Differentiate natural exponentials and logarithms with their inner rates
Use the natural-function derivatives, multiply by the inner derivative and keep each logarithm's original positive-argument domain.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.4 and 3.4 cover exponential, natural logarithmic and trigonometric derivatives, composites, products and quotients, and first derivatives of parametric and implicit curves, including tangents and normals. Paper 3 additionally requires the derivative of inverse tangent; use its separate lesson for that method. Derivatives of inverse sine and inverse cosine are not required in section 3.4. Trigonometric differentiation uses radians. Keep original domains and check zero derivative denominators separately. Written practice is self-checked.
Before you begin
- Apply the chain rule and differentiate polynomials.
- Use natural logarithms with positive arguments and solve exponential equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the derivative of e^(3x)?
What you will learn
- Differentiate eˣ and ln x, including constant multiples, sums, differences and composites.
- State the original real domain and use an evaluated derivative as a rate.
The natural base has a particularly simple derivative
The derivative of eˣ is eˣ, while the derivative of ln x is 1/x for x > 0. For a differentiable inner rule u(x), the chain rule gives d(eᵘ)/dx = u′eᵘ and d(ln u)/dx = u′/u where u > 0. A constant coefficient multiplies the derivative, and sums and differences are differentiated term by term. The inner rate is still needed: e^(2x) differentiates to 2e^(2x), not merely e^(2x).
For y = e^(2x) + ln(3x − 1), the original log requires x > 1/3. The derivative is 2e^(2x) + 3/(3x − 1) on that domain. A displayed derivative formula might give a number at another input, but it is not the derivative of this real original function there. Do not replace the original logarithm by a log of an absolute value unless the problem defines that different function.
A positive inner argument can allow negative inputs
For y = ln(x² − 4), the logarithm needs x² − 4 > 0, so x < −2 or x > 2. The derivative is 2x/(x² − 4) on those two intervals. Negative x is allowed because the entire log argument is positive. At x = −3 the gradient is −6/5. For ln(5x) on x > 0, the derivative is 5/(5x) = 1/x; the inner coefficient cancels rather than leave an extra 5.
In a supplied amount model P(t) = 10e^(0.2t) grams for t ≥ 0 hours, the rate is P′(t) = 2e^(0.2t) grams per hour. When P = 20, the exact time is t = ln 2/0.2 = 5 ln 2, and the rate is 4 grams per hour. A quantity and its instantaneous rate have different units; evaluate the derivative at the time that satisfies the original condition.
Pause and explain
For ln(x² − 4), is x = −3 allowed and what is its gradient?
Worked example
Differentiate y = e^(2x) + ln(3x − 1), stating its original real domain.
Show the worked solution
- The exponential is defined for every real x, while 3x − 1 > 0 requires x > 1/3.
- For the exponential, multiply its unchanged outer output by the inner derivative 2.
- For the logarithm, divide its inner derivative 3 by 3x − 1 and add the two terms.
Answer y′ = 2e^(2x) + 3/(3x − 1), for x > 1/3.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Differentiate y = 5e^(−3x) + 2 ln x, stating the real domain.
Give me a hint
Differentiate each term and multiply by the exponential's negative inner rate.
Compare my reasoning
- The original logarithm requires x > 0.
- The exponential term differentiates to 5(−3)e^(−3x) = −15e^(−3x).
- The log term gives 2/x, so y′ = −15e^(−3x) + 2/x on the stated domain.
y′ = −15e^(−3x) + 2/x, for x > 0.
Look for these in your work
- Use the inner derivative and each constant multiplier.
- Retain the original positive log-argument domain.
Differentiate y = ln(x² − 4), state all real input intervals and evaluate its gradient at x = −3.
Give me a hint
Solve x² − 4 > 0 before applying the chain derivative u′/u.
Compare my reasoning
- The original domain is x < −2 or x > 2.
- The inner derivative is 2x, giving y′ = 2x/(x² − 4).
- At the permitted input −3, y′ = −6/5; a negative input is valid because its whole log argument is positive.
y′ = 2x/(x² − 4), for x < −2 or x > 2; y′(−3) = −6/5.
Look for these in your work
- Find both intervals where the log argument is positive.
- Evaluate the chain derivative only at a permitted original input.
A supplied amount model is P(t) = 10e^(0.2t) grams for t ≥ 0 hours. Find its rate when the amount reaches 20 grams.
Give me a hint
Find the exact time from the amount equation, then evaluate the derivative rather than the original amount.
Compare my reasoning
- The condition gives e^(0.2t) = 2, so t = 5 ln 2 hours.
- Differentiate to get P′(t) = 2e^(0.2t) grams per hour.
- At the required time the exponential is 2, so the rate is 4 grams per hour.
At t = 5 ln 2 hours, the amount increases at 4 grams per hour.
Look for these in your work
- Use the original condition to locate the evaluation time.
- Distinguish the amount's units from its instantaneous rate units.
Common mistakes
- Both natural-function chain rules need the inner derivative.
- Check the whole log argument, rather than assume every negative input is forbidden.
- A derivative formula does not extend the original function's real domain.
What can you explain now?
Differentiate ln(5x) for x > 0.
Compare with the explanation
1/x.
The chain rule gives 5/(5x), so the constant factor cancels.
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Make it stick
Write the natural exponential and logarithmic chain rules. Explain why a composite logarithm may allow negative inputs and why its derivative formula cannot change the original domain.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.