Differentiation from first principles · Differentiation

In General Maths you differentiated with the power rule (see calculus): $x^n$ becomes $nx^{n - 1}$ . Further Maths asks why. A question that says “from first principles” wants the derivative worked out from the definition, without the rules.

The idea: a chord that shrinks to a tangent

The gradient of a curve at a point $P$ is the gradient of its tangent there. You can’t measure a tangent’s gradient directly, but you can measure a chord’s. Take a second point $Q$ a small step $h$ further along the curve. The chord $PQ$ has gradient

\frac{f(x + h) - f(x)}{h}

As $h$ shrinks, $Q$ slides towards $P$ and the chord turns into the tangent:

Chord to tangentAs Q slides towards P (h → 0), the chord's gradient becomes the tangent's

So the derivative is the limit of the chord’s gradient as $h \to 0$ . Shrink $h$ and watch:

Chord to tangentPick a function, shrink h

6chord gradient, h = 15f′(1) = 2x + 3 at x = 1

[(x + h)² + 3(x + h) − x² − 3x] ÷ h = 2x + 3 + h. With h = 1 the chord's gradient is 6. As h → 0, the terms with h vanish, leaving f′(x) = 2x + 3, which is 5 at x = 1.

Evaluating limits

$\displaystyle\lim_{x \to a} f(x)$ means the value $f(x)$ gets close to as $x$ gets close to $a$ . Usually you just put $x = a$ in. But if that gives $\frac00$ , the formula has a gap at $x = a$ : factorise, cancel the factor $(x - a)$ , and then put $x = a$ in:

\begin{aligned} \lim_{x \to 3}\frac{x^2 - 9}{x - 3} &= \lim_{x \to 3}\frac{(x - 3)(x + 3)}{x - 3} \\ &= \lim_{x \to 3}\,(x + 3) = 6 \end{aligned}

When $x \to \infty$ , divide the top and the bottom by the highest power of $x$ on the bottom. Every term like $\frac1x$ or $\frac{1}{x^2}$ then tends to 0:

\lim_{x \to \infty}\frac{2x + 1}{x + 1} = \lim_{x \to \infty}\frac{2 + \frac1x}{1 + \frac1x} = \frac{2}{1} = 2

A gap in the graph(x² − 9) ÷ (x − 3) is x + 3 with a gap at x = 3: the limit is the height of the gap, 6

A limit at infinityAs x grows, (2x + 1) ÷ (x + 1) gets as close to 2 as you like

Worked example

Evaluate

\displaystyle\lim_{x \to 2}\frac{x^2 + x - 6}{x^2 - 4}

Try x = 2
- Top: ${4 + 2 - 6 = 0}$ .
- Bottom: ${4 - 4 = 0}$ .
- $\frac00$ has no value, so both must have a factor $(x - 2)$ .
Think first. Put x = 2 into the top and the bottom. What do you get?
Factorise
- Top: ${x^2 + x - 6 = (x + 3)(x - 2)}$ .
- Bottom: ${x^2 - 4 = (x + 2)(x - 2)}$ .
Think first. Factorise the top and the bottom.
Cancel
- Cancel $(x - 2)$ : ${\frac{(x + 3)(x - 2)}{(x + 2)(x - 2)} = \frac{x + 3}{x + 2}}$ .
- This is allowed because $x$ only gets close to 2, so $x - 2$ is never 0.
Think first. Which factor goes?
Now put x = 2 in
- ${\frac{2 + 3}{2 + 2} = \frac54}$ .
- So the limit is $\frac54$ .

More: limits

WAEC 2022 · Paper 1 · Q38Evaluate: $\displaystyle\lim_{x \to -2}\left(\frac{x^3 + 8}{x + 2}\right)$ .

