The natural exponential as the function that is its own derivative and why that singles out e, the natural logarithm's derivative by implicit differentiation, general bases and their stray logarithm factor, logarithmic differentiation for awkward products and powers, and the completed proof of the power rule for every real exponent.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Derivatives of Exponential and Logarithmic Functions
Objectives
Five outcomes, and the first explains a choice made back in Section 1.5: why calculus uses e rather than any friendlier base.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 275-287 — the section these objectives are drawn from
Warm-up
Section 1.5 introduced e through continuous compounding, and Section 3.3 warned that the power rule does not apply to 2 to the x because the exponent is the variable.
Discussion prompt
The base-2 exponential doubles every time x increases by 1. Guess what its derivative looks like in shape, and say why no rule so far can produce it.
Hint: Ask what the graph's steepness does as the height grows.
Answer:
The steeper the curve gets, the taller it already is — the slope grows in proportion to the height. So the derivative should look like the function itself, up to a constant factor.
\[ \frac{d}{dx}\left[b^{x}\right] = k\,b^{x} \quad \text{for some constant } k \text{ depending on } b \]
No rule so far can produce it, because the exponent is the variable rather than the base. This section finds that constant, and discovers that for exactly one base it equals 1 — which is why that base gets its own name.
Concept
Every exponential's derivative is a constant multiple of itself. For base e the constant is exactly one, so the natural exponential is its own derivative — a property no other function of this kind has.
the natural exponential — The exponential with base e, which equals its own derivative at every input. Its slope at each point equals its height there, and this property characterises it among all exponentials.
\[ \frac{d}{dx}\left[e^{x}\right] = e^{x} \]
This is the answer to a question Section 1.5 left hanging. The number e is not chosen for elegance but because it is the base at which the constant of proportionality becomes one, and every other base carries a logarithm around forever.
Figure (svg): The natural exponential with tangents whose slopes equal the heights at the same points
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 275-278
Section
Section 1
Concept
The derivative of e to the x is e to the x. At the origin its value is 1 and its slope is 1; at x equal to 1 both are e. The graph's steepness at any point is exactly its height there.
self-derivative — A function equal to its own derivative. Up to a constant multiple, the natural exponential is the only such function, which is why it appears in every model of growth proportional to size.
\[ y = e^{x} \;\Longrightarrow\; y' = y \]
The property has real consequences. Any quantity growing at a rate proportional to its current size satisfies exactly this equation, which is why the natural exponential describes populations, compound interest and radioactive decay alike.
Figure (svg): The natural exponential with tangents whose slopes equal the heights at the same points
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 275-279 — the derivative of the natural exponential
Picture it
The natural exponential with tangents at 0 and at 1.
Figure (svg): The natural exponential with tangents whose slopes equal the heights at the same points
At the origin the height is 1 and the tangent rises at 45 degrees; at x equal to 1 both height and slope are about 2.718. The coincidence is not a coincidence — it is the defining property.
Worked example
Example 3.66. The rule plus Section 3.6.
\[ \text{Differentiate } y = e^{3x^{2}} \text{ and } y = xe^{x}. \]
For the first, apply the exponential rule
Why: The exponential is unchanged.
\[ e ^{3 x ^{2}} \]
Multiply by the exponent's derivative
Why: Chain rule.
\[ \times 6 x \]
State the first answer
Why: Collect.
\[ 6 x e ^{3 x ^{2}} \]
For the second, use the product rule
Why: Two factors.
\[ 1 \times e ^{x} + x \times e ^{x} \]
Factor
Why: Common factor.
\[ e ^{x}(1 + x) \]
Figure (svg): The solution to Worked example the exponential with the chain rule shown as a ladder of expressions, one row per legal move
\[ 6xe^{3x^{2}}, \qquad e^{x}(1+x) \]
Verify: check the second at a landmark
Why: At x equal to negative 1 the second derivative expression gives zero, so x times e to the x has a horizontal tangent there — and indeed that function has a minimum at negative 1, since it is negative to the left and rising to the right. For the first, the exponent's derivative 6x vanishes at 0, giving a horizontal tangent at the origin, which the even function e to the 3x squared must have. Both check against the graphs.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 277-278
Sorting
Look at where the variable is.
Sort into buckets
Sort each expression.
x to the e is worth pausing on: e is just a number, about 2.718, so this is an ordinary power and the power rule applies. Comparing it with e to the x, which sits in a different bucket, is the sharpest test of whether the distinction has landed.
