Euler's number e, the limit (a^h - 1)/h and the derivative of e^x and Ae^(kx)

1. The limiting rate at the origin

For the exponential curve y=a^x, where a>0, the point at x=0 is always (0,1). A secant from that point to (h,a^h) has gradient (a^h−1)/h. This quotient is defined for nonzero h. Making h approach zero investigates the tangent gradient at the origin: it does not mean substituting zero into a fraction with both numerator and denominator zero. The limit describes what nearby secant gradients approach.

Use technology to compare small positive and negative increments for a selected base. Record the base, the h values and the resulting quotients in a labelled table. For example, the positive-h quotient for a=2 approaches about 0.693, while the corresponding value for a=3 approaches about 1.099. The base that produces a limiting gradient of exactly 1 lies between these bases. These decimal observations support a conjecture; they are not an exact proof from a finite table.

Taking h smaller is useful only while the arithmetic remains reliable. When a^h is extremely close to 1, subtracting 1 can lose significant digits in a calculator’s finite-precision representation. An apparent string of zeros or unstable final digits need not describe the mathematical limit. Compare a sensible sequence of increments and look for stable behaviour on both sides.

This investigation connects the slope of an exponential at the origin to its base. The output being studied is a gradient, not the value a⁰, which is 1 for every positive base. Clearly distinguishing those two quantities explains why only one base is special even though all the curves pass through the same point.

Worked limit. For a=3, take h=0.01 and h=−0.01. The quotients are (3^0.01−1)/0.01≈1.10467 and (3^−0.01−1)/(−0.01)≈1.09259. They approach ln3≈1.09861 from opposite sides. The calculation supports, but does not define, the limiting gradient: h remains nonzero in each quotient.

For a=1 the numerator is always zero, so the limit is 0; for 0<a<1 it is negative. These signs agree with the local direction of the corresponding exponential graphs through (0,1).

A useful numerical check halves |h|. With h=0.005 the base-3 quotient is about 1.10163, closer to ln3. Reporting the increment beside the value prevents a calculator decimal from being mistaken for the exact limit.

2. Why the base e is special

The number e is the unique positive base for which the limit of (a^h−1)/h as h approaches zero is 1. Its decimal value is approximately 2.71828, but the defining property is exact. Replacing e by a rounded decimal in a numerical calculation can introduce approximation; retaining the symbol e preserves an exact answer. The curve y=e^x has tangent gradient 1 at (0,1).

The origin calculation extends to a general point by the index law e^x+h=e^xe^h. The difference quotient becomes [e^x+h−e^x]/h=e^x(e^h−1)/h. During the limit h changes while x is held fixed, so e^x is a constant multiplier. The bracketed quotient approaches 1, giving d(e^x)/dx=e^x.

The derivative equalling the original function has a geometric meaning: at every x, the numerical tangent slope equals the function’s height. Since e^x is positive for every real x, the curve is increasing everywhere. Differentiating again gives the same positive expression, so its slope increases and the graph is concave up. Neither positivity nor increase alone would establish the other for an arbitrary function.

As x decreases without bound, e^x approaches zero without attaining it. The x-axis is a horizontal asymptote, not an intercept. These properties make a coherent sketch: domain all real x, range positive y, intercept (0,1), positive slope and upward concavity. Use the derivative to explain the shape rather than relying only on a remembered drawing.

Worked derivation. Let f(x)=e^x. At x=2, [f(2+h)−f(2)]/h=e²(e^h−1)/h. Since the second factor tends to 1, f′(2)=e². The same factorisation works at an arbitrary x, giving f′(x)=e^x rather than merely verifying the slope at the origin.

The tangent at x=2 is therefore y−e²=e²(x−2). Both its height and gradient contain e²; replacing e by 2.718 too early weakens an exact answer.

For g(x)=e^x−4, g′(x)=e^x−4. The translation changes the point at which the value equals 1, but the exponent has derivative 1, so no additional multiplier appears.

The identity e^x+c=e^ce^x also shows why a horizontal translation can be rewritten as a constant vertical multiplier.

