AP Calculus AB and BC

Derivative of ln x: Answer, Proof, and Mistakes

Q: Is the derivative of ln x always 1/x?

On the domain of \ln x, yes: for x > 0, \frac{d}{dx}[\ln x] = \frac{1}{x}. If you instead use \ln|x|, its derivative is \frac{1}{x} for every x \neq 0. The rule only changes when the base is not e or the inside is not just x.

Q: What is the derivative of ln(u)?

Use the chain rule: \frac{d}{dx}[\ln(u)] = \frac{u'}{u}, the derivative of the inside divided by the inside. For instance, \frac{d}{dx}[\ln(3x^2+1)] = \frac{6x}{3x^2+1}.

Q: Why is the derivative of ln x equal to 1/x?

Because \ln x is the inverse of e^x. Rewrite y = \ln x as e^y = x, differentiate both sides to get e^y \cdot \frac{dy}{dx} = 1, then solve: \frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}.

Q: What is the derivative of log base b of x?

\frac{d}{dx}[\log_b x] = \frac{1}{x \ln b}. For base 10 that is \frac{1}{x \ln 10}. Only the natural log, where b = e, gives the clean \frac{1}{x}, since \ln e = 1.

The derivative of ln x is 1/x, valid for x > 0. In prime notation, if f(x) = ln x then f'(x) = 1/x. For a composite ln(u), the chain rule gives u'/u, not 1/u. This follows because ln x is the inverse of e^x: writing e^y = x and differentiating gives dy/dx = 1/x.

\frac{d}{dx}\left[\ln x\right] = \frac{1}{x}

How to prove the derivative of ln x

Start from $y = \ln x$ . Because $\ln x$ is the inverse of the natural exponential, this is the same statement as $e^y = x$ , which already forces $x > 0$ (the exponential $e^y$ is always positive).

e^y = x

Differentiate both sides with respect to $x$ . The right side is just 1; the left side needs the chain rule, since $y$ depends on $x$ .

e^y \cdot \frac{dy}{dx} = 1

Solve for $\frac{dy}{dx}$ and substitute $e^y = x$ back in.

\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}

Watch the domain

$\ln x$ exists only for $x > 0$ , so the derivative $\frac{1}{x}$ is stated on that domain. The related function $\ln|x|$ has derivative $\frac{1}{x}$ for every $x \neq 0$ , and that is the version that appears when you antidifferentiate $\frac{1}{x}$ .

Where ln x's derivative shows up on the AP exam

The rule $\frac{d}{dx}[\ln x] = \frac{1}{x}$ is introduced in Unit 2, Topic 2.7 (Derivatives of $\cos x$ , $\sin x$ , $e^x$ , and $\ln x$ ). It is one of the basic derivatives the College Board expects you to recall without deriving, and Unit 2 carries a weighting of 10 to 15 percent on AB and 5 to 10 percent on BC.

It most often appears inside a composite $\ln(u)$ , where the chain rule (Unit 3, Topic 3.1) applies and you differentiate the inside and divide by it. It also turns up bare, or folded into a product or quotient such as $x \ln x$ .

\frac{d}{dx}\left[\ln(u)\right] = \frac{u'}{u}

The proof above is implicit differentiation (Topic 3.2): you write $e^y = x$ , differentiate both sides, and solve for $\frac{dy}{dx}$ . It works because $\ln x$ is the inverse of $e^x$ , the relationship that Topic 3.3 (Differentiating Inverse Functions) is built on. That topic's own rule, $g'(x) = \frac{1}{f'(g(x))}$ , reaches the same answer directly: with $f(x) = e^x$ and $g(x) = \ln x$ , $g'(x) = \frac{1}{e^{\ln x}} = \frac{1}{x}$ . So this one derivative ties four CED topics together.