Worked example
Checkpoint 3.66. The constant, computed.
\[ \text{Show that } \frac{d}{dx}\left[b^{x}\right] = b^{x}\ln b, \text{ and find the base making the constant } 1. \]
Rewrite the base in terms of e
Why: Any positive base is a power of e.
\[ b = e ^{\ln b} \]
Substitute
Why: So b to the x is an exponential of a multiple.
\[ b ^{x} = e ^{x \ln b} \]
Differentiate with the chain rule
Why: The exponent's derivative is the constant ln b.
\[ e ^{x \ln b} \times \ln b \]
Rewrite back
Why: Recognise the exponential.
\[ b ^{x} \ln b \]
Set the constant to 1
Why: The logarithm equals 1 only at e.
\[ \ln b = 1\text{ gives } b = e \]
Figure (svg): Three exponentials with their tangent slopes at the origin, showing e as the base where the slope is exactly one
\[ \frac{d}{dx}\left[b^{x}\right] = b^{x}\ln b; \quad \ln b = 1 \iff b = e \]
Verify: check the constant for two familiar bases
Why: For base 2 the constant is ln 2, about 0.693, so 2 to the x rises less steeply than its own height; for base 4 it is ln 4, about 1.386, so that curve rises more steeply than its height. Only at e do the two coincide, and the graph confirms it: the dashed line of slope 1 through the point (0,1) is tangent to e to the x alone. This is the whole reason calculus prefers e — it is the base at which the stray constant disappears.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 278-279
Trap
\[ y = e^{x} \]
Bring the exponent down
Why: The student treats it as a power of the variable.
\[ y' = xe^{x-1} \quad \text{(wrong)} \]
The power rule applies to a variable base with a constant exponent. Here the base is the constant e and the exponent is the variable — the opposite situation.
\[ \frac{d}{dx}\left[e^{x}\right] = e^{x}, \qquad \frac{d}{dx}\left[x^{e}\right] = ex^{e-1} \]
Check which of the base and exponent carries the variable
Why: Variable base with constant exponent is the power rule; constant base with variable exponent is the exponential rule.
The two expressions above look almost identical and are entirely different functions with entirely different derivatives. When the variable appears in BOTH places, as in x to the x, neither rule applies and logarithmic differentiation is needed — which is the last idea of this section.
Fill the middle
Differentiating a general exponential by rewriting it in base e.
Fill in the blanks
b^\ln b = e^___ \;\Longrightarrow\; \frac______ = b^___\cdot___
Why: The exponent x times ln b has derivative ln b, and the chain rule brings it out front. That constant is 1 only when b is e, which is exactly why e is the natural base.
Two truths and a lie
All three are about the exponential.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it applies the power rule to an exponential. The base here is the constant e and the exponent is the variable, which is the opposite of what the power rule handles. Testing at x equal to 0 refutes it: the true derivative is e to the 0, which is 1, while the false one gives 0.
Prediction
Commit before reasoning.
Predict first
A quantity grows at a rate proportional to its current size. What kind of function is it?
Correct: An exponential — the only family whose derivative is proportional to itself.
\[ y' = ky \;\Longrightarrow\; y = Ce^{kt} \]
Why: The condition that the rate be proportional to the size is exactly the equation y prime equals k y, and the exponentials are its solutions. This is why compound interest, population growth and radioactive decay all produce exponentials: in each case the change per unit time is proportional to how much is currently there. A linear function grows at a constant rate regardless of size, and a quadratic's rate grows with the input rather than with the output.
Section
Section 2
Concept
Writing y equal to the natural logarithm of x and exponentiating gives e to the y equals x. Differentiating that implicitly and solving produces one over x, with the original equation supplying the final substitution.
the logarithm's derivative — The derivative of the natural logarithm of x is one over x, for positive x. For the logarithm of the absolute value it is also one over x, now valid for every non-zero x.
\[ \frac{d}{dx}\left[\ln x\right] = \frac{1}{x}, \quad x > 0 \]
The extension to the absolute value matters more than it looks. It makes one over x have an antiderivative on both sides of zero, which Section 5.6 will need constantly.
Figure (svg): Deriving the logarithm's derivative by implicit differentiation of the exponential form
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 279-282 — the derivative of the logarithm
Picture it
The logarithm's derivative, derived.
Figure (svg): Deriving the logarithm's derivative by implicit differentiation of the exponential form
This is Section 3.8's method applied to the equation defining the logarithm, and the last step substitutes the original equation exactly as that section's procedure prescribes.
