G1 - Derivatives from first principles

G1 - Derivatives from first principles

Learn how a chord gradient approaches a tangent gradient, and use that idea to derive derivatives of polynomial powers. Connect the resulting gradient function with instantaneous rates and with the shape of a curve.

Tangent gradient as a limit

How can a curve have a gradient at one point when the gradient formula needs two points? On y=x2y=x^2, the chord from (3,9)(3,9) to (3.1,9.61)(3.1,9.61) has gradient 6.16.1. Moving the second input to 3.013.01 gives gradient 6.016.01. The approaching value, 66, describes the tangent at the first point.

You already know

A straight line has gradient (y2−y1)/(x2−x1)(y_2-y_1)/(x_2-x_1). Expanding (x+h)2(x+h)^2 gives x2+2xh+h2x^2+2xh+h^2. These are recall results, not supplied in the formula booklet.

For P=(x,f(x))P=(x,f(x)) and Q=(x+h,f(x+h))Q=(x+h,f(x+h)), the change in input is hh, also written Δx\Delta x or δx\delta x. It may be positive or negative. The chord gradient is [f(x+h)−f(x)]/h[f(x+h)-f(x)]/h, with h≠0h\ne0. As QQ approaches PP, those chord gradients may approach a common value.

Two chords of the parabola approach the tangent at P as the increment shrinks

Derivative

The derivative at an interior input is the common finite limit of the chord gradients from both sides, when it exists. It is the tangent gradient there. For y=f(x)y=f(x), write it as f′(x)f'(x), dy/dxdy/dx or d(f(x))/dxd(f(x))/dx.

A limit is an approached value; the expression need not be defined at the limiting input. A corner illustrates the boundary: for y=∣x∣y=|x| at zero the left chord gradients are −1-1 and the right ones are 11, so there is no derivative there.

Differentiation from first principles

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

Simplify while h≠0h\ne0, then take the limit.
In the formula booklet.

This definition turns the opening numerical pattern into an exact calculation.

Worked example 1: a tangent gradient at a fixed input

Question. For f(x)=x2f(x)=x^2, use a chord gradient and a limit to find the tangent gradient at x=3x=3.

Plan. Form the quotient at 33, expand, cancel the nonzero increment and take its limit.

m(h)=(3+h)2−9hchange in height over input=6h+h2hexpand and subtract=6+h(h≠0)cancel hf′(3)=lim⁡h→0(6+h)=6take the limit\begin{aligned} m(h)&=\frac{(3+h)^2-9}{h} && \text{change in height over input}\\ &=\frac{6h+h^2}{h} && \text{expand and subtract}\\ &=6+h\quad(h\ne0) && \text{cancel }h\\ f'(3)&=\lim_{h\to0}(6+h)=6 && \text{take the limit} \end{aligned}

Why this step? Cancelling is legal before taking the limit because the approaching increments are nonzero.

Answer. The tangent gradient is 66.

Check. At h=−0.01h=-0.01 and h=0.01h=0.01, the chord gradients are 5.995.99 and 6.016.01, approaching the same value.

Now change only the fixed input, from 33 to −2-2. A negative gradient should emerge because this side of the parabola falls as you move right.

Worked example 2: approaching from either side

Question. For f(x)=x2f(x)=x^2, find the tangent gradient at x=−2x=-2 from first principles.

Plan. Keep the negative input in brackets and repeat the difference-quotient method.

m(h)=(−2+h)2−4hform the chord gradient=−4h+h2h=−4+h(h≠0)expand and cancelf′(−2)=lim⁡h→0(−4+h)=−4take the limit\begin{aligned} m(h)&=\frac{(-2+h)^2-4}{h} && \text{form the chord gradient}\\ &=\frac{-4h+h^2}{h}=-4+h\quad(h\ne0) && \text{expand and cancel}\\ f'(-2)&=\lim_{h\to0}(-4+h)=-4 && \text{take the limit} \end{aligned}

Why this step? The square applies to the whole input −2+h-2+h; keeping the brackets prevents the middle term's sign being lost.

Answer. The tangent gradient is −4-4.

Check. Increments −0.01-0.01 and 0.010.01 give −4.01-4.01 and −3.99-3.99, consistent with a falling tangent.

