This section answers a fundamental question: how do we differentiate logarithms? Since $\ln x$ is the inverse of $e^x$, and we already know $\frac{d}{dx}(e^x) = e^x$, we can find the derivative of $\ln x$ using implicit differentiation. The result, $\frac{d}{dx}(\ln x) = \frac{1}{x}$, is one of the cleanest formulas in calculus.
This matters because logarithms appear everywhere: in exponential growth and decay problems, in integration (where $\int \frac{1}{x}\,dx = \ln\vert x\vert + C$ fills a gap in the power rule), and in the powerful technique of logarithmic differentiation that simplifies products, quotients, and variable exponents.
| Function | Derivative | Domain |
|---|---|---|
| $\ln x$ | $\displaystyle\frac{1}{x}$ | $x > 0$ |
| $\ln\|x\|$ | $\displaystyle\frac{1}{x}$ | $x \neq 0$ |
| $\ln(g(x))$ | $\displaystyle\frac{g'(x)}{g(x)}$ | $g(x) > 0$ |
| $\log_a x$ | $\displaystyle\frac{1}{x \ln a}$ | $x > 0$, $a > 0$, $a \neq 1$ |
Integration Counterpart: $$\int \frac{1}{x}\,dx = \ln\vert x\vert + C$$
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ID["Implicit<br/>Differentiation"]
LP["Logarithm<br/>Properties"]
DE["Derivative of e^x"]
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IYL["<a href='integral-yielding-ln.html'>Integrals<br/>Yielding ln</a>"]
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GROWTH["Exponential<br/>Growth Models"]
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| Skill | Description | Difficulty | Time |
|---|---|---|---|
| Derivative of Natural Log | The fundamental formula $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and its proof | Intermediate | 20 min |
| Skill | Description | Difficulty | Time |
|---|---|---|---|
| Derivative of General Logarithms | Differentiating $\log_a x$ for any base $a$ | Intermediate | 15 min |
| Integrals Yielding ln | Recognizing when $\int f(x)\,dx = \ln\vert \cdot\vert + C$ | Intermediate | 25 min |
| e as a Limit | Understanding $e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$ | Intermediate | 20 min |
| Skill | Description | Difficulty | Time |
|---|---|---|---|
| Logarithmic Differentiation | Using logs to differentiate products, quotients, and variable exponents | Advanced | 25 min |
If you're new to this material:
If you're reviewing:
| Mistake | Correction |
|---|---|
| Writing $\frac{d}{dx}(\ln x) = \frac{1}{\ln x}$ | The derivative is $\frac{1}{x}$, not $\frac{1}{\ln x}$ |
| Forgetting the chain rule: $\frac{d}{dx}[\ln(2x)] = \frac{1}{2x}$ | Must include derivative of inside: $\frac{2}{2x} = \frac{1}{x}$ |
| Writing $\int \frac{1}{x}\,dx = \ln x + C$ | Need absolute value: $\ln\|x\| + C$ (valid for $x \neq 0$) |
| Confusing $(\ln x)^2$ with $\ln(x^2)$ | $(\ln x)^2$: the log squared. $\ln(x^2) = 2\ln x$: log of a square |
By the end of Section 6.4, you should be able to:
| ← Section 6.3 | Skills Index | Section 6.5 → |