3. Constant multiples and linear exponents

For y=Ae^kx, where A and k are constants, differentiation gives y′=Ake^kx. The constant multiplier A stays outside. The exponential differentiates to itself with respect to its exponent, and the derivative k of that exponent multiplies the result. Although the chain rule is developed more generally in a later note, this linear-exponent case is central to practical exponential models.

For y=7e^−2x, the derivative is −14e^−2x. The original function is positive, but its rate of change is negative: positive output and increasing output are different ideas. For y=4e^3x+6, the derivative is 12e^3x; the vertical shift disappears. For y=e^x−4, the inner derivative is 1, so the derivative is simply e^x−4. Do not multiply by −4 because it is an additive shift, not the coefficient of x.

Repeated differentiation supplies another useful check: y″=Ak²e^kx. When A>0 and k≠0 this is positive for both growth and decay. A positive exponential decay curve can therefore be decreasing and concave up simultaneously: its negative slope becomes less negative as the curve flattens towards zero.

Retain the independent variable used by the model. If time is t, write dN/dt rather than silently changing the variable to x. Include rate units: differentiating an amount measured in grams with respect to hours gives grams per hour. A symbolic derivative without its sign, units or contextual interpretation may not answer an applied question completely. Check each multiplier before using the derivative in a numerical calculation.

Worked derivative. If y=5e^−0.4x, then y′=5(−0.4)e^−0.4x=−2e^−0.4x. At x=3 the value is 5e^−1.2≈1.506 and the rate is −2e^−1.2≈−0.602 per x-unit. The negative derivative describes decay even though y itself remains positive.

For y=5e^−0.4x+7, the derivative is unchanged, while the horizontal asymptote moves from y=0 to y=7. A vertical shift affects values but contributes zero to the rate.

The ratio y′/y equals −0.4 only before an additive shift is introduced. For the shifted model, dividing by the whole output does not recover the exponential component’s constant relative rate.

At x=0, the derivative −2 agrees with A k=5(−0.4), a fast coefficient check.

4. Absolute and relative rates in growth and decay

Consider N(t)=120e^0.03t for t≥0, with N measured in model population units and t in years. The initial value is N(0)=120. Differentiation gives N′(t)=3.6e^0.03t population units per year. The absolute rate increases as the model population increases. Dividing by the positive amount gives N′(t)/N(t)=0.03 per year, a constant continuous relative rate.

A relative rate of 0.03 per year does not say that the amount increases by exactly 3% over a whole year. The one-year multiplier is e^0.03, so the finite annual proportional increase is e^0.03−1, approximately 0.030455 or 3.0455%. The distinction is between an instantaneous rate relative to the current amount and accumulated change over a finite interval. They are close for small rate parameters, but they are not identical.

For A(t)=80e^−0.2t, the initial amount is 80 and the initial rate is −16 amount-units per time-unit. At later times the amount is smaller and the magnitude of the loss rate is smaller. The model remains positive for every finite time; it does not predict a negative amount or a finite time at which the exponential alone becomes exactly zero.

Parameters must be read with their units and scope. A laboratory decay fit over several hours need not remain valid indefinitely. A population model can ignore resource limits, changing conditions or discrete individuals. Use the equation within its stated interval, identify whether a question asks for amount or rate, and qualify extrapolation rather than treating the formula as a universal description.

Worked rate comparison. For M(t)=250e^0.06t, M′(4)=15e^0.24≈19.07 units per year and M(4)=250e^0.24≈317.81. Hence M′(4)/M(4)=0.06 per year. Over one complete year the multiplier is e^0.06≈1.06184, an increase of about 6.184%, not exactly 6%.

For D(t)=90e^−0.15t, the half-life solves e^−0.15t=1/2, giving t=ln2/0.15≈4.62 time-units. The initial amount cancels from this calculation.

An absolute rate has amount-per-time units; a relative rate has inverse-time units. Writing −0.15 as ‘15 units lost per year’ confuses these two quantities and ignores the current amount.

Doubling time similarly solves e^0.06t=2, so t=ln2/0.06≈11.55 years.

5. Tangents and local interpretation

A tangent calculation requires both a point and a gradient. For y=f(x) at x=a, use y−f(a)=f′(a)(x−a). The function value supplies the point, while the derivative evaluated at the same input supplies the slope. Substituting a into the derivative alone does not produce an equation of a line. Keep these two evaluations visibly separate in written working.

