In Leaving Cert Maths we are often asked to differentiate from first principles. This means that we must use the definition of the derivative — which was defined by Newton/ Leibniz — the principles underpinning this definition are these first principles. You can follow the argument at the start of Chapter 8 of these notes:

https://jpmccarthymaths.files.wordpress.com/2010/07/lecture-notes.pdf,

to see where this definition comes from, namely:

f'(x)\equiv \frac{dy}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}. (*)

If you are asked to differentiate from first principles you must use this formula. There are a number of basic functions which you are asked to differentiate. Chapter 9 of

https://jpmccarthymaths.files.wordpress.com/2010/07/lecture-notes.pdf,

details the ones that are examinable.

Derivative of Powers

I will present two proofs of the differentiation of x^n here. One is inductive and is the one they usually ask for; i.e.

Using induction prove that the derivative of x^n is nx^{n-1}.

They could also ask you:

Using induction, or otherwise, prove that the derivative of x^n is nx^{n-1}.

The “or otherwise” refers to the second, alternative, proof, that uses the Binomial Theorem (brutally cut from the Project Maths curriculum but in the tables.)

Proof by Induction

Let f_n(x)=x^n for n\in\mathbb{N}. Let P(n) be the proposition that:

f'_n(x)=nx^{n-1}.

Consider P(1). In this case f_1(x)=x. Now f'_1(x)=1 (usually just saying this is O.K. but to be more careful say that f_1(x) is a line of slope 1 hence the slope of the tangent is 1 — or maybe throw f_1(x) into *). Also 1x^{1-1}=1x^0=1 hence P(1) is true.

Assume P(k) is true; that is, if f_k(x)=x^k then

f'_k(x)=kx^{k-1}.

P(k+1)? Let f_{k+1}(x)=x^{k+1}. Now f_{k+1}(x)=x^k\cdot x. Now using the product rule (where we use P(k)):

f'_{k+1}(x)=x^k(1)+kx^{k-1}x=x^k+kx^k=(k+1)x^{k}=(k+1)x^{(k+1)-1}.

That is P(k) implies P(k+1) is true.

Hence by induction P(n) is true for all n\in\mathbb{N} and we are done \bullet

Alternative Proof

Let f(x)=x^n. Now f(x+h)=(x+h)^n; so expanding using the Binomial Theorem:

(x+h)^n=\sum_{k=0}^n{n\choose k} x^{n-k}h^k=x^n+nx^{n-1}h+\mathcal{O}(h^2) terms,

where \mathcal{O}(h^2) means that the terms have a h^2 or h^3 … or h^n part. Now

f(x+h)-f(x)=x^n+nx^{n-1}h+\mathcal{O}(h^2)-x^n,

\Rightarrow \frac{f(x+h)-f(x)}{h}=nx^{n-1}+\mathcal{O}(h).

Now taking the limit as h goes to o, all the \mathcal{O}(h) terms contain a h (or a higher power of h so all go to 0.), and we have

f'(x)=nx^{n-1}  \bullet