − 3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 12
If g(x) = x15 (x − 1)
35 , find g ′(x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
g ′(x) = x15 · 3
5(x − 1)−
25 + (x − 1)
35 · 1
5x−
45
=3x
15
5(x − 1)25
+(x − 1)
35
5x45
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 13
f (x) = x2(1− 3x3)13 , f ′(x) =?
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f ′(x) = x2d
dx(1− 3x3)
13 + (1− 3x3)
13
(d
dx(x2)
)= x2
[1
3(1− 3x3)−
23 )
d
dx(1− 3x3)
]+ (1− 3x3)
13 (2x)
=1
3x2(1− 3x3)−
23 )(−9x2) + 2x(1− 3x3)
13
= −3x4(1− 3x3)−23 + 2x(1− 3x3)
13
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
= x(1− 3x3)−23
[−3x2 + 2(1− 3x3)
]= x(1− 3x3)−
23 (2− 9x3)
=x(2− 9x3)
(1− 3x3)−23
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
If f (x) =u(x)
v(x), where u and v are differentiable functions of x , then
f ′(x) =v(x)u′(x)− u(x)v ′(x)
[v(x)]2
where v(x) 6= 0
Thus if u and v are differentiable function of x and y =u
v, where v 6= 0,
then
dy
dx= y ′ =
vdu
dx− u
dv
dxv2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 1
If y =x + 1
x − 1, find
dy
dx
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
(x − 1)d
dx(x + 1)− (x + 1)
d
dx(x − 1)
(x − 1)2
=(x − 1)(1)− (x + 1)(1)
(x − 1)2=
x − 1− x − 1
(x − 1)2
=−2
(x − 1)2= − 2
(x − 1)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 2
Finddy
dxif y =
x3
x2 + 1
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
(x2 + 1)d
dx(x3)− x3
d
dx(x2 + 1)
(x2 + 1)2
=(x2 + 1)(3x2)− x3(2x)
(x2 + 1)2
=3x4 + 3x2 − 2x4
(x2 + 1)2
=x4 + 3x2
(x2 + 1)2
=x2(x2 + 3)
(x2 + 1)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 3
Calculate y ′ when y =2x5 − x3
3x4 − 2x
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
(3x4 − 2x)d
dx(2x5 − x3)− (2x5 − x3)
d
dx(3x4 − 2x)
(3x4 − 2x)2
=(3x4 − 2x)(10x4 − 3x2)− (2x5 − x3)(12x3 − 2)
(3x4 − 2x)2
=30x8 − 9x6 − 20x5 + 6x3 − 24x8 + 4x5 + 12x6 − 2x3
(3x4 − 2x)2
=6x8 + 3x6 − 16x5 + 4x3
(3x4 − 2x)2
=x3(6x5 + 3x3 − 16x2 + 4)
(3x4 − 2x)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 4
Finddy
dxif y =
(3x2 − x)4
(x3 − 2x4)3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
(x3 − 2x4)3d
dx(3x2 − x)4 − (3x2 − x)4
d
dx(x3 − 2x4)3
(x3 − 2x4)3·2
=(x3 − 2x4)34(3x2 − x)3
d
dx(3x2 − x)− (3x2 − x)43(x3 − 2x4)
d
dx(x3 − 2x4)
(x3 − 2x4)6
=4(x3 − 2x4)3(3x2 − x)3(6x − 1)− 3(3x2 − x)4(x3 − 2x4)(3x2 − 8x3)
(x3 − 2x4)6
=(x3 − 2x4)2(3x2 − x)3
[4(6x − 1)(x3 − 2x4)− 3(3x2 − 8x3)(3x2 − x)4
](x3 − 2x4)6
=(3x2 − x)3
(x3 − 2x4)4(24x5 − 19x4 + 5x3)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 5
Finddy
dxif y =
(2x − 3
5x2 + 1
)7
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx= 7
(2x − 3
5x2 + 1
)6
(5x2 + 1)d
dx(2x − 3)− (2x − 3)
d
dx(5x2 + 1)
(5x2 + 1)2
= 7
(2x − 3
5x2 + 1
)6 [(5x2 + 1)(2)− (2x − 3)(10x)
(5x2 + 1)2
]= 7
(2x − 3
5x2 + 1
)6 [10x2 + 2− 20x2 + 30x
(5x2 + 1)2
]= 7
(2x − 3
5x2 + 1
)6(2 + 30x − 10x2
(5x2 + 1)2
)=
7(2x − 3)6(2 + 30x − 10x2)
(5x2 + 1)6(5x2 + 1)2
=14(1 + 15x − 5x2)(2x − 3)6
(5x2 + 1)8
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 6
Calculated
dx
[x3 − 1
5√
4x3 + 3
]
