Differentiation from first principles

Differentiation

A-level Maths (9MA0)

ORIGINAL

Write the limit definition, expand f(x + h), cancel and divide by h, then let h tend to 0.

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Question

Given that

f(x) = 3x^2 - 5x

, prove from first principles that

f'(x) = 6x - 5

Every step worked, with the reasoning.

1
$f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$
Start from the definition of the derivative: the limit of the gradient of a chord.
2
$f(x+h) = 3(x+h)^2 - 5(x+h) = 3x^2 + 6xh + 3h^2 - 5x - 5h$
Replace every $x$ with $x + h$ and expand the bracket.
3
$f(x+h) - f(x) = 6xh + 3h^2 - 5h$
Subtract $f(x) = 3x^2 - 5x$ : the $3x^2$ and $-5x$ terms cancel.
4
$\dfrac{f(x+h) - f(x)}{h} = 6x + 3h - 5$
Divide every term by $h$ (allowed, because $h \ne 0$ as it tends to 0).
5
As $h \to 0$ , $3h \to 0$ , so $f'(x) = \displaystyle\lim_{h \to 0}(6x + 3h - 5) = 6x - 5$
Take the limit and say that the term in $h$ tends to 0 — writing the limit step is what makes this a proof.

Answer: $f'(x) = 6x - 5$ , as required.