AP Calculus AB · September 30, 2026

Logarithmic Differentiation

Use inverse functions to differentiate logarithms, carry domain restrictions carefully, and unlock functions whose variable appears in both the base and exponent.

52-minute edited lesson10 chaptersCaptions + corrected transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Where logarithmic derivatives fit 2:0602Start with the natural logarithm 7:1103Logarithms as inverse functions 10:3404Derive the derivative of ln x 14:3505Differentiate logarithms with any base 19:1806Apply the chain rule inside logarithms 25:1507Compare ln x and ln absolute x 35:3908Differentiate x to the 3x 43:5809A variable base and exponent 49:0710When logarithmic differentiation is optional

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02 · Understand

Learning targets

Use inverse-function reasoning to derive $\frac{d}{dx}\ln x=\frac1x\qquad(x>0)$ and $\frac{d}{dx}\log_b x=\frac{1}{x\ln b}\qquad(b>0,\ b\ne1)$ .
Apply the chain rule to natural and other-base logarithmic functions.
Distinguish the domains of $\ln x$ and $\ln|x|$ and carry those restrictions to their derivatives.
Use logarithmic differentiation when a variable appears in both the base and exponent.

03 · Connect

Core rules and domain

Each formula keeps its original function's domain.

01Natural logarithm: $\frac{d}{dx}\ln x=\frac1x\qquad(x>0)$ The natural logarithm is the inverse of the exponential function with base $e$ .
02Any logarithmic base: $\frac{d}{dx}\log_b x=\frac{1}{x\ln b}\qquad(b>0,\ b\ne1)$ The extra $\ln b$ factor comes from differentiating an exponential with base $b$ .
03Logarithmic chain rule: $\frac{d}{dx}\ln|u(x)|=\frac{u'(x)}{u(x)}\qquad(u(x)\ne0)$ Differentiate the inside and divide by that same inside expression.
04Absolute value: $\frac{d}{dx}\ln|x|=\frac1x\qquad(x\ne0)$ The algebraic derivative matches $1/x$ , but its domain is all nonzero real numbers.
05Logarithmic differentiation: $\ln y=3x\ln x$ Take logarithms, move the exponent in front, differentiate implicitly, solve for the derivative, and substitute the original function.

04 · Apply

Worked examples

Examples from the recorded lesson, rewritten with precise typeset mathematics.

Inverse functions

Derive the natural-log rule

y=\ln x\iff e^y=x

Rewrite the logarithmic equation in exponential form.

Differentiate implicitly, then use the inverse identity to replace the exponential expression with x.

\frac{d}{dx}\ln x=\frac1x\qquad(x>0)

Any base

Differentiate log base b

b^y\ln b\,\frac{dy}{dx}=1\quad\Longrightarrow\quad\frac{dy}{dx}=\frac{1}{x\ln b}

The derivative of b to the y includes the constant factor ln b.

Solve for dy/dx and replace b to the y with x.

\frac{d}{dx}\log_b x=\frac{1}{x\ln b}\qquad(b>0,\ b\ne1)

Chain rule

A logarithm containing a trigonometric-polynomial sum

f(x)=\log_4(\sin x+x^3)

Use one over the inside expression times ln 4.

Multiply by the derivative of the inside expression.

f'(x)=\frac{\cos x+3x^2}{(\sin x+x^3)\ln4}

Log property

A power inside a natural logarithm

f(x)=\ln(x^7)

Either apply the chain rule directly or rewrite ln(x to the seventh) as 7 ln x on its valid domain.

Both approaches simplify to the same derivative.

f'(x)=\frac7x

Domain

Natural log of an absolute value

\ln|x|=\begin{cases}\ln x,&x>0,\\\ln(-x),&x<0,\end{cases}

Treat the absolute-value function piecewise on the positive and negative intervals.

The derivative is the same algebraic expression on both intervals, but zero remains outside the domain.

\frac{d}{dx}\ln|x|=\frac1x\qquad(x\ne0)

Variable base and exponent

Differentiate x to the 3x

y=x^{3x}

Take the natural logarithm of both sides and move the exponent in front.

Differentiate implicitly with the product rule, solve for y prime, and replace y with the original function.

y'=(3\ln x+3)x^{3x}

Logarithmic differentiation

Differentiate sine x raised to cosine x

y=(\sin x)^{\cos x}

Take logarithms so the variable exponent becomes a coefficient.

Differentiate the resulting product implicitly, solve for y prime, and substitute the original function.

y'=(\sin x)^{\cos x}\left[-\sin x\ln(\sin x)+\frac{\cos^2x}{\sin x}\right]

05 · Avoid

Common mistakes

Forgetting the logarithm's inside derivative.

$\frac{d}{dx}\ln(5x+6)=\frac{5}{5x+6}$ includes the factor 5.

Dropping the base factor.

$\frac{d}{dx}\log_bx=\frac{1}{x\ln b}$ . Only the natural logarithm has $\ln e=1$ .

Ignoring the original domain.

The derivative formula may look valid at points where the original logarithmic function does not exist.

Using the power rule on a variable exponent.

$x^{3x}$ is not a power function because its exponent is not constant.

Using the exponential rule on a variable base.

$x^{3x}$ is not an ordinary exponential function because its base is not constant.

Forgetting to substitute for y.

After implicit differentiation, replace $y$ with the original function before finishing.

06 · Check yourself

Try it first

These are new practice problems, not repeats of the worked examples above. Reveal a solution only after trying each one.

1Differentiate

f(x)=\ln(2x^2+5)

$f'(x)=\frac{4x}{2x^2+5}$ Use $u'(x)/u(x)$ with $u(x)=2x^2+5$ .

2Differentiate

g(x)=\log_3(x^2+1)

$g'(x)=\frac{2x}{(x^2+1)\ln3}$ Differentiate the inside and keep the base factor $\ln3$ in the denominator.

3Differentiate

h(x)=\ln|x^2-9|

and state its domain.

$h'(x)=\frac{2x}{x^2-9},\qquad x\ne-3,3$ The absolute-value logarithm rule gives $u'/u$ wherever $u\ne0$ .

4Use logarithmic differentiation on

y=(x+1)^{2x}

$y'=(x+1)^{2x}\left[2\ln(x+1)+\frac{2x}{x+1}\right],\qquad x>-1$ From $\ln y=2x\ln(x+1)$ , use the product rule and solve for $y'$ .

5Differentiate

p(x)=\ln(4-x)

and keep the correct domain.

$p'(x)=-\frac{1}{4-x},\qquad x<4$ The inside derivative is negative one, and the derivative's domain cannot exceed the original logarithm's domain.

07 · Revisit

Transcript and practice

Search the corrected transcript for a rule, derivation, or worked example. Captions follow the edited video timeline.

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Return to Wednesday 9/30 in eKadence for Stewart Section 3.6 practice.