Both calculations keep a distinction between a small increment and an increment equal to zero.

Watch out

Wrong: substitute h=0h=0 into the original quotient and call 0/0=00/0=0. Division by zero is undefined. Right: simplify for h≠0h\ne0, then take the limit of the simplified expression.

Try one more input, retaining the limit notation in your conclusion.

First principles for polynomial powers

Keeping the input as xx gives every tangent gradient at once: [(x+h)2−x2]/h=2x+h[(x+h)^2-x^2]/h=2x+h, so f′(x)=2xf'(x)=2x. This gradient function can then be evaluated at any required input. The result d(x2)/dx=2xd(x^2)/dx=2x is not in the formula booklet: learn it.

The same method works for other small positive integer powers. Form f(x+h)f(x+h) by replacing every xx by x+hx+h, subtract the whole of f(x)f(x), expand and cancel, divide by h≠0h\ne0, then take the limit. Keep xx fixed while hh changes.

Worked example 3: the cubic power

Question. Differentiate f(x)=x3f(x)=x^3 from first principles.

Plan. Expand the cube inside the difference quotient and identify which terms survive the limit.

f′(x)=lim⁡h→0(x+h)3−x3hdefinition=lim⁡h→03x2h+3xh2+h3hexpand and subtract=lim⁡h→0(3x2+3xh+h2)h≠0=3x2terms containing h vanish\begin{aligned} f'(x)&=\lim_{h\to0}\frac{(x+h)^3-x^3}{h} && \text{definition}\\ &=\lim_{h\to0}\frac{3x^2h+3xh^2+h^3}{h} && \text{expand and subtract}\\ &=\lim_{h\to0}(3x^2+3xh+h^2) && h\ne0\\ &=3x^2 && \text{terms containing }h\text{ vanish} \end{aligned}

Why this step? Only the term originally containing exactly one factor of hh remains after division and the limit.

Answer. f′(x)=3x2f'(x)=3x^2.

Check. At x=2x=2, the chord quotient is 12+6h+h212+6h+h^2, which tends to 12=3(2)212=3(2)^2.

Change only the function's exponent, from 33 to 44. There is an extra binomial term, but the same cancellation decides the answer.

Worked example 4: a fourth power

Question. Differentiate g(x)=x4g(x)=x^4 from first principles.

Plan. Expand (x+h)4(x+h)^4, cancel x4x^4 and divide before taking the limit.

g′(x)=lim⁡h→0(x+h)4−x4hdefinition=lim⁡h→04x3h+6x2h2+4xh3+h4hbinomial expansion=lim⁡h→0(4x3+6x2h+4xh2+h3)h≠0=4x3take the limit\begin{aligned} g'(x)&=\lim_{h\to0}\frac{(x+h)^4-x^4}{h} && \text{definition}\\ &=\lim_{h\to0}\frac{4x^3h+6x^2h^2+4xh^3+h^4}{h} && \text{binomial expansion}\\ &=\lim_{h\to0}(4x^3+6x^2h+4xh^2+h^3) && h\ne0\\ &=4x^3 && \text{take the limit} \end{aligned}

Why this step? The powers of hh show exactly why the extra terms disappear; quoting a power rule would skip the requested derivation.

Answer. g′(x)=4x3g'(x)=4x^3.

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

For a small positive integer nn, the binomial expansion begins (x+h)n=xn+nxn−1h+⋯(x+h)^n=x^n+nx^{n-1}h+\cdots. The binomial expansion is supplied in the formula booklet. After subtraction and division, the surviving term is nxn−1nx^{n-1}. For n=1n=1, the quotient is exactly 11. Constants have zero change. Thus the first-principles calculation for 2x3−x2x^3-x gives quotient 6x2+6xh+2h2−16x^2+6xh+2h^2-1 and limit 6x2−16x^2-1. These derived power and constant results are not supplied in the formula booklet.

Watch out

Wrong: (x+h)3=x3+h3(x+h)^3=x^3+h^3. The missing middle terms contain the gradient information. Right: (x+h)3=x3+3x2h+3xh2+h3(x+h)^3=x^3+3x^2h+3xh^2+h^3.

The next question changes the exponent once more; its expansion is longer, but only one term survives.

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