Differentiation

Question 6 & Question 7

IntroductionWhat does the term derivative mean?

A derivative simply refers to the rate of change of one object in relation to another.

Where do we encounter rates of change in everyday life? If we look at the position of a sprinter we see that it changes over time, hence

speed is a real life example of a derivative. How quickly a disease, such as Swine Flu spreads over a given time is another

example of a derivative in everyday life. The change in the volume oil that leaks from a shipwrecked vessel over time

could also be calculated through the use of derivatives.

Rates of Change in MathematicsWe have also encountered rates of change in other strands of mathematics.

If you recall from co-ordinate geometry the formula used to find the slope of the line is: .

This formula represents another rate of change as we can see it is the change in our y values over the change in our x values.

⇒ The derivative can also allow us to find the slope of a straight line.

Slopes of Curves• Differentiation can also help us to find the slopes of curves.

• The slope of a curve is constantly change (unlike the slope of a straight line). Hence we cannot find a single slope and state that this is the slope of our entire curve.

• Instead we can only find the slope of a curve at a given point so what we do is find the slope of the tangent (a straight line) at this point. The derivative allows us to find an expression for the slope of a tangent at a given point.

• How is this possible?

Slope of Curves

http://www3.ul.ie/cemtl/Applets/Feb2008/diff1.html

What can we conclude from this?We know as h tends towards zero that the line PQ is the tangent to the curve. However we also know that the derivative is another way of writing the slope of the tangent to the curve.⇒

Differentiating from First Principles

This is the formula we use to differentiate from first principles.Hence in order to differentiate from first principles we must:

1. Find f(x + h) and subtract f(x) – the original function2. Divide this expression by h (You must divide every term)3. Apply the limit h → 0

Example 12001: Question 6 (b) (ii)Find from first principles the derivative of with respect to x.

Solution Multiply above and below by

Solution 1

Example 22004: Question 6 (b) (ii)Find from first principles the derivative of cos (x) with respect to x.

Solution From the trigonometric identities in our log tables we know that cos A – cos B =

Solution 2 (Dividing top & bottom by 2) We know = 1

Basic Differentiation• Differentiation from first principles is a long way of doing

something that is relatively straightforward..• Another way of finding the derivative of the function y = xn is:

• In English what we are doing is multiplying each term by the power and reducing the power by 1.

• But how do we prove that this short method works? We will look at this proof in a few minutes!!

Example2001: Question 6 (b) (ii)Give that y = what is Solution

Three Important RulesDifferentiation can often be slightly more complicated than the example we have just looked at and as a result we have three different rules that will help us out.Rule 1: The Chain RuleThis is the rule we use when we have a function within a functionExamples: sin(5x); ln (3x2+1); (x2 + 5x + 6)The rule states:

Three Important RulesRule 2: The Product RuleThis is the rule we use when we have two expressions multiplied by each other.Examples: (x +1)(3x2 – 6); e2xsin xThe rule states:

Given y = uvRule 3: The Quotient RuleThis is the rule we use when we have one expression divided by another Examples: The rule states:

Given y =

Example 12006: Question 6(a)Differentiate ( with respect to xSolutionIn this example we have two expressions multiplied together hence we know we must use the product rule.

Now that we have worked out these four we can apply the rule.

u = v = 1

Example 22005: Question 6(b)Let y = . Show that = t + t3, where t = tanSolutionWe can see that y involves us dividing one expression by another and hence we must use the Quotient Rule to find .

=

u = 1 – cos x v = 1 + cos x sin x - sin x

Example 2Hence we must prove that = t + t3 = tan + tan3 x = + = and 1 + cos x = From the log tables we know that:sin 2A = 2sin A cos A and cos 2A = cos2 A - sin2 A LHS = 1 - sin2 A = cos2 A LHS = = =

Example 2 = tan () and = LHS = = LHS = = = t + t3

Example 32003: Question 6(a)Differentiate with respect to x.SolutionIn this instance we have a function within a function hence we use the Chain Rule. We have the function 1 + 4x raised to the power of a half.y = (1 +4x)1/2

= 4 = = =

Let u = 1 + 4x v = u1/2

4

Proving the Differentiation Rule2006: Question 6 (c)Prove by induction that .Solution In order to prove this using induction we must first prove that it is true for n = 1If n = 1 then f (x) = x1 = x

Proof Ctd. When n = 1 we know that nxn-1 = (1)x1-1 = 1x0 = 1L.H.S. = R.H.S and so we know the statement is true for n = 1Now we assume that it is true for n = k i.e. Finally we must prove it is true for n = k + 1We are looking to prove that From the product rule and the assumption we have made we know: = = =

Proof Ctd.