Worked example
Example 3.69. Exponentiate, differentiate, solve.
\[ \text{Find } \frac{d}{dx}\left[\ln x\right]. \]
Name the function and remove the logarithm
Why: Exponentiate both sides.
\[ y = \ln x,\text{ so } e ^{y} = x \]
Differentiate implicitly with respect to x
Why: The chain rule on the left.
\[ e ^{y}(\,dy / \,dx) = 1 \]
Solve for the derivative
Why: Divide.
\[ \,dy / \,dx = 1 / e ^{y} \]
Substitute the original equation
Why: e to the y is x.
\[ = \frac{1}{x} \]
State the domain
Why: The logarithm needs a positive input.
\[ \text{for } x > 0 \]
Figure (svg): The natural logarithm above its derivative, showing a shrinking positive slope
\[ \frac{d}{dx}\left[\ln x\right] = \frac{1}{x} \]
Verify: check the sign, the value at 1, and the long-run behaviour
Why: The derivative is positive throughout the domain, so the logarithm always rises — which it does. At x equal to 1 it is exactly 1, and the logarithm does cross the axis there at 45 degrees. And it tends to 0 as x grows, matching a curve that rises without bound but ever more slowly. All three agree with the graph, and the derivation used nothing but Section 3.8's technique.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 280-281
Fill the middle
The last step of the logarithm derivation.
Fill in the blanks
\fracx___ = \frac______} = \frac______}
Why: The equation from the second line says e to the y is x, so substituting gives one over x. Using the original equation at the end is standard practice in implicit differentiation.
Worked example
Checkpoint 3.69. The inner derivative over the inner function.
\[ \text{Differentiate } y = \ln(x^{2}+3) \text{ and } y = \ln|x|. \]
For the first, apply the logarithm rule
Why: One over the inside.
\[ \frac{1}{x ^{2} + 3} \]
Multiply by the inside's derivative
Why: Chain rule.
\[ \times 2 x \]
State
Why: Collect.
\[ 2 x / (x ^{2} + 3) \]
For the second, split by sign
Why: For positive x the absolute value does nothing.
\[ \frac{1}{x} \]
Check the negative side
Why: For negative x, ln(-x) has derivative -1/(-x).
\[ \text{also } \frac{1}{x} \]
Figure (svg): The solution to Worked example logarithms with the chain rule shown as a ladder of expressions, one row per legal move
\[ \frac{2x}{x^{2}+3}, \qquad \frac{d}{dx}\ln|x| = \frac{1}{x} \]
Verify: check the second result on the negative side explicitly
Why: For x negative, the absolute value is minus x, and differentiating ln of minus x gives one over minus x times minus 1, which is one over x. So the same formula holds on both sides — a genuinely useful fact, since it means one over x has an antiderivative wherever it is defined rather than only on the positive half. Section 5.6 relies on this constantly, and the absolute value is not decoration there.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 281-282
Error analysis
A student differentiates a logarithm of a composite argument.
Annotate
On: \( \frac{d}{dx}\left[\ln(x^{2}+3)\right] = \frac{1}{x^{2}+3} \)
A reduction check catches it: with a bare x inside, the extra factor is 1 and the familiar rule is recovered. Whenever the argument is anything else, the factor is doing real work — and here it changes the sign of the answer on the negative half.
Matching
Inner derivative over inner function.
Match the pairs
Why: The general pattern is the inner function's derivative over the inner function itself. The last row shows it producing something unexpectedly tidy: cosine over sine is the cotangent, which is why logarithms of trigonometric functions turn up in integration tables.
Two truths and a lie
All three are about the logarithm.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one: the chain rule's factor of 2x is missing. The general pattern is the inner derivative over the inner function, and here that is 2x over x squared plus 3. Note that the false version is positive everywhere while the true one is negative for negative x — the missing factor changes the answer's sign, not merely its size.
Prediction
Commit before reasoning.
Predict first
Why is the derivative usually stated for ln|x| rather than ln x?
Correct: So the formula covers negative x, giving one over x an antiderivative wherever it is defined.
\[ x < 0: \; \frac{d}{dx}\ln(-x) = \frac{1}{-x}\cdot(-1) = \frac{1}{x} \]
Why: The natural logarithm accepts only positive inputs, so ln x has a derivative only on the positive half — but one over x is defined for every non-zero x. The absolute value extends the antiderivative to the negative half, where differentiating ln of minus x gives one over x again by the chain rule. The absolute value does NOT change the derivative; that is precisely the point, and Section 5.6 depends on it.
Section
Section 3
Concept
Rewriting any base as a power of e shows that a general exponential's derivative picks up a factor of the natural logarithm of the base, and a general logarithm's derivative picks up the same factor in its denominator.
general base rules — The derivative of b to the x is b to the x times the natural logarithm of b; the derivative of the logarithm of x to base b is one over x times that same logarithm. Both reduce to the natural case when b is e.
\[ \frac{d}{dx}\left[b^{x}\right] = b^{x}\ln b, \qquad \frac{d}{dx}\left[\log_{b}x\right] = \frac{1}{x\ln b} \]
Rather than memorising these, convert to base e first. Any exponential is e to the x times ln b, and any logarithm is a natural logarithm divided by ln b, after which only the natural rules are needed.
