Calculus Unit 1: Basic Derivatives
1-1 The Power Rule
The derivative of a single algebraic term is obtained by using the Power Rule.
Given a basic algebraic term: axn
The derivative of the term is: naxn-1
Exception: the derivative of a constant term is zero
A constant term is zero-degree: ax0
The derivative of a constant term is: 0•ax0-1 = 0
Examples:
a) f (x) = 3x4 f ' (x) = 12x3
b) g (x) = -15 • 5Öx3 f ' (x) = -9 x-2/5
c) y = -13 y'=0
Exercises:
Find the derivative of each of the following functions.
a) f(x) = 32 b) f(x) = x4
c) y = x12 d) y = -3.724
e) f(x) = x f) f(x) = xe
g) f(x) = x43 h) f(x) = 25
i) g(x) = x-2 j) g(x) = x3/2
k) f(x) = 8x12 l) f(x) = -9x0
m) f(t) = 3t4/3 n) g(t) = 8t-3/4
o) y = x-4 p) y = 2/x2
q) g(t) = (2t)3 r) h(y) = (y/3)3
s) f(x) = 3Öx t) f(x) = 3Öx2
u) f(x) = 1/Öx v) y = 3/ 4Öx
Answers:
a) 0 b) 4x3 c) 12x11 d) 0 e) 1 f) exe-1 g) 43x42
h) 0 i) -2x-3 j) 3x1/2/2 k) 96x11 l) 0 m) 4t1/3
n) -6t-7/4 o) -4/x5 p) -4/x3 q) 24t2 r) y2/9 s) x-2/3/3
t) 2x-1/3/3 u) -x-3/2/2 v) -3x-5/4/4
Calculus Unit 1: Basic Derivatives
1-2 The Sum and Difference Rules
The derivative of a function that is a combination of sums or differences of
algebraic terms is obtained by using the following rules:
1. Sum Rule: f(x) = p(x) + q(x) Þ f '(x) = p ' (x) + q ' (x)
2. Difference Rule: f(x) = p(x) - q(x) Þ f '(x) = p ' (x) - q ' (x)
Examples:
a) f (x) = 3x4 + 5x7 f ' (x) = 12x3 + 35x6
b) g(x) = 3x2 ( 4x - 12x3 ) g ' (x) = 36x2 - 180x4
Exercises:
Use the sum and difference rules to find the derivatives of the
following functions.
a) f(x) = x2 + 4x b) f(x) = 3x5 - 6x4 + 2
c) g(x) = x10 + 25x5 - 50 d) g(x) = x2 - 2x-2
e) h(x) = x1/2 - 5x4 f) h(x) = (x -1)(x + 6)
g) y = (x + 1)x-1/2 h) y = t5 - 6t-5
Answers:
a) 2x + 4 b) 15x4 - 24x3 c) 10x9 + 125x4 d) 2x + 4x-3
e) 1/(2x1/2) - 20x3 f) 2x + 5 g) x-1/2/2 - x-3/2/2 h) 5t4 + 30t-6
Calculus Unit 1: Basic Derivatives
1-3 The Product Rule
The derivative of a function that is a product of algebraic terms is obtained by
using the following rule:
Product Rule: f(x) = p(x) • q(x) f '(x) = p ' (x) • q (x) + p (x) • q ' (x)
Example: f (x) = (2x5 + 6)(3x7 - 13x)
f ' (x) = (10x4)(3x7 - 13x) + (2x5 + 6)(21x6 - 13)
= 72x11 + 126x6 - 156x5 - 78
Verify this result by expanding f(x) and then using the sum and difference
rules. Compare your answer with f ' (x) in its expanded form.
Exercises:
Use the Product Rule to find the derivatives of the following functions.
a) y = x3(x2 + 2x + 3) b) y = x-2(x3 - 3x2 + 6)
c) f(x) = (1 - x2)(2 - x3) d) f(x) = (3x3 + 4)(1 - 2x3)
e) f(t) = (6 + t-2)(8t10 - 5t3) f) f(t) = (at + b)(ct2 - d)
g) g(u) = Öu (2 - u2 + 5u4) h) g(v) = (v - Öv)(v2 + Öv)
Answers:
. a) 5x4 + 8x3 + 9x2 b) 1 - 12x-3 c) 5x4 - 3x2 - 4x d) - 36x5 - 15x2
e) 480t9 + 64t7 - 90t2 - 5 f) 3act2 + 2bct - ad
g) 45u7/2/2 - 5x3/2/2 + u-1/2 h) 3v2 - 5v3/2/2 + 3v1/2/2 - 1
Calculus Unit 1: Basic Derivatives
1-4 The Quotient Rule
The derivative of a function that is a quotient of algebraic terms is obtained by
using the following rule:
Product Rule: f(x) = p(x) f '(x) = p' (x) • q (x) - p (x) • q ' (x)
q(x) [ q(x) ] 2
Example
f (x) = (x5 - 3x4) f ' (x) = (x2 + 7x)(5x4 - 12x3) - (x5 - 3x4)(2x + 7)
(x2 + 7x) (x2 + 7x)2
Verify this result by expressing f(x ) as (x5 -3x4)(x2 + 7x)-1 and then using the
product rule. Simplify your answer by finding a common denominator. Compare
your answer with f ' (x) in its expanded form.
