1.07g, 1.07i - First principles and the power rule

1.07g, 1.07i - First principles and the power rule

Derive polynomial gradients from the definition before using the faster power rule. In 1.07g you prove the small positive integer cases from first principles; in 1.07i you differentiate rational powers, constant multiples, sums and differences, keeping real domains in view.

Deriving a derivative from its definition

Why does squaring an input lead to the gradient 2x2x? At x=3x=3, the chord to 3+h3+h has gradient 6+h6+h, so its limit is 66. Doing that algebra with a general xx proves the rule at every point, rather than checking a few chosen inputs.

You already know

A derivative is a limiting chord gradient. Expand (x+h)2=x2+2xh+h2(x+h)^2=x^2+2xh+h^2, and subtract the whole function value using brackets. Cancelling a factor is valid only when that factor is nonzero.

The definition supplies a starting point even when you do not yet know a differentiation rule.

Differentiation from first principles

Where the finite two-sided limit exists,

f′(x)=lim⁡h→0f(x+h)−f(x)h.\mathrm f'(x)=\lim_{h\to0}\frac{\mathrm f(x+h)-\mathrm f(x)}{h}.

Simplify the quotient while h≠0h\ne0, then take the limit.

In the formula sheet.

The picture shows the cancellation behind the square: after subtracting the fixed square, the remaining area has a factor hh.

The change in a square's area when its side increases by h

Worked example 1: deriving the square rule

Question. Use first principles to show that f(x)=x2\mathrm f(x)=x^2 has derivative 2x2x.

Plan. Expand the changed function, subtract the fixed function, cancel the increment and take its limit.

f′(x)=lim⁡h→0(x+h)2−x2huse the definition=lim⁡h→02xh+h2hexpand and subtract=lim⁡h→0(2x+h)cancel for h≠0=2xhold x fixed\begin{aligned} \mathrm f'(x)&=\lim_{h\to0}\frac{(x+h)^2-x^2}{h} && \text{use the definition}\\ &=\lim_{h\to0}\frac{2xh+h^2}{h} && \text{expand and subtract}\\ &=\lim_{h\to0}(2x+h) && \text{cancel for }h\ne0\\ &=2x && \text{hold }x\text{ fixed} \end{aligned}

Why this step? Cancelling hh exposes the part which survives when the increment tends to zero.

Answer. f′(x)=2x\mathrm f'(x)=2x for every real xx.

Check. At x=3x=3, the result is 66, agreeing with the opening chord calculation.

Add just one linear term. Its contribution survives the limit too, and brackets protect its sign.

Worked example 2: a square plus a linear term

Question. Use first principles to show that f(x)=x2+3x\mathrm f(x)=x^2+3x has derivative 2x+32x+3.

Plan. Substitute x+hx+h into both terms before subtracting all of f(x)\mathrm f(x).

f′(x)=lim⁡h→0(x+h)2+3(x+h)−(x2+3x)hsubtract the whole function=lim⁡h→02xh+h2+3hhexpand and collect=lim⁡h→0(2x+h+3)cancel for h≠0=2x+3take the limit\begin{aligned} \mathrm f'(x)&=\lim_{h\to0}\frac{(x+h)^2+3(x+h)-(x^2+3x)}{h} && \text{subtract the whole function}\\ &=\lim_{h\to0}\frac{2xh+h^2+3h}{h} && \text{expand and collect}\\ &=\lim_{h\to0}(2x+h+3) && \text{cancel for }h\ne0\\ &=2x+3 && \text{take the limit} \end{aligned}

Why this step? A first-principles proof must build the derivative from function values; quoting the answer from a rule would miss the argument.

Answer. f′(x)=2x+3\mathrm f'(x)=2x+3 for every real xx.

Check. At x=0x=0, the chord gradient is h+3h+3, tending to 33.

The same definition gives d(x)/dx=1\mathrm d(x)/\mathrm dx=1, since ((x+h)−x)/h=1((x+h)-x)/h=1, and d(c)/dx=0\mathrm d(c)/\mathrm dx=0 for a constant cc. These derivative results are not in the formula sheet: learn them.

