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Differentiation

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Differentiation

Question 6 & Question 7

What 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.

We 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.

- 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?

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.

⇒

This is the formula we use to differentiate from first principles.

Hence in order to differentiate from first principles we must:

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

2001: Question 6 (b) (ii)

Find from first principles the derivative of with respect to x.

Solution

Multiply above and below by

2004: 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 =

(Dividing top & bottom by 2)

We know = 1

- 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!!

2001: Question 6 (b) (ii)

Give that y = what is

Solution

Differentiation 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 Rule

This is the rule we use when we have a function within a function

Examples: sin(5x); ln (3x2+1); (x2 + 5x + 6)

The rule states:

Rule 2: The Product Rule

This is the rule we use when we have two expressions multiplied by each other.

Examples: (x +1)(3x2– 6); e2xsin x

The rule states:

Given y = uv

Rule 3: The Quotient Rule

This is the rule we use when we have one expression divided by another Examples:

The rule states:

Given y =

2006: Question 6(a)

Differentiate ( with respect to x

Solution

In 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

2005: Question 6(b)

Let y = . Show that = t + t3,where t = tan

Solution

We can see that y involves us dividing one expression by another and hence we must use the Quotient Rule to find .

=

u = 1 – cosxv = 1 + cosx

sin x- sin x

Hence we must prove that = t + t3 = tan + tan3

x = +

= and 1 + cosx =

From the log tables we know that:

sin 2A = 2sin A cos A and cos 2A = cos2 A - sin2A

LHS =

1 - sin2A = cos2A

LHS = = =

= tan () and =

LHS = =

LHS =

=

= t + t3

2003: Question 6(a)

Differentiate with respect to x.

Solution

In 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 + 4xv = u1/2

4

2006: 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 = 1

If n = 1 then f (x) = x1 = x

When n = 1 we know that nxn-1 = (1)x1-1 = 1x0 = 1

L.H.S. = R.H.S and so we know the statement is true for n = 1

Now we assume that it is true for n = k i.e.

Finally we must prove it is true for n = k + 1

We are looking to prove that

From the product rule and the assumption we have made we know:

= = =

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:

.

There 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.

2006: Question 6(b)(i)

The parametric equations of the curve are:

x = 3cos –cos3

y = 3sin–sin3

Find and

Solution

x = 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

2006: Question 6(b)(ii)

Hence show

Solution

We know that =

= =

But remember the important trig identity: cos2x + sin2x = 1

Finally again referring back to our trig identities we know:

=

Hence =

- 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.

- 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.

- 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

2006: Question 6(b)

The equation of a curve is y = 3x4 – 2x3 – 9x2 + 8

- Show that the curve has a local maximum at (0, 8)
- Find the co-ordinates of the two local minimum points on the curve.
- Draw a sketch the curve.

Max and min points occur at = 0.

= 0

If (0,8) is a turning point then = 0

= 0 – 0 – 0 = 0

So 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)

To find the other turning points we must solve = 0

= 0

= 0

= 0 or = 0

= 0

(2x – 3)(x + 1) = 0

x = , x = -1

To find the corresponding y values we sub these x values back into our original function.

f () = = -3.8125

f () = = 4

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

- 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: Acceleration

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

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

Speed

= Acceleration

2004 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.

Solution

We know given s that speed =

s = 12 + 24t – 3t2

⇒ = 24 – 6t

⇒ At t = 3: = 24 – 6(3) = 6

Speed after 3 seconds = 6 m/s