Figure (svg): The general-base derivatives, with the stray logarithm factor that base e avoids
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 282-284 — derivatives with a general base
Picture it
The four cases side by side.
Figure (svg): The general-base derivatives, with the stray logarithm factor that base e avoids
When the base is e the factor is the logarithm of e, which is 1, so the clean rules are the general ones specialised. Nothing separate is being remembered.
Worked example
Example 3.71. Convert to base e, or use the rules.
\[ \text{Differentiate } y = 3^{x^{2}} \text{ and } y = \log_{2}(5x+1). \]
For the first, apply the general exponential rule
Why: The exponential, times the base's logarithm.
\[ 3 ^{x ^{2}} \ln 3 \]
Multiply by the exponent's derivative
Why: Chain rule.
\[ \times 2 x \]
State
Why: Collect.
\[ 2 x \ln 3 \times 3 ^{x ^{2}} \]
For the second, apply the general logarithm rule
Why: One over the argument times ln 2.
\[ \frac{1}{(5 x + 1) \ln 2} \]
Multiply by the argument's derivative
Why: Chain rule.
\[ \times 5 \]
Figure (svg): The solution to Worked example a general exponential and logarithm shown as a ladder of expressions, one row per legal move
\[ 2x\ln 3\cdot 3^{x^{2}}, \qquad \frac{5}{(5x+1)\ln 2} \]
Verify: check by converting to base e instead
Why: Rewriting the first as e to the x squared times ln 3 and differentiating with the chain rule gives the same answer, with the factor ln 3 emerging from the exponent's derivative rather than from a memorised rule. Likewise the second is the natural logarithm of 5x plus 1, divided by the constant ln 2, and differentiating that gives 5 over 5x plus 1, all over ln 2 — identical. The conversion route needs no extra rules at all, which is why it is worth preferring.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 283-284
Fill the middle
Differentiating a base-2 logarithm via change of base.
Fill in the blanks
\log_\ln 2x = \frac______ \;\Longrightarrow\; \frac______ = \frac______}}
Why: Change of base makes the position obvious: the constant divides the whole logarithm, so it divides the derivative too and lands in the denominator. Deriving it beats remembering it.
Worked example
Checkpoint 3.71. Compare the two routes.
\[ \text{Compare differentiating } 10^{x} \text{ and } e^{x}. \]
Differentiate the natural one
Why: It is its own derivative.
\[ d / \,dx [e ^{x}] = e ^{x} \]
Differentiate base 10
Why: The rule with the stray factor.
\[ 10 ^{x} \ln 10 \]
Note the constant
Why: About 2.303.
\[ \ln 10 = 2.3026 \]
Differentiate base 10 twice
Why: The factor appears again.
\[ 10 ^{x}(\ln 10) ^{2} \]
Compare
Why: The natural one accumulates nothing.
Figure (svg): The solution to Worked example why base e is preferred shown as a ladder of expressions, one row per legal move
\[ \frac{d^{n}}{dx^{n}}\left[10^{x}\right] = 10^{x}(\ln 10)^{n} \]
Verify: consider what this means for repeated differentiation
Why: After ten differentiations, base 10 has accumulated ln 10 to the tenth power, about 4700, while base e is unchanged. In any calculation involving repeated differentiation — a Taylor series, a differential equation — that accumulation is a permanent nuisance. This is the practical answer to why calculus uses e rather than the base humans count in: it is the only base for which differentiating leaves no residue.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 284-284
Trap
\[ \frac{d}{dx}\left[\log_{2}x\right] = \frac{\ln 2}{x} \quad \text{(wrong)} \]
Put the logarithm in the numerator
Why: The student mirrors the exponential rule, where it multiplies.
The base-2 logarithm rises faster than the natural one, so its derivative must be LARGER than one over x, and dividing by ln 2 does that while multiplying does not.
\[ \frac{d}{dx}\left[\log_{2}x\right] = \frac{1}{x\ln 2} \]
Derive it from the change of base rather than remembering it
Why: Log base 2 is ln x over ln 2, so the constant divides.
Change of base settles the position instantly: the base-2 logarithm IS the natural one divided by ln 2, so its derivative is one over x divided by ln 2. Since ln 2 is less than 1, dividing makes the derivative larger, which matches the base-2 logarithm being the steeper curve.
Matching
Watch where the stray logarithm goes.
Match the pairs
Why: The factor multiplies for exponentials and divides for logarithms, which is exactly what the inverse relationship demands: reciprocal slopes, as Section 3.7 established. Both rows collapse to the clean ones when b is e.
