3.4 - Differentiation and rates of change

3.4 - Differentiation and rates of change

Differentiation finds how quickly a function changes at one input. You will connect this rate to a tangent, then differentiate integer powers, including negative powers. The original function gives a graph's height; its derivative gives the gradient.

Understanding variable rates (Higher)

On y=x2y=x^2, one interval gives an average gradient of 33, and the next gives 55. What is the rate right between them, at x=2x=2? A curve's changing steepness makes a one-point rate different from an interval average.

You already know

For two distinct points, gradient is vertical change divided by horizontal change. A tangent follows a curve's direction at a chosen point. The graph y=x2y=x^2 has height 44 at x=2x=2.

A straight line has constant rate: equal input increases give equal output increases. A curve can have a variable rate of change, with different gradients at different inputs.

Average rate of change

average rate=change in outputchange in input=y2−y1x2−x1,x2≠x1\text{average rate}=\frac{\text{change in output}}{\text{change in input}}=\frac{y_2-y_1}{x_2-x_1},\qquad x_2\ne x_1

This is the gradient of the straight joining line between two curve points. Not in the Higher formulae sheet: learn it. This lesson is Higher only.

The instantaneous rate is the tangent gradient at one input. The diagram deliberately shows three different lines: two interval averages and one tangent at PP.

[DIAGRAM: asset_slug: m055_variable_rates; caption: Two average-rate joining lines and the tangent give different gradients on the same curve; description: Exact graph y=x squared with A (1,1), P (2,4), B (3,9), joining lines AP and PB of gradients 3 and 5, and tangent y=4x-4 at P of gradient 4; file: m055_variable_rates.png; recommended_method: matplotlib]
Two average-rate joining lines and the tangent give different gradients on the same curve

Calculate the average rates first, keeping the output and input differences in the same order.

Worked example 1: the interval before P

Question. For y=x2y=x^2, find the average rate from x=1x=1 to x=2x=2.

Plan. Evaluate both heights, then divide their change by the input change.

y(1)=1,y(2)=4find the heightsaverage rate=4−12−1change over change=3evaluate\begin{aligned} y(1)&=1,\quad y(2)=4 &&\text{find the heights}\\ \text{average rate}&=\frac{4-1}{2-1} &&\text{change over change}\\ &=3 &&\text{evaluate} \end{aligned}

Why this step? The joining line records the total output change over that whole interval.

Answer. Average rate 33.

Check. A horizontal change of 11 accompanies an output increase of 33, matching line APAP.

Change only the interval, moving it one unit to the right while keeping its width.

Worked example 2: the interval after P

Question. For y=x2y=x^2, find the average rate from x=2x=2 to x=3x=3.

Plan. Use the same change-over-change calculation for the new endpoints.

y(2)=4,y(3)=9new heightsaverage rate=9−43−2same input width=5evaluate\begin{aligned} y(2)&=4,\quad y(3)=9 &&\text{new heights}\\ \text{average rate}&=\frac{9-4}{3-2} &&\text{same input width}\\ &=5 &&\text{evaluate} \end{aligned}

Why this step? Equal-width intervals need not have equal output changes on a curve.

Answer. Average rate 55; the rate is not constant.

Check. Line PBPB is steeper than APAP, agreeing with 5>35>3.

To investigate the rate at 22, bring a second input closer to it:

IntervalAverage rate
22 to 2.12.1(4.41−4)/0.1=4.1(4.41-4)/0.1=4.1
22 to 2.012.01(4.0401−4)/0.01=4.01(4.0401-4)/0.01=4.01

The values approach 44, the tangent gradient. In fact, for nonzero hh,

(2+h)2−4h=4h+h2h=4+h\frac{(2+h)^2-4}{h}=\frac{4h+h^2}{h}=4+h

As hh becomes small, this approaches 44. Never substitute zero into the original fraction. Differentiation gives the one-point rate directly, without repeating these interval calculations.

The derivative is the function giving instantaneous rates. Write it as dy/dx\mathrm{d}y/\mathrm{d}x, or f′(x)f'(x) if the function is f(x)f(x). These symbols describe gradient, not graph height. Rate units are output units divided by input units, such as litres per minute.

Watch out

Writing “the rate at 22 is 55” from the interval [2,3][2,3] confuses an average with a one-point rate. The average is 55; the tangent rate at 22 is 44.

Compare two new equal-width intervals.

Differentiating powers and sums (Higher)

For y=x2y=x^2, the derivative is 2x2x, giving the rate 44 at x=2x=2. A general rule lets you find the rate for other integer powers just as quickly.

Integer power rule

ddx(axn)=anxn−1,n≠0\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=anx^{n-1},\qquad n\ne0

Here aa is constant and nn is an integer. Apply the rule where the original power is defined. A constant has derivative zero. Not in the Higher formulae sheet: learn it.

Multiply the coefficient by the old exponent, then reduce the exponent by one. Thus x4x^4 becomes 4x34x^3. For a linear term, ax=ax1ax=ax^1, so its derivative is aa. A constant's value does not change with the input, so its rate is zero; write this directly rather than 0x−10x^{-1}.

Differentiate sums and differences term by term, keeping every sign. The answer is usually an expression because the rate depends on the input. Substitute a particular input only after differentiating.

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