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




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





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





Solution Let  


Example 4Edit

Show that if




Example 5Edit

Differentiate from first principle





1. Differentiate


2. Differentiate


3. Differentiate from first principle


4. Differentiate


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




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




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)


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




where g is a function of z, we get


which confirmed the result derived using a counting argument.



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.