Rates of change, first principles and derivatives of power functions

1. Average rates and the slope of a secant

A rate of change compares a change in one quantity with the corresponding change in another. If a quantity Q is measured in litres and time t in minutes, the rate has units litres per minute. For a function f over an interval [a,b], where a≠b, the average rate is [f(b)−f(a)]/(b−a). Both differences must refer to the same ordered pair of inputs; reversing only one difference changes the sign incorrectly.

For f(x)=x²+2x, the values at x=1 and x=4 are 3 and 24. The average rate over [1,4] is (24−3)/(4−1)=7. Graphically this is the gradient of the secant through (1,3) and (4,24). It describes the whole interval, not the slope of the curve at every point in it. The secant equation is y−3=7(x−1).

A linear function f(x)=mx+c has the same average rate m over every nonzero interval. In a table, equal input increments therefore produce equal output increments. Unequal input increments require comparing ratios, not raw output differences. A table with outputs increasing by 6 while the input steps change from 1 to 3 does not show a constant rate: the corresponding rates are 6 and 2.

For a nonlinear function, the chosen interval matters. With f(x)=x²+2x, the average rate over [1,2] is 5, whereas over [2,4] it is 8. An average rate of zero means the endpoint outputs agree; the function may still rise and fall between them. In a motion problem, average velocity uses signed displacement divided by elapsed time. Average speed instead uses total distance divided by elapsed time, so it cannot always be read from the same endpoint calculation.

A useful diagnostic reverses the calculation. If f(2)=5 and the average rate over [2,6] is −3, then f(6)−5=−3(6−2), giving f(6)=−7. The rate determines the endpoint change, but it does not determine all intervening values. A straight line and a curved path can share these endpoints and the same average rate. When a graph uses thousands of dollars vertically and months horizontally, a displayed gradient of −3 means a decline of $3,000 per month, not $3. Label both axes before forming the quotient. For measured data, the quotient is an average over the actual sampling interval; claiming that it equals the instantaneous rate at an endpoint requires further information about the model.

Unequal interval. For f(x)=x²−3x, the average rate from x=1 to x=4 is [f(4)−f(1)]/3=(4−(−2))/3=2. The secant slope uses the change in input 3, not the final input 4.

2. From nearby secants to an instantaneous rate

To study a rate at x=a, compare f(a) with f(a+h) for a nonzero increment h. The difference quotient [f(a+h)−f(a)]/h is a secant gradient. If these gradients approach one finite value as h tends to zero from both sides within an open domain interval, that value is the derivative f′(a). It is also the gradient of the tangent at (a,f(a)). The notation dy/dx describes the same rate when y=f(x).

For f(x)=x²+2x, expand f(a+h)=(a+h)²+2(a+h). Subtracting f(a)=a²+2a gives 2ah+h²+2h. Dividing by h, which is allowed because h≠0, gives 2a+h+2. Taking the limit gives f′(a)=2a+2. At a=1 the instantaneous rate is 4, although the average rate over [1,4] was 7. The interval result and point result answer different questions.

Substituting h=0 into the original quotient gives 0/0, which is undefined. The argument works by simplifying for nonzero h and then finding the limit. At a=1, increments h=0.1, 0.01 and −0.01 give secant gradients 4.1, 4.01 and 3.99. These numbers illustrate the approach to 4, while the algebra establishes it exactly.

A derivative need not exist everywhere. For f(x)=|x| at zero, the quotient |h|/h equals 1 for positive h and −1 for negative h. The two sides disagree, producing a corner rather than one tangent gradient. At an endpoint, a one-sided rate may be meaningful, but it should not be confused with an ordinary two-sided derivative. A numerical derivative also needs caution near a corner or domain boundary: one calculator value does not establish the required limit.

A second first-principles example exposes a domain issue: for f(x)=1/x and a≠0, [1/(a+h)−1/a]/h=−1/[a(a+h)], provided h≠0 and a+h≠0. Its limit is −1/a². The reciprocal function is decreasing on each side of zero, but this does not make it decreasing across the discontinuity as one interval. Never bridge a missing domain point in a derivative sign argument. Numerically, an extremely small h can cause subtraction of almost equal rounded values, making the estimated quotient less reliable. Agreement over several sensible increments is a useful check, while the simplified limit gives the exact result.

One-sided check. For f(x)=|x|, the right difference quotient at zero is 1 and the left quotient is −1. Since the limits disagree, continuity at zero does not produce differentiability there.

