Differentiation from First Principles
Write the chord gradient as [f(x + h) - f(x)] / h, simplify it algebraically, and let h approach zero to get the derivative. This is the definition every rule later rests on.
What a learner can do afterwards
- Differentiate f(x) = x² from first principles, showing the cancellation of h
- Differentiate f(x) = x³ from first principles
- Say why the h in the denominator can only be cancelled while h is not zero
1 · Read
Last stop you shrank chords with a calculator and found the gradient of y = x² at x = 2, then again at x = 5. That works, but it answers one point at a time, and the Ridgeway crew wants the steepness all along the ramp. Every shortcut you meet later is proved by running this calculation once. There is a better move, and it is the one algebra is for: do the shrinking once with a letter instead of a number. The answer that comes out is not a number at all. It is a rule that gives the gradient at every x.
Set it up in function notation. Writing f(x + h) means putting x + h everywhere an x appears in the rule, and if no x appears, nothing changes. The chord now runs from the point (x, f(x)) to the point (x + h, f(x + h)), so the rise is f(x + h) - f(x) and the run is h. That makes the chord gradient [f(x + h) - f(x)] / h, which is last stop's calculation with letters in place of numbers. The value it settles on as h goes to zero is written f'(x) and is called the derivative.
Take f(x) = x². Then f(x + h) = (x + h)², which expands to x² + 2xh + h². Subtracting f(x) = x² leaves 2xh + h², so the chord gradient is (2xh + h²) / h. Every term on top carries an h, so factorise it out: h(2x + h) / h. Cancel the h and the fraction is gone, leaving 2x + h. Now let h go to zero and what is left is 2x. So f'(x) = 2x. Check it against last stop: at x = 2 that gives 4, and at x = 5 it gives 10.
The cancelling step is the one to be careful about, because it is where the whole method could go wrong. While the chord exists at all, the two points are apart, so h is not zero and dividing top and bottom by h is allowed. Setting h = 0 first would leave 0 / 0, which is nothing. So the order is fixed: cancel while h is still alive, and only then let it go to zero. When f is itself a fraction, put its two fractions over one denominator first. And when the limit comes, x + h becomes x.
Now f(x) = x³. Expanding (x + h)³ takes one more line: (x + h)(x² + 2xh + h²) gives x³ + 3x²h + 3xh² + h³. Subtracting x³ leaves 3x²h + 3xh² + h³, so the chord gradient is that over h. Every term on top has an h in it, so cancelling one h from each gives 3x² + 3xh + h². Letting h go to zero kills the last two terms, because both still carry an h. So f'(x) = 3x².
First principles turns last stop's shrinking chords into algebra. Write the chord gradient as [f(x + h) - f(x)] / h, expand and simplify until every term on top carries an h, cancel that h while it is still non-zero, and only then let h go to zero. What survives is f'(x), the derivative: a rule giving the gradient at every x. For x² it comes out as 2x, and for x³ as 3x².
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24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.