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
d
dx
[x3 − 1
5√
4x3 + 3
]=
5√
4x3 + 3d
dx(x3 − 1)− d
dx(4x2 + 3)
15
( 5√
4x3 + 3)2
=
5√
4x3 + 3(3x2)− (x3 − 1) · 1
5(4x2 + 3)−
45d
dx(4x2 + 3)
( 5√
4x3 + 3)2
=3x2 5√
4x3 + 3− 1
5(x3 − 1)(4x2 + 3)−
45 (8x)
( 5√
4x3 + 3)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
=3x2 5√
4x3 + 35√
4x3 + 3)2− 8x(x3 − 1)
5· 1
5√
4x3 + 3)2· 1
(4x2 + 3)
4
5
=3x2
5√
(4x3 + 3)− 8x(x3 − 1)
5· 1
5√
4x3 + 3)2· 1
5√
(4x3 + 3)4
=3x2
5√
(4x3 + 3)− 8x(x3 − 1)
5 5√
(4x3 + 3)6
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 7
Find y ′ when y =
√x − 1
x3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
dy
dx=
x3d
dx(x − 1)
12 − (x − 1)
12d
dx(x3)
(x3)2
=x3 · 1
2(x − 1)−
12d
dx(x − 1)− (x − 1)
12 (3x2)
x6
=
1
2x3(x − 1)−
12 (1)− (x − 1)
12 (3x2)
x6
=
1
2x3(x − 1)−
12 − (x − 1)
12 (3x2)
x6
=x3 − 6x2(x − 1)
2x6(x − 1)12
=6x2 − 5x3
2x6√x − 1
=x2(6− 5x)
2x6√x − 1
=(6− 5x)
2x4√x − 1
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 8
If g(t) =t3
1− t4find g ′(t)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
g ′(t) =(1− t4)(3t2)− t3(−4t3)
(1− t4)2
=3t2 − 3t6 + 4t6
(1− t4)2
=3t2 + t6
(1− t4)2
=t2(3 + t4)
(1− t4)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 9
Given that f (x) =2x + 3
x2 + 1, find f ′(x)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f ′(x) =(x2 + 1)(2)− (2x + 3)(2x)
(x2 + 1)2
=2x2 + 2− 4x2 − 6x
(x2 + 1)2
=2x2 − 6x + 2
(x2 + 1)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
EXERCISES
1.f (x) = x2(3x2 − 1)
2.y = 6x4(x3 + 2x)
3.f (x) = (1− x2)(1 + x2)
4.
g(x) =1
x3(x2 − x)
5.
y =
(2
x+ 1
)(1
x2− 3
)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
EXERCISES
6.g(x) = (3x4 − 1)(5x2 − 6x + 11)
7.s = (t2 + 1)(2t2 − t + 3)
8.y = (1−
√x)(2 +
√x)
9.
f (x) =2
x3
10.y =
x
x + 1
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
EXERCISES
11.
y =2x
x + 1
12.
f (x) =x − 3
2x + 5
13.
g(x) =5x + 1
x2 − 1
14.
y =x2
x3 + 4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
15.
y =x2 − x + 1
x2 + 2x − 3
16.
g(x) =7x4 + 11
x − 2
17.
f (x) =
(1 +
1
x
)(1 +
1
x2
)
18.
f (x) =(9x8 − 8x9
)(x +
1
x
)
19.
g(x) =2x2 + 1
x + 5
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
20.
f (x) =x4
4− x3
3+
x2
2
21.
y =4x3 + 1
x3 − 1
22.
y =
(x
1 + x
)(2− x
3
)
23.
f (x) = (2x − 3)
(2x − 3
x
)
24.
f (x) = (7x3 − 4x2 + 2)
1
4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
For each negative integer n ,f (x) = xn has the derivativef ′(x) = nxn−1
In particular
f (x) = x−1; f ′(x) = (−1)x−2 = −x−2
f (x) = x−2; f ′(x) = (−2)x−3 = − 2
x3
f (x) = x−3; f ′(x) = (−3)x−3 = − 3
x4
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example
If f (x) =5
x2− 6
x, find f ′(x) and f ′(
1
2)
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution
f (x) = 5x−2 − 6x−1
f ′(x) = 5(2)x−3 − 6(−1)x−2
= −10
x3+
6
x2
f ′(1
2) = − 10
(1
2)3
+6
(1
2)2
= −80 + 24 = −56
Note: Because y =1
xncan be expressed as y = x−n, all problems where
quotient rule is applicable, can be converted to products so that youapply product rule. Here, it is a matter of choice or preference.