For f=x², secants from 2 to 2+h have slope 4+h, visibly tending to 4.

3. Power functions and rational exponents

The power rule is d(xⁿ)/dx=nxⁿ⁻¹ wherever the real-valued power function is differentiable. The exponent n may be rational, not just a positive integer. Multiply by the old exponent and subtract one from it. For a constant multiple, retain that coefficient: d(5x³)/dx=15x². A constant term has derivative zero, and derivatives of sums are found term by term.

First principles explain the rule rather than merely providing a shortcut. For x³, [(x+h)³−x³]/h simplifies to 3x²+3xh+h², whose limit is 3x². For √x at x>0, multiply [√(x+h)−√x]/h by the conjugate. The quotient becomes 1/[√(x+h)+√x], tending to 1/(2√x). This agrees with the power-rule result (1/2)x⁻¹ᐟ².

Rewrite radicals and reciprocals before differentiating. On x>0, y=3√x−2/x²+7 becomes 3x¹ᐟ²−2x⁻²+7, giving y′=(3/2)x⁻¹ᐟ²+4x⁻³. At x=4 this derivative is 3/4+1/16=13/16. The positive derivative of −2x⁻² comes from multiplying two negative numbers; overlooking that sign is a common error.

Domain restrictions survive the calculation. The real square-root function exists for x≥0 but has no finite ordinary derivative at zero. A negative integer power excludes x=0. An even-root expression requires a non-negative radicand for its value and may require strict positivity for its derivative formula. Odd roots can admit negative inputs, but fractional-power calculator conventions should be checked against the intended real root. State the domain before using an algebraic derivative, especially when solving an equation obtained by multiplying through by a denominator.

For an odd-root example, write y=x^(2/3) as the square of the real cube root. At x≠0, y′=(2/3)x^(−1/3), which is negative for negative x and positive for positive x. At zero the function exists but its derivative formula is undefined, and the two sides have unbounded slopes. That point must be considered separately if an optimisation question includes zero. By contrast, differentiating x^0=1 gives zero everywhere, including zero when the constant function is the stated model. The original function and its domain take priority over a symbolic expression that appears to contain a meaningless intermediate factor such as 0x⁻¹.

Negative exponent. d(x⁻³)/dx=−3x⁻⁴=−3/x⁴ for x≠0. At x=−2 the derivative is −3/16, so the reciprocal-power curve is decreasing on both sides of its excluded point.

4. Equations of tangents and normals

A tangent at x=a passes through the curve point (a,f(a)) and has gradient f′(a), provided the derivative exists. Its point-slope equation is y−f(a)=f′(a)(x−a). Finding the derivative alone is not enough: the function supplies the point and the derivative supplies the slope. Keeping those roles separate prevents using (a,f′(a)) as a point on the original curve.

For f(x)=x²+2x at x=1, the point is (1,3) and the tangent gradient is 4. Hence y−3=4(x−1), or y=4x−1. A normal is perpendicular to the tangent. When the tangent slope m is finite and nonzero, the normal slope is −1/m. Here the normal is y−3=−(1/4)(x−1), or y=−x/4+13/4. Substituting x=1 into both equations returns y=3, and the slopes multiply to −1.

The negative-reciprocal rule has a special case. If f′(a)=0, the tangent is horizontal, y=f(a), and the normal is vertical, x=a. Do not attempt to divide by zero or describe the normal as having slope zero. For f(x)=x² at a=0, these lines are y=0 and x=0. A corner without a unique tangent also does not supply a unique normal by this derivative method.

A tangent gives a local linear approximation f(a+h)≈f(a)+hf′(a). For x²+2x near a=1, the estimate at x=1.02 is 3+4(0.02)=3.08, while the actual value is 3.0804. The discrepancy is h² in this example. It becomes small as h approaches zero but is not identically zero. A tangent may cross the curve, so defining it as a line that touches a graph only once is unreliable. Use the derivative and point, and restrict approximation claims to a suitable neighbourhood.

A question may specify a tangent slope rather than an input. For y=x³−3x, tangents parallel to y=9x+4 satisfy 3x²−3=9, so x=±2. The curve points are (2,2) and (−2,−2), giving tangent lines y=9x−16 and y=9x+16. Both are required. Their intercepts differ from the reference line because parallel lines share a gradient, not necessarily a point. If a normal slope is supplied instead, first convert it to the tangent slope where the reciprocal rule applies. Substitution into the final line and the derivative equation independently checks the location and orientation.