Sorting
Multiplying, or dividing?
Sort into buckets
Sort each derivative.
The multiplying and dividing pattern is what reciprocal slopes require: if the exponential's derivative is scaled up by a factor, its inverse's must be scaled down by the same one. Section 3.7's theorem predicts it exactly.
Prediction
Commit before reasoning.
Predict first
What practical advantage does base e have over base 10?
Correct: Differentiating in base e leaves no stray constant.
\[ \frac{d^{10}}{dx^{10}}\left[10^{x}\right] = 10^{x}(\ln 10)^{10} \approx 4700\cdot 10^{x} \]
Why: Every differentiation of a base-10 exponential multiplies by ln 10, so after n differentiations the factor is ln 10 to the nth — a growing nuisance in any calculation involving repeated derivatives. Base e leaves the function unchanged every time. Computability is not the issue, since both are equally easy for a machine, and irrationality is shared by many numbers that lack this property. The advantage is entirely about what differentiation does.
Section
Section 4
Concept
For a complicated product, quotient or power, taking the natural logarithm of both sides converts products into sums and exponents into factors. Differentiating implicitly then gives the derivative over the function, and multiplying back finishes it.
logarithmic differentiation — The technique of taking natural logarithms of both sides of y equals f of x, differentiating implicitly, and multiplying by y to recover the derivative. The log laws flatten the structure before any differentiation happens.
\[ \ln y = \ln f(x) \;\Longrightarrow\; \frac{y'}{y} = \frac{d}{dx}\big[\ln f(x)\big] \]
The left side always produces y prime over y, because differentiating the logarithm of y gives one over y times the implicit factor. That is Section 3.8's technique doing the work.
Figure (svg): Logarithmic differentiation turning a product and a power into a sum
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 284-286 — logarithmic differentiation
Picture it
A product over a root, flattened by logarithms.
Figure (svg): Logarithmic differentiation turning a product and a power into a sum
The log laws of Section 1.5 turn a three-rule problem into a sum of four simple derivatives. Neither the product rule nor the quotient rule is needed at any point.
Worked example
Example 3.74. Logs first, then differentiate.
\[ \text{Differentiate } y = \frac{x^{2}(3x+1)^{5}}{\sqrt{x-2}}. \]
Take natural logarithms of both sides
Why: The log laws split it.
\[ \ln y = 2 \ln x + 5 \ln(3 x + 1) - (\frac{1}{2}) \ln(x - 2) \]
Differentiate both sides implicitly
Why: The left gives y prime over y.
\[ y' / y = \frac{2}{x} + \frac{15}{3 x + 1} - \frac{1}{2(x - 2)} \]
Multiply both sides by y
Why: Recover the derivative.
\[ \text{y' } = y \times\text{ that bracket} \]
Substitute y back
Why: The original expression.
\[ \text{y' } = (x ^{2}(3 x + 1) ^{5} / \sqrt{x - 2}) \times\text{ the bracket} \]
Figure (svg): Logarithmic differentiation turning a product and a power into a sum
\[ y' = y\left(\frac{2}{x} + \frac{15}{3x+1} - \frac{1}{2(x-2)}\right) \]
Verify: count the rules avoided and check one term
Why: Done directly this would need the quotient rule, the product rule inside it, and the chain rule three times. Logarithmic differentiation needed only the logarithm rule, applied four times to simple arguments. Checking the middle term: 5 ln of 3x plus 1 differentiates to 5 times 3 over 3x plus 1, which is 15 over 3x plus 1 — the exponent became a coefficient and the chain rule supplied the 3. Each term is independently checkable, which is another advantage over one long quotient-rule computation.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 285-285
Ranking
Logarithmic differentiation from start to finish.
Put in order
Why: Step c is where the work is saved and step e is where it is most often left unfinished. The left side always produces y prime over y, because differentiating the logarithm of y gives one over y times the implicit factor.
Worked example
Checkpoint 3.74. The case where no other method works.
\[ \text{Differentiate } y = x^{x} \text{ for } x > 0. \]
Note that neither earlier rule applies
Why: The variable is base and exponent.