CED topic	How the derivative of ln x appears
2.7 Derivatives of $\cos x$ , $\sin x$ , $e^x$ , and $\ln x$	The rule itself, memorized in Unit 2
3.1 The Chain Rule	Composites such as $\ln(x^2+1)$
3.2 Implicit Differentiation	Differentiating $e^y = x$ to prove the rule
3.3 Differentiating Inverse Functions	The inverse-function theorem gives the same $\frac{1}{x}$

Common mistakes with the derivative of ln x

Dropping the chain rule. $\frac{d}{dx}\left[\ln(x^2+1)\right] = \frac{2x}{x^2+1}$ , not $\frac{1}{x^2+1}$ . The inside's derivative belongs on top.
Using it where $\ln x$ is undefined. There is no value of $\ln x$ for $x \le 0$ , so " $\frac{1}{x}$ for all $x$ " is wrong; write $x > 0$ , or $x \neq 0$ if you mean $\ln|x|$ .
Confusing $\ln x$ with $e^x$ . The exponential keeps itself, $\frac{d}{dx}[e^x] = e^x$ , while $\frac{d}{dx}[\ln x] = \frac{1}{x}$ turns into a power. They are inverses, not the same rule.
Treating every logarithm like the natural log. For a general base, $\frac{d}{dx}\left[\log_b x\right] = \frac{1}{x \ln b}$ . Only base $e$ gives $\ln b = 1$ , which is why $\ln x$ produces the clean $\frac{1}{x}$ .
Half-applying the chain rule on $(\ln x)^2$ . The power rule and chain rule combine to $2\ln x \cdot \frac{1}{x} = \frac{2\ln x}{x}$ ; keep both the leftover $\ln x$ and the $\frac{1}{x}$ .

Worked chain-rule examples

Example 1. Differentiate $\ln(x^2+1)$ . Let $u = x^2+1$ , so $u' = 2x$ ; apply $\frac{u'}{u}$ .

\frac{d}{dx}\left[\ln(x^2+1)\right] = \frac{2x}{x^2+1}

Example 2. Differentiate $\ln(\sin x)$ . Now $u = \sin x$ and $u' = \cos x$ , so the quotient simplifies to $\cot x$ .

\frac{d}{dx}\left[\ln(\sin x)\right] = \frac{\cos x}{\sin x} = \cot x

Example 3. Differentiate $\ln(5x)$ . Here $u = 5x$ and $u' = 5$ , and the constant cancels.

\frac{d}{dx}\left[\ln(5x)\right] = \frac{5}{5x} = \frac{1}{x}

Why the constant vanished

$\ln(5x) = \ln 5 + \ln x$ , and $\ln 5$ is a constant, so $\ln(5x)$ and $\ln x$ share the same derivative $\frac{1}{x}$ . Any positive constant multiplier inside the log drops out.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

Is the derivative of ln x always 1/x?

On the domain of $\ln x$ , yes: for $x > 0$ , $\frac{d}{dx}[\ln x] = \frac{1}{x}$ . If you instead use $\ln|x|$ , its derivative is $\frac{1}{x}$ for every $x \neq 0$ . The rule only changes when the base is not $e$ or the inside is not just $x$ .

What is the derivative of ln(u)?

Use the chain rule: $\frac{d}{dx}[\ln(u)] = \frac{u'}{u}$ , the derivative of the inside divided by the inside. For instance, $\frac{d}{dx}[\ln(3x^2+1)] = \frac{6x}{3x^2+1}$ .

Why is the derivative of ln x equal to 1/x?

Because $\ln x$ is the inverse of $e^x$ . Rewrite $y = \ln x$ as $e^y = x$ , differentiate both sides to get $e^y \cdot \frac{dy}{dx} = 1$ , then solve: $\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$ .

What is the derivative of log base b of x?

$\frac{d}{dx}[\log_b x] = \frac{1}{x \ln b}$ . For base 10 that is $\frac{1}{x \ln 10}$ . Only the natural log, where $b = e$ , gives the clean $\frac{1}{x}$ , since $\ln e = 1$ .