Normal equation. For y=x³ at x=2, the point is (2,8) and tangent slope is 12. The normal slope is −1/12, giving y−8=−(x−2)/12.

Substitution of x=2 into both the curve and tangent verifies that they share (2,8).

5. Derivative signs and local turning behaviour

The sign of a derivative describes the direction in which a differentiable function changes. Where f′ is positive throughout an interval, f increases; where it is negative throughout an interval, f decreases. A stationary point has f′(x)=0. To classify it, inspect the derivative on both sides rather than assuming that every zero produces a turning point.

Let f(x)=x³−3x²+2. Then f′(x)=3x²−6x=3x(x−2), so the stationary inputs are 0 and 2. For x<0, both x and x−2 are negative and their product is positive. For 0<x<2 the product is negative, and for x>2 it is positive. The function therefore increases, decreases and then increases across these three intervals.

The change from positive to negative at x=0 gives a local maximum at (0,2). The change from negative to positive at x=2 gives a local minimum at (2,−2). Report coordinates by substituting the stationary inputs into f, not into f′. The derivative sign diagram justifies the classification; a plot can support it but may hide details if the viewing window is poorly chosen.

For g(x)=x³, g′(x)=3x² is positive on either side of zero and equals zero at zero. The curve keeps increasing, so the stationary point is not a local maximum or minimum. On a restricted closed interval, an absolute maximum or minimum may occur at an endpoint even though the derivative there is nonzero. For example, on [−1,3], the earlier cubic has endpoint values −2 and 2, tying its local minimum and maximum values. Compare all eligible stationary points and endpoints, and distinguish local behaviour from the largest or smallest value on the entire permitted domain.

A derivative sign table should use intervals separated by every zero and every excluded point. For f(x)=x+1/x on x≠0, f′(x)=1−1/x²=(x²−1)/x². The denominator is positive, so the sign is positive outside [−1,1] and negative inside it, with zero excluded. At x=−1 the sign changes positive to negative, producing a local maximum (−1,−2); at x=1 it changes negative to positive, producing a local minimum (1,2). These labels are local within the separate domain components. The smaller y-value being a local maximum is not contradictory because the graph has a break between the two points.

Sign table. If f′=(x+1)(x−3), then f increases for x<−1, decreases on (−1,3), and increases for x>3. Thus −1 is a local maximum and 3 a local minimum.

6. Rates in motion and a bounded optimisation model

When s(t) is displacement along a line, its derivative v(t)=s′(t) is instantaneous velocity. If displacement is in metres and time in seconds, velocity is in metres per second. A positive velocity means motion in the chosen positive direction, while a negative velocity means motion in the opposite direction. Speed is |v(t)|. A negative displacement alone does not determine the direction of motion.

For s(t)=t²−6t+8 on 0≤t≤5, v(t)=2t−6. The object moves in the negative direction for 0≤t<3 and in the positive direction for 3<t≤5, with zero velocity at t=3. Its positions at t=0,3,5 are 8,−1,3 metres. Over the full interval its average velocity is (3−8)/5=−1 m/s. It travels 9 metres and then 4 metres, so its average speed is 13/5=2.6 m/s. The reversal explains why speed and signed average velocity differ.

For an optimisation example, suppose a rectangle has perimeter 24 metres. If one side is x, the other is 12−x and the area is A(x)=x(12−x), with 0<x<12 for positive side lengths. Differentiating gives A′(x)=12−2x, which is zero at x=6. The derivative is positive before 6 and negative after 6, proving the largest area occurs at side lengths 6 by 6 metres, with area 36 m².

The mathematical domain is part of the solution. The limiting area tends to zero as x approaches either boundary, and A(x)=36−(x−6)² supplies a second exact check that no feasible area exceeds 36. If a practical problem imposed an additional side-length limit, the permitted maximum could change. Name the quantity being optimised, eliminate the dependent variable using the constraint, justify the stationary point within the feasible interval and return the answer with the original dimensions and units.

For a constrained variation, require the rectangular side x to lie between 2 and 5 metres while the perimeter remains 24 metres. The stationary value x=6 is then infeasible. Since A′(x)=12−2x remains positive throughout [2,5], the largest permitted area is A(5)=35 m², with sides 5 and 7 metres. This demonstrates why solving A′=0 cannot be the whole optimisation method. A boundary can determine the answer. In motion, likewise, a calculated stopping time outside the specified observation interval must not be inserted as a turning point when splitting the distance calculation.

