BA201 ENGINEERING MATHEMATICS 2 2012 28 CHAPTER 2 DIFFERENTIATION 2.1 FIRST ORDER DIFFERENTIATION What is Differentiation? 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. Notation for the Derivative IMPORTANT: The derivative (also called differentiation) can be written in several ways. This can cause some confusion when we first learn about differentiation. The following are equivalent ways of writing the first derivative of y = f(x): ' ' or ( ) or dy f x y dx 2.1.1 RULES OF DIFFERENTIATION A. Derivative of Power Function n y ax So 1 n dy nax dx Examples: 1. Find the derivative of y = -7x 6 Note: We can do this in one step: We can write: OR y' = -42x 5
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BA201 ENGINEERING MATHEMATICS 2 2012
28
CHAPTER 2 DIFFERENTIATION
2.1 FIRST ORDER DIFFERENTIATION
What is Differentiation?
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.
Notation for the Derivative
IMPORTANT: The derivative (also called differentiation) can be written in several ways. This can cause some confusion when we first learn about differentiation.
The following are equivalent ways of writing the first derivative of y = f(x):
' ' or ( ) or dy
f x ydx
2.1.1 RULES OF DIFFERENTIATION
A. Derivative of Power Function
ny ax
So 1ndynax
dx
Examples:
1. Find the derivative of y = -7x6
Note: We can do this in one step:
We can write: OR y' = -42x5
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Example 1 Find the derivative for each of the following function.
a) 3y x b) 52y x c) xy 4 d) 2
1y
x
e) 2y x f)
2
3y x g) 3
2
3y
x h)
5 3
2y
x
i) 1
yx
j) 3
5y
x
k) y x l)
2
3
3
2
y
x
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B. Derivative of a Constant Function
y a
So 0dy
dx
Example Find the derivative for each of the following functions.
a) 1y b) y c) 40y d) 1
3y
2.1.2 THE DERIVATIVE OF SUMMATION AND SUBSTRACTION
If ( ) and ( ) are differentiable functions, the derivative of
y f x g x y f x g x
and
' 'dy
f x g xdx
' 'dy
f x g xdx
Examples:
1. Find the derivative of y = 3x5 - 1
y = 3x5 − 1
Now,
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And since we can write:
So,
2. Find the derivative of
Now, taking each term in turn:
(using )
(using )
(since -x = -(x1) and so the derivative will be -(x0) = -1)
(since )
So
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Example
Find the following derivatives;
a) 42 6y x x
b) 3 32y x x
x c) 33 4s t t d) 2 2 3p q q
e) 8 41 1
34 2
y x x
f)
4 29 5y x x x
f)
1 4 2y x
x g) 3 21 2y x x
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Exercise
Find the derivative of the following function;
i. 4 2f x x x
ii. 116 9f x x
iii. 8 5y x x
iv. 3 1y x
x
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2.1.3 THE DERIVATIVES OF COMPOSITE FUNCTION
Chain Rule
If y f u , where, u is a function of x, so:
dy dy du
dx du dx
This means we need to
1. Recognise u (always choose the inner-most expression, usually the part inside brackets, or under the square root sign).
2. Then we need to re-express y in terms of u. 3. Then we differentiate y (with respect to u), then we re-express everything in
terms of x.
4. The next step is to find du
dx.
5. Then we multiply dy
du and
du
dx.
Example 1:
Differentiate each the following function with respect to x.
i. y = (x2+ 3)5
In this case, we let u = x2 + 3 and then y = u5.
We see that:
u is a function of x and y is a function of u.
For the chain rule, we firstly need to find and .
So
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ii.
In this case, we let u = 4x2 − x and then .
Once again,
u is a function of x and y is a function of u.
Using the chain rule, we firstly need to find:
and
So
i. 3
4y x
ii. 1
4 9y
x
iii. 2 4y x
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The Extended Power Rule
An extension of the chain rule is the Power Rule for differentiating. We are finding the derivative of un (a power of a function):
k
ny ax b
1
11
kn n
kn n
dy dyk ax b ax b
dx dx
dykan ax b
dx
Example:
1.
In the case of we have a power of a function.
If we let u = 2x3 - 1 then y = u4.
So now
y is written as a power of u; and u is a function of x [ u = f(x) ].
To find the derivative of such an expression, we can use our new rule:
where u = 2x3 - 1 and n = 4.
So
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We could, of course, use the chain rule, as before: dx
du
du
dy
dx
dy*
a) 3
4y x
b) 3
5 4y x c) 2 4y x
d) 8
3 6 7y x
e)
5
12
xy
f)
1
7
xy
g) 5
1 3y
x
h)
5
2
9 4y
x
i)
7 23 24 3 2y x x x
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2.1.4 DERIVATIVE OF A PRODUCT FUNCTION
If u and v are two functions of x, then the derivative of the product uv is given by...
In words, this can be remembered as:
"The derivative of a product of two functions is the first times the derivative of the second, plus the second times the derivative of the first."
Example:
If we have a product like
y = (2x2 + 6x)(2x3 + 5x2)
we can find the derivative without multiplying out the expression on the right.
We use the substitutions u = 2x2 + 6x and v = 2x3 + 5x2.
We can then use the PRODUCT RULE:
We first find: and
Then we can write:
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Exercise:
a) 2 34 2 3 5y x x x
b) 2
1 1y x x c) 1 2y x x
d) 10 3y x x
e) 4
3 28 1y x x f) 5 6
5 1 5y x x
2.1.5 DERIVATIVE OF A QUOTIENT FUNCTION
(A quotient is just a fraction.)
If u and v are two functions of x, then the derivative of the quotient u/v is given by...
In words, this can be remembered as:
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"The derivative of a quotient equals bottom times derivative of top minus top times derivative of the bottom, divided by bottom squared."
Example:
1. We wish to find the derivative of the expression: