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Differentiation
An important topic in economics is the study of how quickly quantities change over time e.g. GDP
growth, inflation etc.
The mathematical concept used to describe the rate of change of a function is the derivative.
Differentiating a Linear Function
Functions of the form y = nx + b are called linear, as their pictorial representation is given by a
straight line.
The rate of change (derivative) for a linear function is simply the slope of a straight line.
If
Example
Differentiate y = 5x + 2
Interpreting the Slope
Example
Data was collected on the depth of a dive of penguins and the duration of the dive. The following
linear model is a fairly good summary of the data, where x is the duration of the dive in minutes and y
is the depth of the dive in yards.
The equation for the model is y = 2.915x + 0.015
This means that for every 1 minute increase in the duration of the dive, we predict the depth of the
dive will increase by approximately 2.915 yards.
Calculating the Slope
A straight line has a constant slope, or in other words, the rate of change of y with respect to x is a
constant. It can be calculated using the coordinates of two points on it.
Let point A on a line be (x1, y1) and point B be (x2, y2). The slope (m) is given by:
If the slope (m) and the coordinates of one point (x0, y0) are known, then the equation of a straight line
can be given as:
If two lines (let line 1 = ε1and line 2 = ε2) are parallel to each other (ε1 || ε2), then there slopes are
equal:
mε1 = mε2
If two lines (ε1and ε2) are perpendicular to each other (ε1 ⊥ ε2), then:
mε1 * mε2 = -1
Example
Find the equation of the line A, which goes through point (1, 2) and is perpendicular to the line Z: y =
2x + 1
Slope of line Z = 2
mA * 2 = -1
mA = -1/2 = -0.5
Equation of Line A
y = mx + (y0 – mx0)
y = -0.5x + (2 – (-0.5*1)
y = -0.5x + 2.5
Example
Find the equation of the line A, which goes through point (-1, 3) and is parallel to the line Z: y = -0.5x
+3
Slope of Z = -0.5
Slope of A = -0.5
Equation of Line A
y = mx + (y0 – mx0)
y = -0.5x + (3 – (-0.5*-1)
y = -0.5x + 2.5
Example
Find the equation of the line which goes through the points (1, -1) and (2, -3)
Equation of Line
y = mx + (y0 – mx0)
y = -2x + (-1 – (-2*1)
y = -2x + 1
However, not all functions are linear.
Differentiating Powers of X
How do you differentiate functions of the form y = xn? The number n can be a positive integer, a
negative integer, a fraction or even zero. If
Example: When n is a positive integer.
Differentiate y = x6
Example: When n is a fraction.
Differentiate y = x1/2
Example: When n is a negative integer.
Differentiate y = x-1
Example: When n = 1
Differentiate y = x
We obtain the result that the derivative of y = x is simply 1, as we would expect since y = x is the
equation of a straight line with a slope of 1.
Differentiation of a Function Multiplied by a Constant
If
y = kf(x)
where k is a constant
then
This means that we can differentiate a constant multiple of a function, simply by differentiating the
function and multiplying by the constant.
Example
Suppose we want to differentiate y = 6x3 − 12x4 + 5
Example
Suppose we want to differentiate y = x − 5x5 + 6x7 + 25.
The Quotient Rule
Functions often come as quotients, by which we mean one function divided by another function.
For example,
We write this as
where we identify u as x2 + 6 and v as 2x – 7
There is a formula we can use to differentiate a quotient - it is called the quotient rule.
The quotient rule: If
Example
Suppose we want to differentiate
u = x2 + 6
v = 2x – 7
The Chain Rule
Suppose we wish to differentiate y = (2x − 5)10
Immediately we note that this is different from the straightforward function, 2x – 5 or 2x - 510.
We call such an expression a function of a function.
Suppose, in general, that we have two functions, f(x) and g(x). Then
y = f(g(x))
is a function of a function. In our case, the function f is the outer function i.e. ()10 and the function g is
the inner function i.e (2x – 5)
In order to differentiate a function of a function, y = f(g(x)), we need to do two things:
1. Substitute u = g(x). This gives us
y = f(u)
2. Use the Chain Rule
Example
Suppose we wish to differentiate y = (2x − 5)10
u = 2x – 5
y = f(u)
y = u10
Example
Suppose we wish to differentiate y = (3x + 1)2
f = ()2
g = 3x + 1
u = 3x + 1
y = f(u)
y = u2
Example
Suppose we wish to differentiate
f = ()1/2
g = 13x2 – 5x + 8
u = 13x2 – 5x + 8
y = f(u)
y = u1/2
The Product Rule
Sometimes we are given functions that are actually products of other functions. This means, two
functions multiplied together. So, an example would be
So here we have one function, x3, multiplied by a second function, (4 – x)1/2
We can write this as y = uv where u = x3 and v = (4 – x)1/2
The product rule: if
Example
Suppose we want to differentiate
u = x3
v = (4 – x)1/2
v is a function of a function so we need to use the chain rule on that first
f = ()1/2
u=4–x
v = f(u)
v = u1/2
u = x3
v = (4 – x)1/2
Tidying up
We now put everything over a common denominator. To do this, multiply the second term, top
and bottom, by 2(4 x)1/2:
We present the solution in a tidy form so that if we need to work with it somewhere else we can.
Differentiating Logarithm and Exponential Functions
Rule 1
If y = ln x then
Rule 2
If y = ln f(x) then
Rule 3
If y = ex then
Note: The exponential function (and multiples of it) is the only function which is equal to its
derivative.
Laws of logarithms
logA + logB = logAB
logA − logB = logA/B
mlogA = logAm
We can use logarithms to simplify certain functions before we differentiate them.
Example
Suppose we want to differentiate y = ln(3x4 + 7)5.
Using the laws of logarithms (mlogA = logAm ) we can write y as
y = 5 ln(3x4 + 7)
This is more straightforward to differentiate because we have y = ln f(x) with f(x) = 3x4 + 7.
Rule 2 above tells us that
If y = ln f(x) then
Example
Suppose we want to differentiate
y = ln (1 − 3x)/(1 + 2x)
Using the laws of logarithms (logA − logB = logA/B) we can write y as
y = ln (1 – 3x) – ln (1 + 2x)
This is more straightforward to differentiate because we have y = ln f(x) - ln f(x) with f(x) = 1-3x and
f(x) = 1 + 2x
Rule 2 above tells us that
If y = ln f(x) then
We can write the answer as a single term by writing them both over a common denominator: