PowerPoint PresentationFEPS Mathematics Support Framework
11.1 Introduction
The aim of this unit is to review some of the basic rules of differentiation.
While studying these slides you should attempt the ‘Your Turn’ questions in the slides.
After studying the slides, you should attempt the Consolidation Questions.
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11.1 Learning objectives
At the end of the following lecture and seminar, you should be able to
11.1.1 Differentiate (by applying standard results) functions containing , ln, cos, sin, tan, including sums, differences and composites using the chain rule, e.g., = 2 cos
11.1.2 Differentiate products of different types of functions using the product rule, e.g., = cos.
11.1.3 Differentiate quotients of different types of functions using the
quotient rule, e.g., =
2
The straight lines below all have different gradients.
The gradient is a measure of how steep the slope is and in what direction.
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Consider one point on a curve:
At the point , the gradient of the curve equals the gradient of the tangent to the curve.
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Finding the gradient of a curve
We need a method to find the gradient function, given the function for the curve. This method is called differentiating, or finding the derivative.
Notation: If we use to denote the function of the curve, then ′ or
denotes the gradient function.
If we use () to denote the curve, then we use ′() to denote the gradient function.
Once we have found the derivative we can find a numerical value for the rate at which is changing with respect to , for any specific value of .
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If () = , then the derivative ′ = −1
E.g. given = 3 3 5 −
4
2 − 2 + 3, find ′ and the rate at which is
changing, w.r.t , at the point where = 1, hence find the equation of the tangent to the curve at = 1.
Step 1: covert all fractions/roots to index form
= 35/3 − 4−2 − 2 + 3
Step 2: differentiate term-by-term
′() = 5 2 3 + 8−3 − 2
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Step 3: Evaluate ′ at = 1, too get the rate of change (gradient of the curve) at = 1.
′ 1 = 5 1 2
3 + 8 1 −3 − 2 = 11
Step 4: The equation of the tangent at = 1 has the form = + . We have already found the gradient, = 11. We also need to find the corresponding −value (value of ()) at = 1
1 = 3(1)5/3−4 1 −2 − 2 1 + 3 = 0
Step 5: We now know the tangent line passes through (1, 0) and has a gradient of 11.
Using = + , at 1, 0 ; 0 = 11 1 + ⇒ = −11
∴ = 11 − 11 is the equation of the tangent at = 1
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Stationary points
A function () has stationary points at values of where ′() = 0. This means that at a stationary point, the gradient of the curve representing () is zero. The tangent to the curve at a stationary point is parallel to the axis.
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Maximum Point
Minimum Point
point which is non-
sketch
A
B
C
Example
Find the coordinates of the stationary points of () = 3 – 32 + 2
() = 3 – 32 + 2
′() = 32 – 6
At a stationary point ′() = 0
32 – 6 = 3 ( − 2) = 0, so = 0 or = 2
We need the coordinates of the points; (0) = 2 and (2) = −2
The coordinates of the stationary points are (0,2) and (2, −2)
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We can distinguish between the different types of points using the second derivative test
If
If
() = 3 – 32 + 2
′() = 32 – 6 and ′() = 0 when = 0 or = 2
′′() = 6 – 6
At = 0 ′′ 0 = −6 ∴ (0, 2) is a local maximum
At = 2 ′′ 2 = 6 ∴ (2, −2) is a local minimum
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11.1.1 Differentiation of exponential, log and trig functions and their composites.
We can define the derivative from the geometrical construct in the figure below. In the limiting case, as → 0, we have the derivative (or derived function) which gives us the gradient of the tangent at = .


Definition of the derivative
Derivatives of trigonometric functions
We are mostly concerned with applying standard results, as opposed to obtaining the standard results from first principles. will revisit these fundamental results in a little while and see how we can obtain further results by using some of the rules of differentiation.
() ′()
The Chain Rule for differentiation
Just to recap, the chain rule enables us to differentiate a composite
function (a function of a function).
If = then
= ′( ) ′()
If and are both differentiable at then
Alternative notation: If = and = () then
dy dy du
dx du dx =
= ln implies = and we know
= =
Using the chain rule, we obtain a useful relationship
=
1
Applying the chain rule to solving problems
We now have four new results which we can extend quite easily to a variety of problems, making further use of the chain rule. Here are some examples:
Differentiate = 3
let = 3, then =
d
d =
d
d
d
3
Don’t forget to write your final answer in terms of (the original independent variable)!
Another example
() = ln() , where = 3 − 4
′() = ′
let = 2 + , then = 4sin
d
d =
d
d
d
d
Your turn! (1)
2. Differentiate = ln( + + 5) w.r.t.
3. Differentiate = sin 2
2 +

Solutions
2.
11.1.2 Differentiation using the product rule
If and are functions of , and = , then
d[]
d =
′ = ′ + ′
This is the product rule for differentiation and you must learn how to apply it.
Example
1. Use the product rule to differentiate = ln cos
Let = ln, = cos
Product rule: ′ = ′ + ′
′ = 1
d
′ = , ′ = cos
Example
3. Given = 2 ln 2 + 5 , find ′
This is a bit more tricky as we have to use both the product rule and the chain rule.
= 2 , = ln 2 + 5
′ = 2 2 , ′ =
Example
4. Find the gradient of the tangent to the curve, given by () = 2cos(2) at (0,1).
Gradient of the tangent to the curve at (0,1) is given by ′ 0 .
= 2, = cos(2)
′ = −22 sin 2 + 22cos(2)
′ 0 = 0 + 20 cos 0 = 2
Your turn! (2)
1. = ( − 2)
2. = 2cos(4)
4. Find the gradient of the graph of the function
= ( + 3) at point (0,3).
Answers
2
3.
4. ′ = + 3 + , ′ 0 = 4
11.1.3 The quotient rule for differentiation
The quotient rule is useful for finding the derivatives of different types of
function expressed in fractional form, =
where and are functions of .
It is useful to know that a quotient can also be expressed in product form and
the product rule applied, since =
= −1.
where and are functions of ,
d
d

′ = ′ − ′
find ′

Note, we could have just written = −cos and applied the product rule.
Differentiating tan using the quotient rule
sin [tan ] [ ]
d
(cos 2)2
2.
+3 2
Additional questions
1. Find any stationary points for the curve = − 2
2. Use the second derivative test to determine the nature of the stationary
point of the curve = ln
2 > 0
1+cos then
2 ,
2
2 < 0 at = 1/2
So, the graph of attains its relative maximum at = 1/2.
Answers

= (1 − cos )(1 + cos ) + sin ( − sin )
(1 + cos )2
Then using the trig identity 1 − cos2 = sin2

=