Take logarithms
Why: The power law brings the exponent down.
\[ \ln y = x \ln x \]
Differentiate both sides
Why: Product rule on the right.
\[ y' / y = \ln x + 1 \]
Multiply back
Why: By y.
\[ y' = y(\ln x + 1) \]
Substitute
Why: The original function.
\[ y' = x ^{x}(\ln x + 1) \]
Figure (svg): A function with the variable in both the base and the exponent, needing logarithmic differentiation
\[ y' = x^{x}(\ln x + 1) \]
Verify: find where the derivative vanishes and check it
Why: The derivative is zero when ln x equals negative 1, that is at x equal to one over e, about 0.368. So x to the x has a horizontal tangent there — and it does: the function decreases to a minimum of about 0.692 at that point and rises thereafter. A numerical check near 0.368 confirms the turning point. Note that neither the power rule nor the exponential rule could have produced this, which makes logarithmic differentiation genuinely necessary here rather than merely convenient.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 286-286
Trap
\[ \frac{y'}{y} = \ln x + 1 \]
Report this as the derivative
Why: The student stops after differentiating.
\[ y' = \ln x + 1 \quad \text{(wrong)} \]
The left side is the derivative DIVIDED by the function, so recovering the derivative needs a multiplication by y.
\[ y' = y(\ln x + 1) = x^{x}(\ln x + 1) \]
Multiply both sides by y and substitute the original expression
Why: Two steps, and the second is also easy to forget.
A quick check catches both omissions: at x equal to 1 the true derivative should be 1 times ln 1 plus 1, which is 1, and the function x to the x does pass through (1,1) rising at 45 degrees. The unmultiplied version happens to agree there by coincidence, so test at x equal to 2 instead, where the true value is 4 times ln 2 plus 1, about 6.77.
Fill the middle
The last step for x to the x, after differentiating.
Fill in the blanks
\fracx^x___ = \ln x + 1 \;\Longrightarrow\; y' = ___(\ln x + 1)
Why: Multiplying by y and substituting the original expression gives x to the x times the bracket. Both steps are needed, and stopping before either leaves an answer that is not the derivative.
Sorting
Ask how many rules the direct route would need.
Sort into buckets
Sort each expression.
The bottom bucket is where the technique stops being an optimisation. For anything raised to a variable power there is simply no other route, which is why the method has to be learned rather than merely appreciated.
Prediction
Commit before reasoning.
Predict first
Differentiating ln y with respect to x gives what?
Correct: y prime over y — the chain rule attaches the implicit factor.
\[ \frac{d}{dx}\big[\ln y\big] = \frac{1}{y}\cdot\frac{dy}{dx} \]
Why: The logarithm's derivative is one over its argument, and since y is a function of x the chain rule multiplies by dy by dx. So the result is one over y times y prime. This is exactly Section 3.8's implicit factor, and it is what makes the whole technique work: the left side always yields the derivative divided by the function, so multiplying by the function recovers it.
Section
Section 5
Concept
Logarithmic differentiation proves the power rule for an arbitrary real exponent, including irrational ones. Section 3.3 stated it, Section 3.7 earned the rational case, and this closes the gap.
the completed power rule — For any real number n and positive x, the derivative of x to the n is n times x to the n minus 1. The proof for irrational n requires logarithmic differentiation, since no algebraic argument reaches those exponents.
\[ \frac{d}{dx}\left[x^{n}\right] = nx^{n-1} \quad \text{for every real } n \]
The three-stage history is worth noticing. A rule stated in one line took a binomial argument, an inverse function argument and a logarithmic argument to establish across its full range.
Figure (svg): The power rule for an irrational exponent, proved by logarithmic differentiation
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 286-287 — the power rule for irrational exponents
Picture it
The power rule for an arbitrary real exponent.
Figure (svg): The power rule for an irrational exponent, proved by logarithmic differentiation
The proof is short because logarithmic differentiation is exactly suited to it: the power law turns the exponent into a coefficient, and everything after that is routine.
Worked example
Example 3.76. Logs make the exponent a coefficient.
\[ \text{Prove that } \frac{d}{dx}\left[x^{n}\right] = nx^{n-1} \text{ for any real } n \text{ and } x > 0. \]
Set y equal to the power and take logarithms
Why: The power law brings n down.
\[ \ln y = n \ln x \]
Differentiate both sides
Why: Implicit on the left, ordinary on the right.
\[ y' / y = \frac{n}{x} \]
Multiply back by y
Why: Recover the derivative.
\[ y' = y(\frac{n}{x}) \]
Substitute the original expression
Why: y is x to the n.
\[ = x ^{n}(\frac{n}{x}) \]
Simplify
Why: Subtract the exponents.
\[ = n x ^{n - 1} \]
Figure (svg): The solution to Worked example proving the general power rule shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[x^{n}\right] = nx^{n-1} \]
Verify: confirm the proof genuinely covers irrational exponents
Why: Nothing in the argument assumed n was a whole number or even rational — the power law for logarithms holds for every real exponent, so the proof does too. That matters because no binomial expansion exists for an irrational power and no inverse-function argument reaches it either. Testing at n equal to pi: the derivative of x to the pi is pi times x to the pi minus 1, and a numerical difference quotient at x equal to 2 confirms it to several digits.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 286-287
Fill the middle
The general power rule, after multiplying back by y.