For y=e^x at x=0, f(0)=1 and f′(0)=1, so the tangent is y=x+1. The tangent gives the local approximation e^h≈1+h for small h. At h=0.02, the approximation is 1.02, compared with e^0.02≈1.020201. This is a nearby estimate, not an exact identity for arbitrary h. The positive second derivative explains why the curve lies above its tangent in this example.

For N(t)=120e^0.03t, the tangent at t=0 is N≈120+3.6t. This straight-line estimate uses the initial rate. It is useful locally, but the exponential rate subsequently grows, while the tangent retains a fixed slope. Extending that line far into the future discards the model’s proportional-growth mechanism and can materially underestimate the predicted amount.

When interpreting a tangent on an applied graph, identify its horizontal and vertical units. Its gradient measures an instantaneous rate at the point of contact; the secant between two distinct times measures an average rate over an interval. A calculator-drawn tangent can support a result, but the written equation and derivative provide reproducible justification. Give exact expressions where requested, and round only the final contextual quantities to a suitable precision.

Worked local estimate. For f(x)=3e^2x, f(0)=3 and f′(0)=6, so the tangent at zero is y=3+6x. It estimates f(0.01) by 3.06, while the model gives 3e^0.02≈3.060604. The small positive gap is consistent with f″(x)=12e^2x>0.

At x=1 the tangent becomes y−3e²=6e²(x−1). The contact point must be evaluated from f and the slope from f′ at the same input.

A tangent approximation is local. At x=0.5 the zero-based tangent gives 6, whereas 3e≈8.155; the error grows because the straight line cannot reproduce the increasing exponential gradient.

For a negative step x=−0.01, the tangent gives 2.94 while the curve gives 3e^−0.02≈2.940596; convexity again places the curve above its tangent.

6. Working accurately with models and technology

Begin an exponential problem by listing the defined variable, the model’s domain and the meanings of its constants. Separate the requested quantity into value, rate, tangent or qualitative behaviour. Substitute into the original model for an amount and into its derivative for a rate. Check that the requested time belongs to the model domain before accepting a calculator output.

For a calculator-free task, leave values such as 6e² in exact form unless instructed otherwise. For a calculator-assumed task, enter brackets around the entire exponent and retain sufficient internal precision. The expressions 4e^3t and 4e³t are different functions. A derivative returned by technology should be checked against the constant multiplier and the inner derivative, then interpreted rather than simply copied.

Use simple inputs as diagnostic checks. At t=0, Ae^kt returns A and its derivative returns Ak. With A>0, positive k should give growth and negative k decay. If an entered model produces a negative amount or a positive rate for a positive decay model, inspect the signs and parentheses. These checks catch errors but do not replace a general derivation.

Evidence for a practical model is limited by the data and assumptions. Units constrain which parameter values are meaningful: changing years to months changes the numerical rate parameter. A fitted equation’s apparent accuracy at a few points does not establish unlimited validity. A complete response presents the derivative, necessary working, correct units and a conclusion tied to the stated setting, while distinguishing exact mathematics from rounded calculations and uncertain modelling assumptions.

Technology check. For Q(t)=42e^−0.08t, entering t=12 gives Q≈16.082 and Q′=−3.36e^−0.96≈−1.287 units per hour. Directly evaluating −0.08Q(12) gives the same rate. This second calculation checks both the exponent brackets and the multiplier.

Solving Q(t)=20 gives t=−ln(20/42)/0.08≈9.27 hours. Substitution returns approximately 20, so the logarithmic solution is consistent with the original model and its t≥0 domain.

Round only the reported time. Using 9.3 inside later calculations would compound rounding error; retain the calculator value when a following part asks for the rate at that instant.

7. Connected reasoning across Exponential functions

Synthesis example. A culture follows P(t)=80e^0.025t, with t in hours. Its initial amount is P(0)=80, its derivative is P′(t)=2e^0.025t, and its relative rate is P′/P=0.025 per hour. At t=10, P≈102.72 and P′≈2.568 units per hour. These values answer different questions: one is an amount and one is the instantaneous change in that amount.

