High School Mathematics Extensions/Supplementary/Differentiation

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Differentiate from first principle(otherwise known as differentialisation)Edit

This section and the *differentiation technique* section can be skipped if you are already familiar with calculus/differentiation.

In calculus, differentiation is a very important operation applied to functions of real numbers. To differentiate a function f(x), we simply evaluate the limit

 

where the   means that we let h approach 0. However, for now, we can simply think of it as putting h to 0, i.e., letting h = 0 at an appropriate time. There are various notations for the result of differentiation (called the derivative), for example

 

and

 

mean the same thing. We say, f'(x) is the derivative of f(x). Differentiation is useful for many purposes, but we shall not discuss why calculus was invented, but rather how we can apply calculus to the study of generating functions.

It should be clear that if   then   the above law is important. If g(x) a closed-form of f(x), then it is valid to differentiate both sides to obtain a new generating function.

Also if   This can be verified by looking at the properties of limits.

Example 1Edit

Differentiate from first principle f(x) where

 

Firstly, we form the difference quotient

 

We can't set h to 0 to evaluate the limit at this point. Can you see why? We need to expand the quadratic first.

 
 

We can now factor out the h to obtain now

 

from where we can let h go to zero safely to obtain the derivative, 2x. So

 

or equivalently:

 

Example 2Edit

Differentiate from first principles, p(x) = xn.

We start from the difference quotient:

 

By the binomial theorem, we have:

 

The first xn cancels with the last, to get

 

Now, we bring the constant 1/h inside the brackets

 

and the result falls out:

 

Important Result

If

 

then

 

As you can see, differentiate from first principle involves working out the derivative of a function through algebraic manipulation, and for that reason this section is algebraically very difficult.

Example 3Edit

Assume that if

 

then

 

Differentiate  

Solution Let  

 

Example 4Edit

Show that if

 
 

Solution

 

Example 5Edit

Differentiate from first principle

 

Solution

 

ExercisesEdit

1. Differentiate

 

2. Differentiate

 

3. Differentiate from first principle

 

4. Differentiate

 

5. Prove the result assumed in example 3 above, i.e. if

f(x)=g(x)+h(x)

then

f′(x)=g′(x)+h′(x).

Hint: use limits.

Differentiating f(z) = (1 - z)^nEdit

We aim to derive a vital result in this section, namely, to derive the derivative of

 

where n ≥ 1 and n an integer. We will show a number of ways to arrive at the result.

Derivation 1Edit

Let's proceed:

 

expand the right hand side using binomial expansion

 

differentiate both sides

 

now we use  

 

and there are some cancelling

 

take out a common factor of -n, and recall that 1! = 0! = 1 we get

 

let j = i - 1, we get

 

but this is just the expansion of (1 - z)n-1

 

Derivation 2Edit

Similar to Derivation 1, we use instead the definition of a derivative:

 

expand using the binomial theorem

 

factorise

 

take the limit inside (recall that [Af(x)]' = Af'(x) )

 

the inside is just the derivative of zi

 

exactly as derivation 1, we get

 

Example Differentiate (1 - z)2

Solution 1

f(z) = (1 - z)2 = 1 - 2z + z2
f'(z) = - 2 + 2z
f'(z) = - 2(1 - z)

Solution 2 By the result derived above we have

f'(z) = -2(1 - z)2 - 1 = -2(1 - z)

ExercisesEdit

Imitate the method used above or otherwise, differentiate:

1. (1 - z)3

2. (1 + z)2

3. (1 + z)3

4. (Harder) 1/(1 - z)3 (Hint: Use definition of derivative)

Differentiation techniqueEdit

We will teach how to differentiate functions of this form:

 

i.e. functions whose reciprocals are also functions. We proceed, by the definition of differentiation:

 


 

Example 1Edit

 

by

 

where g is a function of z, we get

 

which confirmed the result derived using a counting argument.


ExercisesEdit

Differentiate

1. 1/(1-z)2

2. 1/(1-z)3

3. 1/(1+z)3

4. Show that (1/(1 - z)n)' = n/(1-z)n+1

Differentiation applied to generating functionsEdit

Now that we are familiar with differentiation from first principle, we should consider:

 

we know

 

differentiate both sides

 


 

therefore we can conclude that

 

Note that we can obtain the above result by the substituion method as well,

 

letting z = x2 gives you the require result.

The above example demonstrated that we need not concern ourselves with difficult differentiations. Rather, to get the results the easy way, we need only to differentiate the basic forms and apply the substitution method. By basic forms we mean generating functions of the form:

 

for n ≥ 1.

Let's consider the number of solutions to

 

for ai ≥ 0 for i = 1, 2, ... n.

We know that for any m, the number of solutions is the coefficient to:

 

as discussed before.

We start from:

 

differentiate both sides (note that 1 = 1!)

 

differentiate again

 

and so on for (n-1) times

 

divide both sides by (n-1)!

 

the above confirms the result derived using a counting argument.