CA1–CA3 - Differentiation and the gradient function

CA1–CA3 - Differentiation and the gradient function

A straight line has one gradient; a curve can have a different gradient at every point. CA2 and CA3 explain what that changing gradient means, while CA1 gives you a quick way to calculate it for polynomials. You will find rates of change and tangent gradients without drawing tiny triangles on a graph.

The gradient at a point

The graph y=x2y=x^2 rises by 55 between x=2x=2 and x=3x=3. Does that mean its gradient at x=2x=2 is 55? Move the second point closer and the answer changes. What single number describes the curve's direction exactly at (2,4)(2, 4)?

You already know

A line's gradient is vertical change divided by horizontal change. A tangent gives a smooth curve's direction at a chosen point; a chord joins two distinct points on the curve.

Starting at (2,4)(2, 4), join it to (2+h,(2+h)2)(2+h, (2+h)^2). For any non-zero hh, the chord's gradient is

(2+h)2−4h=4+4h+h2−4hexpand the square=4+hdivide each term by h\begin{aligned} \frac{(2+h)^2-4}{h}&=\frac{4+4h+h^2-4}{h} && \text{expand the square}\\ &=4+h && \text{divide each term by }h \end{aligned}

The chord gradients are 55, 4.14.1 and 4.014.01 when h=1h=1, 0.10.1 and 0.010.01. From the other side, h=−0.1h=-0.1 gives 3.93.9. They approach 44 from both sides. We never divide by zero: we look at the value approached as the horizontal change becomes arbitrarily small. The tangent at (2,4)(2, 4) therefore has gradient 44.

[DIAGRAM: asset_name: Chords approaching a tangent; asset_slug: ca30-chord-tangent; caption: Near the point 2 comma 4, the chord direction approaches the tangent direction; description: The curve y equals x squared is blue. A sage chord joins P at 2 comma 4 to Q at 3 comma 9 and has gradient 5. A clay tangent through P has gradient 4. A second panel enlarges the neighbourhood of P and shows a chord to x equals 2.1 with gradient 4.1, almost matching the tangent.]
Near the point 2 comma 4, the chord direction approaches the tangent direction

Derivative and gradient function

The derivative of yy with respect to xx, written dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}, gives the instantaneous rate of change of yy with xx. For a smooth graph, it is the gradient of the tangent. A derivative expressed in terms of xx is the gradient function. If y=f(x)y=\mathrm{f}(x), it is also written f′(x)\mathrm{f}'(x).

At a general point (x,x2)(x, x^2), the same calculation gives a chord gradient 2x+h2x+h, approaching 2x2x. Thus the gradient function of x2x^2 is 2x2x. That explains a neat fact: the tangent at the mid-point's horizontal coordinate has exactly the gradient of any chord of this parabola. The chord from x=ax=a to x=bx=b has gradient a+ba+b, and the tangent at x=(a+b)/2x=(a+b)/2 also has gradient a+ba+b. This is exact, however far apart the endpoints are.

The relationship between derivative, tangent gradient and instantaneous rate of change is not on the formulae sheet: learn it. A derivative is not generally the gradient of a chord over a whole interval.

Key idea

The gradient function gives a rule for every point; substituting one value of xx gives the tangent gradient at that point.

Use that distinction to diagnose the opening mistake.

Differentiating a power

Differentiation turns a polynomial term into its gradient function, using the power to choose both the new coefficient and the new power.

For x3x^3, the chord calculation starts with (x+h)3−x3(x+h)^3-x^3. Expanding and dividing by hh gives 3x2+3xh+h23x^2+3xh+h^2, approaching 3x23x^2. For x2x^2, it gave 2x2x. In both cases, the old power becomes a multiplier and then drops by one. Multiplying the original curve's heights by a constant multiplies every vertical change, and therefore every gradient, by that constant.

The power rule

ddx(kxn)=knxn−1\frac{\mathrm{d}}{\mathrm{d}x}(kx^n)=knx^{n-1}

Here kk is constant and nn is a positive integer. A constant has derivative 00, so the case n=0n=0 is handled as a constant rather than by writing an unnecessary negative power. In particular, the derivative of kxkx is kk.
Not on the formulae sheet: learn it.

Read the two operations as a pair: multiply by the old power, then reduce that power by one.

Worked example 1: a positive coefficient

Question. Differentiate y=4x3y=4x^3 with respect to xx.

Plan. Multiply the coefficient by 33, then change the power from 33 to 22.

dydx=4×3x3−1apply the power rule=12x2simplify\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=4\times3x^{3-1} && \text{apply the power rule}\\ &=12x^2 && \text{simplify} \end{aligned}

Why multiply by 33? The derivative of x3x^3 is 3x23x^2; scaling the curve by 44 scales its gradients by 44 too.

Answer. dydx=12x2\dfrac{\mathrm{d}y}{\mathrm{d}x}=12x^2.

Check. At x=1x=1, the exact tangent gradient is 1212. The chord from x=1x=1 to x=1.001x=1.001 has gradient 12.01200412.012004, close to 1212.

Change only the coefficient's sign. The power-rule steps stay the same, but the gradient function changes sign.

Worked example 2: a negative coefficient

Question. Differentiate y=−4x3y=-4x^3 with respect to xx.

Plan. Keep the negative coefficient when multiplying by the power.

dydx=(−4)×3x3−1retain the sign=−12x2simplify\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}&=(-4)\times3x^{3-1} && \text{retain the sign}\\ &=-12x^2 && \text{simplify} \end{aligned}

Why does the minus sign remain? Reflecting the curve in the horizontal axis reverses every rise and fall, so it reverses each gradient.

Answer. dydx=−12x2\dfrac{\mathrm{d}y}{\mathrm{d}x}=-12x^2.

Check. At x=1x=1, the corresponding chord gradient is −12.012004-12.012004, approaching −12-12.

The old power affects two places in the answer. Completing only one operation produces a familiar error.

Watch out

Wrong: ddx(5x4)=20x4\dfrac{\mathrm{d}}{\mathrm{d}x}(5x^4)=20x^4. The coefficient changed but the power did not. Right: 20x320x^3. Also, the derivative of 55 is 00, because a constant graph is horizontal.

Change the coefficient once more, this time to a fraction.

Differentiating a polynomial

Differentiate a polynomial one term at a time, keeping each sign with its term.

Continue this lesson.

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