Bounded endpoint. For s(t)=t³−6t²+9t on 0≤t≤5, v=3(t−1)(t−3). Extreme positions occur at t=0,1,3,5; all four values must be compared because the interval is closed.

7. Differentiability, continuity and what a graph can establish

Differentiability at an interior point implies continuity there, but continuity alone does not establish differentiability. A curve can be unbroken and still have a corner. For f(x)=|x−2|, the values approach zero from both sides at x=2, yet the left slope is −1 and the right slope is 1. No single ordinary derivative exists at that point. A jump is an even clearer obstruction: if the function value suddenly changes, nearby secant slopes cannot settle to a finite tangent gradient.

Consider a piecewise function f(x)=x² for x≤1 and f(x)=ax+b for x>1. Continuity at the join requires a+b=1. Matching derivatives requires a=2 because the left derivative of x² at 1 is 2. Therefore b=−1. The choice a=2, b=−1 makes both height and slope agree. Satisfying only a+b=1 can leave a corner; matching a derivative while leaving a jump does not repair continuity. Solve the two conditions in their logical order.

A graphing window may make a small jump or sharp corner appear smooth. A numerical derivative at a join may use points on both sides and return an average that does not represent either limiting slope. Exact left and right formulas provide the decisive evidence. For measured data without a known formula, it is more careful to describe an estimated local rate than to claim differentiability has been proved.

There are also vertical-tangent examples, such as the real cube-root function at zero. The function is continuous, but its difference quotient grows without a finite limit. Distinguish a vertical tangent from a finite derivative. In this course, applying a derivative formula at an excluded point without discussing that limitation is not a valid calculation. When asked to sketch, indicate the relevant corner, break or vertical tangent explicitly.

For an assessment response, state the candidate point, show the one-sided values or slopes that matter, and draw the conclusion that follows. Do not infer differentiability from a connected-looking plot alone. The key question is whether both nearby rates approach the same finite number within the stated domain.

Cusp evidence. The graph y=|x−2| is continuous at 2 but has slopes −1 and 1 on its two sides. A sharp corner is direct evidence that no single tangent slope exists.

The smooth curve x^(2/3) has an undefined derivative at zero, where its cusp is visible.

8. A complete rate problem with independent checks

Suppose a model for the volume in a tank is V(t)=60+12t−t² litres for 0≤t≤8 minutes. The average rate over the first four minutes is [V(4)−V(0)]/4=(92−60)/4=8 L/min. The instantaneous rate is V′(t)=12−2t, so at four minutes it is 4 L/min. These values answer different questions: the first summarises an interval and the second describes the model at one time.

The rate becomes zero at t=6. It is positive before six minutes and negative afterwards, so the maximum volume on the interval is V(6)=96 L. Checking endpoints gives V(0)=60 and V(8)=92, confirming the global maximum is the interior point. The tank is still holding a positive volume at eight minutes even though its volume is decreasing. A negative rate does not mean a negative stored quantity.

The tangent model at t=4 is L(t)=92+4(t−4). At t=4.1 this predicts 92.4 L. The actual quadratic gives V(4.1)=92.39 L, so the tangent overestimates by 0.01 L. Algebraically, V(4+h)=92+4h−h², making the error h². The approximation becomes more accurate near the chosen time and should not be used as an exact model over the entire eight minutes.

If the question instead asks when the instantaneous rate equals the earlier average rate, solve 12−2t=8 to obtain t=2 minutes. This does not say that the instantaneous rate was 8 throughout the first four minutes. It identifies one point where the two numerical values happen to agree. Always attach the correct time or interval to a reported rate.

A final check can be graphical: a downward-opening quadratic should initially rise, flatten at its maximum, and then fall. Its derivative should therefore be a decreasing straight line crossing zero at the same time. Units provide another check: volume is in litres, V′ is in litres per minute, and multiplying a rate by a small time increment estimates a volume change. A coherent answer connects all three representations.

Error bound. For f=x³ near x=2, the linear estimate f(2.01)≈8+12(0.01)=8.12. The exact value 8.120601 differs by 0.000601, consistent with a small second-order remainder.

Using the exact cubic value at 2.01 confirms both the direction and scale of the linear estimate.

For a 0.01 decrease instead, the linear estimate is f(1.99)≈8−0.12=7.88; the exact 1.99³≈7.880599 brackets the tangent errors on the two sides. The cubic’s positive second derivative near 2 makes both exact points lie slightly above its tangent.