Watch out

Wrong: f(x+h)−f(x)=(x+h)2+3(x+h)−x2+3x\mathrm f(x+h)-\mathrm f(x)=(x+h)^2+3(x+h)-x^2+3x. Right: subtract (x2+3x)(x^2+3x), so both terms change sign. A missing bracket can leave a term which cannot cancel with hh.

Change the coefficient and sign of the linear term while keeping the proof structure.

First principles for cubic and quartic powers

For a higher power, the binomial expansion does the hard work. Only its term linear in hh survives after division by hh and taking the limit: every remaining positive power of hh tends to zero.

You already know

The binomial theorem expands small positive integer powers. Keep every term before subtracting f(x)\mathrm f(x), then remove a common factor of hh. The limit is taken with xx fixed.

Work through the cubic first, where the surviving term is especially easy to spot.

Worked example 3: the cubic derivative

Question. Show from first principles that d(x3)/dx=3x2\mathrm d(x^3)/\mathrm dx=3x^2.

Plan. Expand (x+h)3(x+h)^3, subtract x3x^3, then simplify the difference quotient.

(x+h)3=x3+3x2h+3xh2+h3binomial expansion(x+h)3−x3h=3x2+3xh+h2subtract; divide for h≠0lim⁡h→0(3x2+3xh+h2)=3x2terms containing h vanish\begin{aligned} (x+h)^3&=x^3+3x^2h+3xh^2+h^3 && \text{binomial expansion}\\ \frac{(x+h)^3-x^3}{h}&=3x^2+3xh+h^2 && \text{subtract; divide for }h\ne0\\ \lim_{h\to0}(3x^2+3xh+h^2)&=3x^2 && \text{terms containing }h\text{ vanish} \end{aligned}

Why this step? The expansion makes the factor hh visible in every term of the numerator.

Answer. The derivative is 3x23x^2, for all real xx.

Check. At x=1x=1, chord gradients are 3+3h+h23+3h+h^2, approaching 33.

Increase only the exponent from 33 to 44. One extra term appears, but the cancellation mechanism is unchanged.

Worked example 4: the quartic derivative

Question. Show from first principles that d(x4)/dx=4x3\mathrm d(x^4)/\mathrm dx=4x^3.

Plan. Use the quartic binomial expansion inside the same limit.

(x+h)4=x4+4x3h+6x2h2+4xh3+h4expand(x+h)4−x4h=4x3+6x2h+4xh2+h3cancel for h≠0lim⁡h→0(4x3+6x2h+4xh2+h3)=4x3take the limit\begin{aligned} (x+h)^4&=x^4+4x^3h+6x^2h^2+4xh^3+h^4 && \text{expand}\\ \frac{(x+h)^4-x^4}{h}&=4x^3+6x^2h+4xh^2+h^3 && \text{cancel for }h\ne0\\ \lim_{h\to0}(4x^3+6x^2h+4xh^2+h^3)&=4x^3 && \text{take the limit} \end{aligned}

Why this step? The coefficient 44 comes from the term with exactly one increment, not from guessing a pattern.

Answer. The derivative is 4x34x^3, for every real xx.

Check. At x=1x=1, the quotient is 4+6h+4h2+h34+6h+4h^2+h^3, tending to 44.

A constant multiple can stay outside the quotient: replacing x3x^3 by 2x32x^3 doubles every line and gives 6x26x^2; 5x35x^3 gives 15x215x^2. The square, cubic and quartic derivative formulas are not in the formula sheet: learn them. First-principles proof tasks here use small positive integer powers up to 44; powers above 44 are excluded from that requirement. The rule you use afterwards has a wider range.

Key idea

These calculations prove the powers actually expanded. They suggest the general power rule, but they do not prove it for every rational exponent.

Vary only the quartic's constant multiplier and retain the first-principles reasoning.

Continue this lesson.

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