Fill in the blanks
y' = x^n x^{n-1}\cdot\frac______ = ___
Why: Dividing x to the n by x gives x to the n minus 1, leaving n times that. The argument never assumed anything about n, which is why it covers irrational exponents.
Worked example
Checkpoint 3.76. The rule, used.
\[ \text{Differentiate } y = x^{\sqrt{2}} \text{ and } y = (x^{2}+1)^{\pi}. \]
Apply the power rule to the first
Why: Exponent down, reduced by one.
\[ \sqrt{2} x ^{\sqrt{2} - 1} \]
For the second, apply the power form of the chain rule
Why: Exponent down, inside untouched.
\[ \pi(x ^{2} + 1) ^{\pi - 1} \]
Multiply by the inside's derivative
Why: Chain rule.
\[ \times 2 x \]
State
Why: Collect.
\[ 2 \pi x(x ^{2} + 1) ^{\pi - 1} \]
Figure (svg): The solution to Worked example a general power in practice shown as a ladder of expressions, one row per legal move
\[ \sqrt{2}\,x^{\sqrt{2}-1}, \qquad 2\pi x\,(x^{2}+1)^{\pi-1} \]
Verify: check the sign and a landmark on the second
Why: The second derivative expression vanishes at x equal to 0, so the curve has a horizontal tangent there — which it must, since x squared plus 1 is smallest at the origin and any positive power of it is too. It is negative for negative x and positive for positive x, matching a minimum. Note that the irrational exponents behaved exactly like ordinary ones throughout, which is the point of having proved the rule in full generality.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 287-287
Error analysis
A student differentiates a power with an irrational exponent.
Annotate
On: \( \frac{d}{dx}\left[x^{\pi}\right] = x^{\pi}\ln x \)
An unusual-looking exponent does not change which rule applies. The question is always which of the base and the exponent carries the variable, and pi is simply a number like any other.
Sorting
Ask which position carries the variable.
Sort into buckets
Sort each expression.
The pairs a-b and d-e are deliberately confusing, and the test is always the same: is x downstairs or upstairs? Pi and e are just numbers, and their appearance in an exponent makes it a constant exponent.
Two truths and a lie
All three are about the completed rule.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Pi is a constant exponent and x is the variable base, so this is a power and the power rule applies, giving pi times x to the pi minus 1. The exponential rule is for pi to the x, where the roles are reversed. The exotic-looking exponent changes nothing about which rule is appropriate.
Prediction
Commit before reasoning.
Predict first
The power rule was stated in Section 3.3. Why did it take until now to prove?
Correct: Because different classes of exponent need genuinely different arguments.
\[ \text{whole: 3.3} \;\to\; \text{rational: 3.7} \;\to\; \text{real: 3.9} \]
Why: The binomial expansion proves the whole-number case but says nothing about a half power. The inverse function theorem reaches the rational exponents, since roots are inverses of powers. Neither touches an irrational exponent, for which the only route is logarithmic differentiation — because the log power law is the one tool that holds for every real exponent. The rule never changed and the statement was never wrong; only the justification was incomplete.
Comparison
Fill the blanks. Base e is the case where nothing extra appears.
Comparison matrix
| Function | Derivative | Stray factor |
|---|---|---|
| e^x | e^x | none |
| b^x | b^x ln b | ln b, multiplying |
| ln x | 1/x | none |
| log_b x | 1/(x ln b) | ln b, dividing |
The factor multiplies for exponentials and divides for logarithms, which is exactly what Section 3.7's reciprocal-slope theorem requires of a function and its inverse.
Pattern
Given an expression involving exponentials, logarithms or awkward powers.
Step five is where the method is most often left unfinished. The left side gives the derivative divided by the function, so two operations are needed to recover the derivative itself.
Stewart, Calculus: Early Transcendentals 8e, §3.6 Derivatives of Logarithmic Functions §3.6, pp. 218-223
Check
Power or exponential? Look at the variable's position.
Check your understanding
What is the derivative of e^x?
Answer: A
Why: The natural exponential is its own derivative — the property that singles out e.
Check
The logarithm with a composite argument.
Check your understanding
Differentiate ln(x^2 + 3).
Answer: A
Why: The rule gives one over the argument, and the chain rule multiplies by the argument's derivative.
Check
Logarithmic differentiation. Finish the job.
Check your understanding
For y = x^x, what is y'?
Answer: A
Why: Taking logs gives ln y = x ln x; differentiating gives y'/y = ln x + 1; multiplying back gives the answer.
Real world
A radioactive sample decays according to N(t) = N-nought times e to the minus kt, where k is the decay constant. A physicist needs the activity, which is the rate at which nuclei are decaying.