The time to reach 120 solves 80e^0.025t=120. Dividing first gives e^0.025t=1.5, so t=ln(1.5)/0.025≈16.22 hours. Substitution is a direct check: 80e^0.025(16.22) is approximately 120. The positive solution also fits the stated t≥0 domain.

At t=10 the tangent model is L(t)=P(10)+P′(10)(t−10), or L(t)≈102.72+2.568(t−10). It predicts P(11)≈105.29. The exponential gives 80e^0.275≈105.32, so the local estimate is close over one hour. It would become less reliable across a much longer interval because the true slope keeps increasing.

The continuous parameter 0.025 does not mean an exact 2.5% rise over every hour. The one-hour percentage increase is 100(e^0.025−1)≈2.532%. Over ten hours the multiplier is e^0.25≈1.2840, which agrees with 102.72/80. This links the derivative’s relative rate with finite exponential change without treating them as identical.

A change of time unit changes the numerical parameter. Writing s=60t minutes gives P=80e^(0.025/60)s; differentiating with respect to s gives dP/ds=(0.025/60)P. The physical model is unchanged, while the rate is now measured per minute. Units therefore provide an algebraic check on the exponent and derivative.

For an exam response, state the domain, keep e and logarithms exact until evaluation, distinguish value from rate, and test a solved time in the original equation. A graph should show the intercept 80, increasing concave-up behaviour and horizontal asymptote y=0 to the left; those features agree with P>0, P′>0 and P″>0.

A second culture C(t)=80e^0.04t begins at the same amount but grows faster. The ratio C/P=e^0.015t exceeds 1 for t>0 and reaches 1.2 when t=ln1.2/0.015≈12.15 hours. Differentiating gives C′=3.2e^0.04t, so both its absolute and relative rates can be compared with P at the same time. At t=0, C′−P′=1.2 units per hour.

8. Error analysis and final checks for Exponential functions

Error audit through one model. Consider R(t)=150e^−0.12t for t≥0. The correct derivative is R′(t)=−18e^−0.12t. Omitting −0.12 would incorrectly predict a positive rate; differentiating the exponent as −0.12t would incorrectly leave the variable in the multiplier. At t=5, R≈82.32 and R′≈−9.88 units per day.

Direct substitution of h=0 into (e^h−1)/h gives 0/0, which is undefined. The derivative rule comes from the limit as nonzero h approaches zero. With h=0.001, (e^h−1)/h≈1.000500; with h=−0.001 it is about 0.999500. The values support the limiting result 1 from both sides.

The half-life is found from 150e^−0.12t=75, so t=ln2/0.12≈5.776 days. It is not 0.12/2 or 1/0.12. Substituting 5.776 gives approximately 75 and confirms the positive solution. Because an exponential remains positive at finite t, the model never reaches exactly zero.

The tangent at t=0 is R≈150−18t. It gives R(1)≈132, while the exponential gives 150e^−0.12≈133.04. At t=10 the tangent predicts −30, revealing that the linear approximation has been applied far outside its useful neighbourhood; a negative tangent value does not make the exponential amount negative.

The relative rate R′/R=−0.12 per day is constant, but the absolute loss |R′| decreases as R decreases. The finite one-day reduction is 100(1−e^−0.12)≈11.31%, rather than exactly 12%. State which of these quantities is being interpreted and attach the appropriate units.

Final checks should recover R(0)=150, a negative first derivative and a positive second derivative R″=2.16e^−0.12t. Thus the curve decreases while remaining concave up. These three features, together with substitution of any solved time, expose most sign, bracket and interpretation errors.

A shifted decay S(t)=20+150e^−0.12t approaches 20 rather than zero. Its derivative is still −18e^−0.12t, but S′/S is no longer −0.12 because the constant background is included in the denominator. The time to reach 50 solves 150e^−0.12t=30, giving t=ln5/0.12≈13.41 days. This catches the error of applying the unshifted half-life to the whole output. At that time S′=−3.6 units per day; substitution gives S=50.

At t=ln5/0.12, the unshifted exponential component is exactly 30, which independently confirms S=20+30=50. This value-first check agrees with the derivative calculation.