Limits, First Principles and the Rules of Differentiation

≈ 25 min

From average gradient to instantaneous gradient

The gradient between two points on $y = f(x)$ is $\dfrac{f(x+h) - f(x)}{h}$ . Letting the second point slide towards the first ( $h \to 0$ ) gives the gradient of the tangent:

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

A limit first

\lim_{x \to 2}\frac{x^2-4}{x-2} = \lim_{x \to 2}\frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x+2) = 4

We never substitute $x = 2$ into the original fraction (that gives $\tfrac00$ ); we simplify first, then take the limit.

Worked example: first principles

For $f(x) = 2x^2 - 3x$ :

f(x+h) = 2(x+h)^2 - 3(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h

f(x+h) - f(x) = 4xh + 2h^2 - 3h

\frac{f(x+h)-f(x)}{h} = 4x + 2h - 3 \;\;\longrightarrow\;\; f'(x) = 4x - 3

The rules

\frac{d}{dx}\left[ax^n\right] = a\,n\,x^{n-1}, \qquad \frac{d}{dx}[c] = 0

Derivatives of sums are the sums of derivatives. Before differentiating, rewrite every term as a power of $x$ : $\dfrac{2}{x} = 2x^{-1}$ and $\sqrt{x} = x^{\frac12}$ .

For $f(x) = 4x^3 - \dfrac{2}{x} + 5 = 4x^3 - 2x^{-1} + 5$ :

f'(x) = 12x^2 + 2x^{-2} = 12x^2 + \frac{2}{x^2}

Determine $f'(x)$ from first principles if $f(x) = 3x^2 - 1$ . Give the simplified expression.

Determine $D_x\big[(2x-1)(x+3)\big]$ .

The volume of water (in kilolitres) in a reservoir is modelled by $f(x) = 4x^3 - \dfrac{2}{x} + 5$ for $x > 0$ hours. Calculate $f'(1)$ , the instantaneous rate of change after 1 hour.