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:

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

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 $nx^{n-1}$.

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 $f'(x)=nx^{n-1}$ $\bullet$