Discussion prompt
Find the activity, show that it is proportional to the amount remaining, and explain why the half-life does not depend on how much you started with.
Hint: Differentiate, and compare the result with the original function.
Answer:
\[ N(t) = N_{0}e^{-kt} \;\Longrightarrow\; N'(t) = -kN_{0}e^{-kt} = -kN(t) \]
So the activity is proportional to the amount remaining, with constant of proportionality negative k. That is the self-derivative property of the exponential, and it is the physical statement that each nucleus decays independently with a fixed probability per unit time.
For the half-life, set the amount to half its initial value:
\[ \tfrac{1}{2}N_{0} = N_{0}e^{-kt} \;\Longrightarrow\; e^{-kt} = \tfrac{1}{2} \;\Longrightarrow\; t = \frac{\ln 2}{k} \]
The initial amount cancels completely, so the half-life depends only on k — a property of the isotope, not of the sample. A gram and a tonne of the same material halve in exactly the same time.
This cancellation is a direct consequence of the derivative being proportional to the function rather than constant. If decay happened at a fixed number of nuclei per second instead, a larger sample would take proportionally longer to halve, and there would be no such thing as a half-life. The exponential's defining property is what makes the concept well posed.
Commit first
Answer, then rate your confidence honestly.
Predict first
Which of these needs logarithmic differentiation?
Correct: x to the x — the variable is in both places.
\[ \ln y = x\ln x \;\Longrightarrow\; \frac{y'}{y} = \ln x + 1 \;\Longrightarrow\; y' = x^{x}(\ln x + 1) \]
Why: The power rule handles a variable base with a constant exponent, and the exponential rule a constant base with a variable exponent. When the variable occupies both positions neither applies, and taking logarithms is the only route: ln y equals x ln x, which differentiates by the product rule. The other three are all standard single-rule cases, and applying the wrong one to x to the x gives either x times x to the x minus 1 or x to the x times ln x, neither of which is correct.
Explain it
They ask why calculus insists on the number e when 10 would be so much friendlier.
Discussion prompt
In four sentences or fewer, give them the practical reason.
Hint: Have them differentiate a base-10 exponential a few times.
Answer:
Ask them to differentiate 10 to the x. They get 10 to the x times ln 10, about 2.303 times the original — and differentiating again multiplies by that factor a second time, and again a third.
Now try e to the x: the derivative is e to the x, unchanged, however many times you differentiate. The number e is precisely the base at which that stray constant equals 1, so it is the only base that leaves no residue — and since calculus differentiates constantly, that is worth more than the familiarity of counting in tens.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, ask whether x is the base or the exponent, and remember that pi and e are just numbers. For the stray factor, derive it from change of base rather than recalling where it goes. For logarithmic differentiation, remember the left side gives y prime over y, so two more steps remain. For logarithms, the pattern is always the inner derivative over the inner function. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw the natural exponential with tangents at 0 and 1, labelling both heights and both slopes, and write the one sentence that makes e special. Beside it draw base 2 and base 4 exponentials through the same point with a dashed line of slope 1, and write the slope of each at the origin. Below, derive the logarithm's derivative in full by implicit differentiation, marking where the original equation is substituted. Then write the four rules — natural and general, exponential and logarithmic — in a table, circling where the stray factor sits in each. In the lower half, run logarithmic differentiation completely on x squared times the fifth power of 3x plus 1, over the root of x minus 2, and beside it note which rules the direct route would have needed. Then do x to the x in full, marking the two final steps that are easy to omit, and find where its derivative vanishes. In a margin, write the three-stage history of the power rule with the section number that supplied each stage.
If your logarithmic differentiation ends at y prime over y, it is not finished — two steps remain, and the second of them, substituting the original expression, is the one most often left out.
Recap
Five things, and with them every family from Chapter 1 has a differentiation rule.
| If you see | Then |
|---|---|
| A constant base, variable exponent | The exponential rule, with ln of the base |
| A variable base, constant exponent | The power rule, whatever the exponent looks like |
| The variable in both places | Logarithmic differentiation is the only route |
| A logarithm of something | Inner derivative over inner function |
| A base other than e | Convert, or place ln b: up for exponentials, down for logs |
| A pile of products and powers | Take logs first |
| y prime over y | Two steps remain: multiply by y, then substitute |
That completes Chapter 3. Every standard function can now be differentiated, and Chapter 4 stops building rules and starts using them — to find rates in linked quantities, to approximate, to locate extremes, and to sketch what a function actually does.
OpenStax Calculus Volume 1, §3.9 Derivatives of Exponential and Logarithmic Functions §3.9, pp. 275-287 — everything on these slides traces back here
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