01 · A negative exponent
Differentiate 9e−x/3.
Hint
The exponent coefficient is −1/3.
Worked solution
The derivative is −3e−x/3. The exponential stays positive; the derivative is negative.
Understand · explore · practise
Differentiating exponentials and logarithms
Differentiate eˣ, aˣ and natural logarithms with clear domain checks. Compare function and derivative graphs, find tangents and interpret original growth and decay models.
Before you startExponential and logarithm laws; differentiation and tangents
01 / Separate height and rate
An exponential height is positive; its gradient can be negative.
The function eˣ is special because its gradient at every x equals its height. Other positive bases introduce the factor ln a. If 0 < a < 1, then ln a < 0: the graph is positive but falling.
d/dx(eˣ) = eˣ
d/dx(aˣ) = aˣ ln a, for a > 0
Compare the blue function with the green derivative. The green value at the selected x is the slope of the gold tangent, not its height.
Function value and gradientExplore
f(x) = eˣ; f′(x) = eˣ.
x = 0; function value = 1; gradient = 1.
The blue function and green derivative coincide only for the base-e case among these exponentials.
The gold line touches the blue graph. The green height gives that line's gradient. Changing to ln x switches to positive inputs and a positive-x graph window.
02 / Include a linear exponent factor
A constant outside and a constant inside play different roles.
d/dx[Aeᵏˣ] = Ak eᵏˣ
Differentiate 7e⁻²ˣ + 4e³ˣWorked example
d/dx(7e⁻²ˣ) = −14e⁻²ˣ
Multiply the outer 7 by the exponent coefficient −2.
d/dx(4e³ˣ) = 12e³ˣ
The base remains e.
Derivative = −14e⁻²ˣ + 12e³ˣ
Differentiate the sum term by term.
A constant added to the exponent multiplies the whole exponential by a constant. For example, e²ˣ⁺¹ = e·e²ˣ, so its derivative is 2e²ˣ⁺¹.
02 · A constant term
Find f′(0) for f(x) = 5 + 2e⁴ˣ.
Hint
The constant 5 has derivative zero.
Worked solution
f′(x) = 8e⁴ˣ, so f′(0) = 8.
03 / Derive the rule for any positive base
Rewrite the base using a natural logarithm.
Why does ln a appear?Worked example
aˣ = eˣˡⁿᵃ, with a > 0
The exponent coefficient ln a is a constant.
d/dx(aˣ) = (ln a)eˣˡⁿᵃ
Use the linear exponential rule.
d/dx(aˣ) = aˣ ln a
Return to the original base.
d/dx[aᵏˣ⁺ᵇ] = k(ln a)aᵏˣ⁺ᵇ, for a > 0
For a = 1, the function is constant and ln 1 = 0. A negative base does not define a real-valued aˣ for every real x, so it is outside this rule.
Compare exponential bases and their derivatives
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Another base
Differentiate 3·5²ˣ.
Hint
Use both the inner factor 2 and ln 5.
Worked solution
The derivative is 6(ln 5)5²ˣ.
04 · Falling but positive
Find the gradient of y = (1/4)ˣ at x = −1.
Hint
The function value is 4 and ln(1/4) = −ln 4.
Worked solution
The gradient is −4ln 4. A negative gradient does not make the function value negative.
04 / Differentiate ln x on its domain
The reciprocal 1/x comes with the condition x > 0.
d/dx(ln x) = 1/x, for x > 0
One explanation uses the inverse relationship. If y = ln x, then x = eʸ. Differentiating with respect to y gives dx/dy = eʸ = x. Reversing a non-zero local gradient gives dy/dx = 1/x.
The graph ln x increases everywhere on its domain, even where its value is negative. Its derivative is positive but gets smaller as x increases.
05 · Logarithm values and slopes
At x = 1/2, is ln x increasing or decreasing? Is its value positive or negative?
Hint
Use the derivative for the direction and the logarithm for the height.
Worked solution
The value ln(1/2) is negative, but the gradient 1/(1/2) = 2 is positive. It is increasing.
05 / Differentiate ln(kx) carefully
The inner factor cancels, but the domain remains.
For a non-zero real constant k, ln(kx) exists only where kx > 0. Its derivative there is k/(kx) = 1/x. The general chain rule will explain this cancellation in more detail.
Two valid real domainsWorked example
y = ln(6x), x > 0: dy/dx = 1/x
You may also split ln(6x) = ln 6 + ln x.
y = ln(−3x), x < 0: dy/dx = 1/x
You cannot write ln(−3) + ln x using real logarithms.
k = 0: ln(kx) is undefined
There is no derivative to calculate.
Distinguish ln(kx) from k ln x. The latter has derivative k/x for x > 0.
06 · Compare two expressions
Differentiate ln(8x) and 8ln x, stating their domains.