Exercises:
Use the Quotient Rule to find the derivatives of the following functions.
a) f(x) = (x - 1)/(x + 1) b) f(x) = (2x - 1)/(x2 + 1)
c) g(x) = x/(x2 + 2x - 1) d) g(x) = (x3 - 1)/(x2 + x + 1)
e) y = Öx /(x2 + 1) f) y = (Öx + 2)/(Öx - 2)
g) f(t) = (2t + 1)/(t2 - 3t + 4) h) g(t) = (2t2 + 3t + 1)/(t - 1)
Answers:
a) 2/(x + 1)2 b) (-2x2 + 2x + 2)/(x2 + 1)2 c) (-x2 - 1)/(x2 + 2x - 1)2
d) 1 e) (1 - 3x2)/[2x1/2(x2 + 1)2] f) -2/[x1/2(x1/2 - 2)2]
g) (-2t2 - 2t + 11)/(t2 - 3t + 4)2 h) (2t2 - 4t - 4)/(t - 1)2
Calculus Basic Derivatives
1-5 The Chain Rule
The derivative of a composite function is obtained by using the Chain Rule.
Given a composite function: y = f o g ( x ) or f ( g ( x ) )
The derivative is: y ‘ = f ‘ ( g ( x ) )• g ‘ ( x )
Examples:
a) y = (5x3 + 3x)5 note: f(x) = x5 and g(x) = 5x3 + 3x
f ‘ (x) = 5x4 and g ‘ (x) = 15x2 + 3
y ‘ = 5(5x3 + 3x)4 (15x2 + 3)
b) y = ( x3 - 4x )1/2 note: f(x) = (x)1/2 and g(x) = x3 - 4x
f ‘ (x) = x-1/2 and g ‘ (x) = 3x2 - 4
2
y ‘ = (x3 - 4x) -1/2 (3x2 - 4) or y‘= (3x2 - 4)
2 2(x3-4x)1/2
Exercises:
Use the Chain Rule to find the derivatives of the following functions.
a) f(x) = (x2 + 1)10 b) y = (x4 + x2)1/2
c) f(x) = (3x6 + 2x)1/2 d) y = [ x / (x + 1) ]4
e) y = [ (x2 + 1) / (x2 - 1) ]1/2 f) g(x) = [ (2x - 5) / (5x + 2) ]1/4
Answers:
a) 20x(x2 + 1)9 b) (2x3 + x) (x4 + x2)-1/2 c) (9x5 + 1) (3x6 + 2x)-1/2
d) 4x3 / (x + 1)5 e) -2x / [ (x2 + 1)1/2 (x2 - 1)3/2 ]
f) 29 / [ 4 (5x + 2)5/4 (2x - 5)3/4 ]
Calculus Basic Derivatives
1-6 Higher Derivatives
A derivative of a derivative can be taken repeatedly. The notation used is:
the second derivative of f(x) is f''(x) = d2y
dx2
Examples:
a) Find the first ten derivatives of y = x6
y'= 6x5 y'' = 30x4 y'''= 120x3
y''''= 360x2 y'''''= 720x y''''''= 720
y'''''''= 0 y'''''''''' = 0
The tenth derivative is d10y = 0
dx10
Exercises - Find the second derivatives of the following functions:
1 4
1) y = 2t - 2) y= 3) y = (2x + 1)8
t +1 x
4 t
4) y = t3 + 5) y = x2 + 1 6) y=
t3 t -1
Answers:
2
1) - 2) 3x-5/2 3) 224(2x + 1)6
(1+ t)3
2
4) 6t + 12t-5 5) (x2 + 1)-3/2 6)
(t - 1)3
Calculus Basic Derivatives
1-7 Implicit Derivatives
A function may be expressed EXPLICITLY : y = 2x + 3
(dependent variable isolated)
A function may also be expressed IMPLICITLY : 2x – y + 3 = 0
(dependent variable is not isolated)
For implicitly defined functions, the derivative may be found by first isolating the
dependent variable using algebra, or by using implicit differentiation
Example: find y’ if x2 + y2 = 25
i) differentiate both sides: d (x2 + y2) = d (25)
dx dx
d (x2) + d (y2) = 0
dx dx
Use the chain rule 2x + 2y d (y) = 0
dx
2x + 2yy’ = 0
ii) isolate y’ y’ = -x
y
Exercises – find the derivatives of the following functions:
1) xy = 4 2) x3 + y3 = 6 3) x2 + xy + y2 = 1
2x
4) 2x5 + x4y + y5 = 36 5) x + y =1 6) y=
x+y
Answers:
y + 2x
1) -y/x 2) -x2/y2 3) -
x + 2y
4x 3 y + 10x 4 2y
4) - 5) -(y/x)1/2 6)
x 4 + 5y 4 (x + y)2 + 2x