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Outline
1 INTRODUCTION
2 SOLVED EXERCISES
3 THE PRODUCT RULE
4 WORKED EXAMPLES
5 QUOTIENT RULE
6 WORKED EXAMPLES
7 EXERCISES
8 DERIVATIVE OF NEGATIVE POWERS
9 WORKED EXAMPLES
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 1
Differentiate :(i)(x2 − 2)(x + 3)−2 as a product
(ii)x2 − 2
(x + 3)2as a quotient
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution (i)
Let y = (x2 − 2)(x + 3)−2
dy
dx= (x2 − 2)
d
dx(x + 3)−2 + (x + 3)−2
d
dx(x2 − 2)
= (x2 − 2) · −2(x + 3)−3d
dx(x + 3) + (x + 3)−2(2x)
=−2(x2 − 2)
(x + 3)3(1) +
2x
(x + 3)2
=−2x2 + 4 + 2x(x + 3)
(x + 3)3
=−2x2 + 4 + 2x2 + 6x
(x + 3)3
=4 + 6x
(x + 3)3=
2(2 + 3x)
(x + 3)3
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution (ii)
y =x2 − 2
(x + 3)2
dy
dx=
(x + 3)2d
dx(x2 − 2)− (x2 − 2)
d
dx(x + 3)2
[(x + 3)2]2
=2x(x + 3)2 − 2(x + 3)(x2 − 2)(1)
(x + 3)4
=(x + 3)
[2x(x + 3)− 2(x2 − 2)
](x + 3)4
=2x2 + 6x − 2x2 + 4
(x + 3)3=
6x + 4
(x + 3)3=
2(2 + 3x)
(x + 3)3
NB: On comparing the results of (i) and (ii), you will notice that final
results are the same . Hence , y =x2 − 2
(x + 3)2= (x2 − 2)(x + 3)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Example 2
Differentiate :(i)(x − 1)3(x3 − 1)−1 as a product
(ii)(x − 1)3
(x3 − 1)as a quotient
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution (i)
y = (x − 1)3(x3 − 1)−1
dy
dx= (x − 1)3
d
dx(x3 − 1)−1 + (x3 − 1)−1
d
dx(x − 1)3
= (x − 1)3 · (−1)(x3 − 1)−2d
dx(x3 − 1) + (x3 − 1)−1 · 3(x − 1)2
d
dx(x − 1)
= −(x − 1)3(x3 − 1)−2(3x2) + 3(x3 − 1)−1(x − 1)2(1)
=−3x2(x − 1)3
(x3 − 1)2+
3(x − 1)2
(x3 − 1)
=−3x2(x − 1)3 + (x3 − 1)2
(x3 − 1)2
=(x − 1)2
[−3x2(x − 1) + 3(x3 − 1)
](x3 − 1)2
=(x − 1)2(−3x2 + 3x2 + 3x3 − 3)
(x3 − 1)2=
(3x2 − 3(x − 1)2)
(x3 − 1)2=
3(x + 1)(x − 1)3
(x3 − 1)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
Solution (ii)
y =(x − 1)3
(x3 − 1)
dy
dx=
(x3 − 1)d
dx(x − 1)3 − (x − 1)3
d
dx(x3 − 1)
(x3 − 1)2
=(x3 − 1) · 3(x − 1)2
d
dx(x − 1)− (x − 1)3(3x2)
(x3 − 1)2
=3(x3 − 1)(x − 1)2(1)− (x − 1)3(3x2)
(x3 − 1)2
=3(x − 1)2[(x3 − 1)− x2(x − 1)]
(x3 − 1)2
=3(x − 1)2[(x3 − 1)− x2(x − 1)]
(x3 − 1)2
INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES
EXERCISE
Differentiate :(i) as a product(ii) as a quotient
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