Mc5 - First principles and the power rule

Mc5 - First principles and the power rule

Use a limiting difference quotient to explain derivatives of small positive integer powers, then differentiate rational powers much faster with the power rule. Mc5 also covers constant multiples, sums and differences, with roots and reciprocals rewritten carefully and their real domains preserved.

Deriving a derivative from its definition

Can an arbitrarily small change reveal an exact gradient? For y=x2y=x^2 at x=2x=2, an input change hh produces output change 4h+h24h+h^2. Dividing gives 4+h4+h, whose limit is exactly 44. The increment never has to equal zero.

You already know

A tangent gradient is the limit of chord gradients. Expand brackets before collecting terms, and subtract a multi-term function value inside brackets. A factor can be cancelled only while it is nonzero.

Replace the fixed input 22 by a general xx, and the same idea becomes a proof valid along the whole curve.

First-principles definition

At an input 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}.

Form and simplify the quotient for h≠0h\ne0, then take the limit.

In the formula sheet.

The added area in this square picture has two strips and a small corner. Their combined area explains the algebraic numerator.

Two strips and a corner make the change in a square's area

Worked example 1: a rule from a limit

Question. Show from first principles that f(x)=x2\mathrm f(x)=x^2 has derivative 2x2x.

Plan. Expand the changed square and simplify before taking the limit.

f′(x)=lim⁡h→0(x+h)2−x2hstart from the definition=lim⁡h→02xh+h2hexpand and subtract=lim⁡h→0(2x+h)cancel for h≠0=2xlet h tend to zero\begin{aligned} \mathrm f'(x)&=\lim_{h\to0}\frac{(x+h)^2-x^2}{h} && \text{start from 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{let }h\text{ tend to zero} \end{aligned}

Why this step? The expression 2x+h2x+h equals the chord gradient for every nonzero increment, so it has the same limit.

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

Check. At x=2x=2, it gives 44, agreeing with the opening calculation.

Change only a constant multiplier. It must multiply the entire difference of function values.

Worked example 2: a constant multiple

Question. Show from first principles that the derivative of g(x)=3x2\mathrm g(x)=3x^2 is 6x6x.

Plan. Keep the factor 33 throughout the quotient.

g′(x)=lim⁡h→03(x+h)2−3x2hboth values have factor 3=lim⁡h→06xh+3h2hexpand and subtract=lim⁡h→0(6x+3h)=6xcancel; take the limit\begin{aligned} \mathrm g'(x)&=\lim_{h\to0}\frac{3(x+h)^2-3x^2}{h} && \text{both values have factor }3\\ &=\lim_{h\to0}\frac{6xh+3h^2}{h} && \text{expand and subtract}\\ &=\lim_{h\to0}(6x+3h)=6x && \text{cancel; take the limit} \end{aligned}

Why this step? Multiplying all heights by 33 multiplies every height difference, and hence every slope, by 33.

Answer. g′(x)=6x\mathrm g'(x)=6x, for all real xx.

Check. At x=−1x=-1, the quotient is −6+3h-6+3h, approaching −6-6.

A linear function and a constant provide quick checks on the definition:

(x+h)−xh=1,c−ch=0.\frac{(x+h)-x}{h}=1,\qquad \frac{c-c}{h}=0.

Thus d(x)/dx=1\mathrm d(x)/\mathrm dx=1 and d(c)/dx=0\mathrm d(c)/\mathrm dx=0. These results are not in the formula sheet: learn them. For a sum such as x2+3xx^2+3x, expanding its quotient gives 2x+h+32x+h+3, and the derivative is 2x+32x+3.

Watch out

Wrong: put h=0h=0 into the fraction and call 0/00/0 zero. Right: keep h≠0h\ne0 while simplifying. The limit describes nearby values, not division by zero.

Change the multiplier again, including its sign, and justify the given result.

First principles for cubic and quartic powers

The binomial theorem extends the argument to small positive integer powers. After dividing by hh, only the term which originally contained exactly one factor hh survives the limit.

You already know

In a binomial expansion, keep the coefficients and powers in order. For a fixed input xx, every remaining positive power of hh tends to zero as h→0h\to0.

Begin with the cubic and track the cancellation explicitly.

Worked example 3: deriving the cubic rule

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

Plan. Expand the cube, remove the fixed function value and divide by the increment.

(x+h)3=x3+3x2h+3xh2+h3binomial expansion(x+h)3−x3h=3x2+3xh+h2cancel for h≠0lim⁡h→0(3x2+3xh+h2)=3x2take the limit\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{cancel for }h\ne0\\ \lim_{h\to0}(3x^2+3xh+h^2)&=3x^2 && \text{take the limit} \end{aligned}

Why this step? Subtracting x3x^3 removes the only term without an increment, leaving a factor hh throughout.

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

Check. At x=−1x=-1, the quotient is 3−3h+h23-3h+h^2, tending to 33, a positive gradient at a negative height.

Change only the power from 33 to 44. The same method handles the additional term.

Worked example 4: deriving the quartic rule

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

Plan. Put the quartic expansion into the difference quotient before taking its limit.

(x+h)4=x4+4x3h+6x2h2+4xh3+h4expand(x+h)4−x4h=4x3+6x2h+4xh2+h3divide for h≠0lim⁡h→0(4x3+6x2h+4xh2+h3)=4x3discard vanishing terms\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{divide for }h\ne0\\ \lim_{h\to0}(4x^3+6x^2h+4xh^2+h^3)&=4x^3 && \text{discard vanishing terms} \end{aligned}

Why this step? The binomial coefficient 44 becomes the derivative's coefficient because the other changed terms vanish after taking the limit.

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

Check. At x=−1x=-1, the quotient is −4+6h−4h2+h3-4+6h-4h^2+h^3, tending to −4-4.

These derivative formulas are not in the formula sheet: learn them. A constant factor remains throughout the calculation: 5x35x^3 has derivative 15x215x^2. The small-positive-integer derivations explain the pattern, but are not a proof of the rational-power rule which follows.

Key idea

The difference quotient contains a contribution that survives and contributions that vanish. Expanding clearly shows which is which; “put the exponent in front” alone does not establish a first-principles result.

Apply the same proof with a fresh multiplier on the cubic.

Continue this lesson.

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