The four-step method

f′(x) = lim_h→0 [f(x + h) − f(x)] ⁄ h

1. find f(x + h)2. take away f(x)3. divide by h4. let h → 0

From first principlesSimplify until you can divide by h, then let h → 0

The key is step 3: you must be able to cancel $h$ before you let $h \to 0$ . Putting $h = 0$ straight into $\frac{f(x + h) - f(x)}{h}$ gives $\frac00$ , which means nothing. Some books write $\delta x$ for $h$ and $\delta y$ for $f(x + h) - f(x)$ ; the method is the same.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q2

Differentiate from first principles, with respect to $x$ , $(3x^2 + 2x - 1)$ .

Find f(x + h)
- ${f(x + h) = 3(x + h)^2 + 2(x + h) - 1}$ .
- Expand the square: ${3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2}$ .
- So ${f(x + h) = 3x^2 + 6xh + 3h^2}$ ${+\ 2x + 2h - 1}$ .
Think first. Put x + h in place of every x, then expand.
Take away f(x)
- Take away ${3x^2 + 2x - 1}$ : the $3x^2$ , the $2x$ and the $-1$ cancel.
- So ${f(x + h) - f(x) = 6xh + 3h^2 + 2h}$ .
Think first. Which terms cancel?
Divide by h
- ${\frac{f(x + h) - f(x)}{h} = 6x + 3h + 2}$ .
Think first. Every term has an h. Divide it out.
Let h → 0
- The $3h$ term tends to 0.
- So ${f'(x) = 6x + 2}$ . The power rule agrees: ${\frac{d}{dx}(3x^2 + 2x - 1) = 6x + 2}$ .
Think first. Which term disappears?

More: polynomials from first principles

Fractions from first principles

When $f(x)$ has $x$ on the bottom, step 2 means taking away two fractions. Put them over a common denominator; the top then has a factor $h$ that cancels in step 3.

Worked example · WAEC 2012

WAEC 2012 · Paper 2 · Q6

Find, from first principles, the derivative of $\left(3x - \dfrac{1}{2x}\right)$ with respect to $x$ .

f(x + h) − f(x)
- ${f(x + h) = 3(x + h) - \frac{1}{2(x + h)}}$ .
- Take away ${f(x) = 3x - \frac{1}{2x}}$ : ${f(x + h) - f(x) = 3h - \left[\frac{1}{2(x + h)} - \frac{1}{2x}\right]}$ .
Think first. Write f(x + h), then take away f(x).
The fractions over one denominator
- ${\frac{1}{2(x + h)} - \frac{1}{2x} = \frac{x - (x + h)}{2x(x + h)}}$ .
- Simplify the top: ${\frac{-h}{2x(x + h)}}$ .
- So ${f(x + h) - f(x) = 3h + \frac{h}{2x(x + h)}}$ .
Think first. The common denominator is 2x(x + h). What is the top?
Divide by h
- ${\frac{f(x + h) - f(x)}{h} = 3 + \frac{1}{2x(x + h)}}$ .
Think first. Divide both terms by h.
Let h → 0
- ${x + h \to x}$ , so ${f'(x) = 3 + \frac{1}{2x^2}}$ .

More: fractions from first principles

WAEC 2016 · Paper 2 · Q4Find, from first principles, the derivative of $\dfrac{5}{x^2 + 3}$ with respect to $x$ .

Your turn

WAEC 2018 · Paper 2 · Q11 (a)

(a)
Find, from first principles, the derivative of $5x - \dfrac{6}{x^2}$ with respect to $x$ .

Worked solution (try it first)

(a)

$f(x) = 5x - 6x^{-2}$ , so $f(x + h) - f(x) = 5h - 6\left[\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2}\right]$ .
Over one denominator: $\dfrac{1}{(x + h)^2} - \dfrac{1}{x^2} = \dfrac{x^2 - (x + h)^2}{x^2(x + h)^2}$
$= \dfrac{-h(2x + h)}{x^2(x + h)^2}$ .
So $f(x + h) - f(x) = 5h + \dfrac{6h(2x + h)}{x^2(x + h)^2}$ .
Divide by $h$ : $5 + \dfrac{6(2x + h)}{x^2(x + h)^2}$ .
Let $h \to 0$ : $f'(x) = 5 + \dfrac{12x}{x^4}$
$= 5 + \dfrac{12}{x^3}$ .

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