Looking at the R.H.S we see that = Hence we see that the L.H.S. and R.H.S are equal and so our statement is true for k + 1.Since the statement is true for n = 1 and if it is true for n = k we know it is also true for n = k + 1 we have prove inductively that:

.

Parametric DifferentiationThere are times when a curve in the co-ordinate plane is not just expressed as a function of x but rather as two parametric equations. These equations are generally of the form:

x = x(t) and y = y(t) where t is called the parameter.

In order to find the derivative of a curve like this we apply the Chain Rule and say: However this can also be written as (Knowledge of fractions) and this is the method we will use.

Example 12006: Question 6(b)(i)The parametric equations of the curve are:x = 3cos –cos3 y = 3sin –sin3 Find and Solutionx = 3cos –cos3 = 3cos – (cos )3

= -3sin – 3(cos )2(- sin ) = 3cos2sin – 3sin y = 3sin – sin3 = 3sin – (sin )3

= 3cos – 3(sin )2(cos ) = 3cos – 3sin2cos

Example 12006: Question 6(b)(ii)Hence show SolutionWe know that = = = But remember the important trig identity: cos2 x + sin2x = 1

Example 1

Finally again referring back to our trig identities we know: = Hence =

Application of Differentiation # 1• We have already seen how differentiation can allow us to find

the slope of a curve and a line. However it has numerous other applications in real life.

• The first application that we will look at is in relation to maximum and minimum values.

• We often need to find maximum and minimum values in real life. For example businesses are often looking to maximise their profit, product designers often wish to minimise the amount of materials they use while government bodies wish to minimise the cost of projects.

• However to start off we will just look at finding the maximum and minimum turning points of a function.

Maximum and Minimum Turning Points

Maximum and Minimum Turning Points

• From this we can conclude that the slope of the tangent at the max and min points is equal to zero.

• But we also know that the slope of the tangent at the max and min points can be given by

• Hence we know at the max and min points = 0.

Classifying Maximum and Minimum Turning Points

• Once we find our turning points (by letting = 0) we are often asked to classify them i.e. say which one is the max turning point and which one is the min turning point.

• In order to classify turning points we say:

If @ x; then the point is a maximum

If @ x: > 0 then the point is a minimum

Classifying Maximum and Minimum Turning Points

Example

2006: Question 6(b)The equation of a curve is y = 3x4 – 2x3 – 9x2 + 8 (i) Show that the curve has a local maximum at (0, 8)(ii) Find the co-ordinates of the two local minimum points on

the curve.(iii) Draw a sketch the curve.

Solution (i) Max and min points occur at = 0. = 0If (0,8) is a turning point then = 0 = 0 – 0 – 0 = 0So we know (0,8) is a turning point and to find out if it is a maximum we find At x = 0: = -18 - 18 < 0 Max point occurs at (0, 8)

Solution (ii)To find the other turning points we must solve = 0 = 0 = 0 = 0 or = 0 = 0(2x – 3)(x + 1) = 0x = , x = -1To find the corresponding y values we sub these x values back into our original function.f () = = -3.8125

Solution (ii)

f () = = 4

Hence we know our minimum turning points are (1.5, -3.8125) and (-1, 4).

Solution (iii)

Application of Differentiation # 2• If we were to calculate the distance an object moves in a certain

amount of time what would we be calculating?Answer: Speed.

• If we were to calculate the change in speed of an object over a certain amount of time what would we be calculating?

Answer: AccelerationHence we can conclude that speed and acceleration are rates of change and so differentiation can be used when calculating these.

Application of Differentiation # 2

Given that s is the distance an object travels in t seconds we can say that:

Speed = Acceleration

Example2004 Question 7(a)An object’s distance from a fixed point is given by s =12 + 24t − 3t2, where s is in metres and t is in seconds. Find the speed of the object when t = 3 seconds.SolutionWe know given s that speed = s = 12 + 24t – 3t2⇒ = 24 – 6t⇒ At t = 3: = 24 – 6(3) = 6Speed after 3 seconds = 6 m/s