Differentiation. Question 6 & Question 7. Introduction. 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?
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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?
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.
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:
2001: Question 6 (b) (ii)
Find from first principles the derivative of with respect to x.
Multiply above and below by
2004: Question 6 (b) (ii)
Find from first principles the derivative of cos (x)with respect to x.
From the trigonometric identities in our log tables we know that cos A – cos B =
(Dividing top & bottom by 2)
We know = 1
2001: Question 6 (b) (ii)
Give that y = what is
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
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 =
2005: Question 6(b)
Let y = . Show that = t + t3,where t = tan
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
1 - sin2A = cos2A
LHS = = =
= tan () and =
LHS = =
= t + t3
2003: Question 6(a)
Differentiate with respect to x.
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
2006: Question 6 (c)
Prove by induction that .
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
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)
We know that =
But remember the important trig identity: cos2x + sin2x = 1
Finally again referring back to our trig identities we know:
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
Max and min points occur at = 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 or = 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).
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:
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.
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