Hint
In the first, 8 is inside the logarithm.
Worked solution
Both require x > 0. The derivatives are 1/x and 8/x respectively.
07 · A negative input interval
Find the gradient of y = ln(−5x) at x = −2.
Hint
The log argument is 10, so the point is in the real domain.
Worked solution
The derivative is 1/x on x < 0; its value at −2 is −1/2.
06 / Simplify without losing the domain
Even powers allow negative x inside a logarithm.
For x ≠ 0, ln(x²) = 2ln|x| and its derivative is 2/x. Writing 2ln x would accidentally discard the negative half of the original domain. For ln(x³), the real domain is x > 0, so 3ln x is valid there.
Differentiate ln(x⁴) + ln(2x²)Worked example
Domain: x ≠ 0
Both logarithm arguments are positive on either side of zero.
ln(x⁴) = 4ln|x|; ln(2x²) = ln 2 + 2ln|x|
Use absolute values to keep the full real domain.
Derivative = 6/x, x ≠ 0
The constant ln 2 disappears.
08 · Keep both sides
Differentiate ln(x⁶), and state the full real domain.
Hint
x⁶ is positive for every non-zero x.
Worked solution
ln(x⁶) = 6ln|x|. The derivative is 6/x on x ≠ 0.
09 · An odd power
Differentiate ln(x⁵) on its real domain.
Hint
x⁵ must be positive.
Worked solution
The domain is x > 0. There ln(x⁵) = 5ln x, so the derivative is 5/x.
07 / Use exact values for a tangent
Natural logarithms often give convenient coordinates.
Find the tangent to y = e²ˣ at x = ln 2Worked example
y = e²ˡⁿ² = 4
The contact point is (ln 2, 4).
dy/dx = 2e²ˣ, so m = 8
Evaluate the derivative at the same x.
y − 4 = 8(x − ln 2)
Leave the exact logarithm in the line equation.
10 · Logarithmic tangent
Find the tangent to y = ln x at x = e.
Hint
The contact point is (e, 1).
Worked solution
The gradient is 1/e. Thus y − 1 = (x − e)/e, or y = x/e.
11 · Exponential normal
Find the normal to y = 2ˣ at x = 0.
Hint
The tangent gradient is ln 2 and the point is (0, 1).
Worked solution
The normal is y − 1 = −x/(ln 2).
08 / Solve a gradient condition
An exponential term can never equal zero.
Find the stationary point of f(x) = eˣ − 3xWorked example
f′(x) = eˣ − 3
Set f′(x) = 0.
eˣ = 3, so x = ln 3
Taking ln is valid because both sides are positive.
f(ln 3) = 3 − 3ln 3
Return to the original function for y.
f″(x) = eˣ > 0
The stationary point is a minimum.
12 · A prescribed gradient
At what x does y = 4e²ˣ have gradient 24?
Hint
Solve 8e²ˣ = 24.
Worked solution
e²ˣ = 3, so x = (ln 3)/2.
09 / Interpret growth and decay rates
A finite percentage increase is not the instantaneous proportional rate.
A simple growth model is P = 800(1.06)ᵗ, where t is measured in years. One year multiplies the population by 1.06. Treating the model as continuous gives dP/dt = P ln(1.06), measured in individuals per year. The instantaneous proportional rate is ln(1.06), approximately 0.0583 per year.
A decay model: M = 60e−0.2t grams, t in daysWorked example
dM/dt = −12e−0.2t grams/day
The negative sign means mass is decreasing.
At t = 5ln 2 days, M = 30g
The exponential factor is 1/2.
At that time, dM/dt = −6g/day
The amount and its rate of change are different quantities.
13 · Half-life rate
For M = 60e−0.2t, how does the magnitude of the loss rate change when the mass halves?
Hint
dM/dt = −0.2M.
Worked solution
Its magnitude halves too: it is proportional to the current mass. Initially the loss rate is 12g/day; at half the mass it is 6g/day.
14 · A doubling time
For P = 800(1.06)ᵗ, find the doubling time and the instantaneous rate at that time.
Hint
Set (1.06)ᵗ = 2, then use dP/dt = P ln(1.06).
Worked solution
t = ln 2 / ln 1.06 ≈ 11.9 years. At that time P = 1600, so dP/dt = 1600ln(1.06) ≈ 93.2 individuals/year. A fractional rate is a model average, not a fractional person.
10 / Check the base, the exponent and the domain
A short check prevents several common errors.
- The derivative of eˣ is eˣ; the derivative of aˣ includes ln a.
- Multiply by the coefficient of a linear exponent.
- For ln(kx), check kx > 0 before using 1/x.
- Use ln|x| when a log law must retain negative x.
- Give rates the correct units and interpret negative signs.
Section 1 of 10 · Separate height and rate