Why the derivative of ln x is 1/x, and why ln|x| has the same one

Start from what ln x means

$\ln x$ is the logarithm in base $e$, and its whole definition is that it undoes $e^x$:

Two facts come with that sentence, and both are used below.

• $e^y$ is positive for every $y$, so $\ln x$ exists only for $x>0$. Its domain is $(0,\infty )$.
• The graph of $y=\ln x$ is the graph of $y=e^x$ reflected in the line $y=x$: every point $(a,b)$ on one becomes the point $(b,a)$ on the other.

Getting 1/x in three lines

Rewrite $y=\ln x$ in exponential form and differentiate both sides with respect to $x$. This is implicit differentiation: $y$ depends on $x$, so the left side needs the chain rule.

1. Rewrite. $e^y=x$.
2. Differentiate both sides. $e^y{\displaystyle \frac{dy}{dx}}=1$.
3. Solve, then use the definition once more. $e^y$ is $x$, so $$\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}.$$

So ${\displaystyle \frac{d}{dx}}\ln x={\displaystyle \frac{1}{x}}$ for every $x>0$. Step 3 is the one to remember: the answer first comes out in terms of $y$, and the definition $e^y=x$ is what turns it back into $x$. The same move gives the derivatives of the inverse trigonometric functions.

The same fact, on the graph

Reflecting in $y=x$ swaps rise and run, so a tangent of slope $m$ becomes a tangent of slope $\frac{1}{m}$. On $y=e^x$ the point $(2,e^2)$ has slope $e^2$, because $e^x$ is its own derivative. Its mirror image on $y=\ln x$ is the point $(e^2,2)$, where the slope is therefore $\frac{1}{e^2}$: exactly $\frac{1}{x}$ at $x=e^2$. Every point pairs up the same way.

Why ln|x| has the same derivative

$\ln \left|x\right|$ is defined for every $x\ne 0$, because $\left|x\right|$ is positive on both sides of zero. Split it at zero and differentiate each piece.

• For $x>0$: $\left|x\right|=x$, so $\ln \left|x\right|=\ln x$, whose derivative is $\frac{1}{x}$.
• For $x<0$: $\left|x\right|=-x$, so $\ln \left|x\right|=\ln (-x)$, and the chain rule gives $$\frac{d}{dx}\ln (-x)=\frac{1}{-x}·(-1)=\frac{1}{x}.$$

The two minus signs cancel, so both pieces give the same formula:

Notice that $\frac{1}{x}$ is negative when $x<0$. The graph says the same thing: $y=\ln \left|x\right|$ is the graph of $\ln x$ together with its mirror image in the $y$-axis, and a mirror in the $y$-axis reverses the sign of every slope. The right branch rises, and the left branch falls as $x$ moves towards zero.

Why the integral of 1/x is ln|x| + C

An antiderivative of $\frac{1}{x}$ has to work wherever you need it, and $\frac{1}{x}$ is perfectly well defined for negative $x$.

• $\ln x+C$ has derivative $\frac{1}{x}$ on $(0,\infty )$ only. For negative $x$ it does not exist at all.
• $\ln \left|x\right|+C$ has derivative $\frac{1}{x}$ on $(0,\infty )$ and on $(-\infty ,0)$.

That is the whole reason for the absolute value:

It is still used on one interval at a time. Zero is not in the domain of $\frac{1}{x}$, so the formula applies on an interval of positive numbers or on an interval of negative numbers, never on one that contains $0$.

Two worked examples

1. Differentiate $f(x)=\ln |x^2-9|$

Combine the chain rule with $\frac{d}{du}\ln \left|u\right|=\frac{1}{u}$, taking $u=x^2-9$:

The absolute value is doing real work here. On $(-3,3)$ the inside is negative, so on that interval $f(x)=\ln (9-x^2)$. Differentiating that piece directly:

The same formula. $\frac{d}{dx}\ln \left|u\right|=\frac{u^{\prime }}{u}$ holds whatever the sign of $u$, so you never need to split into cases to differentiate. You only need the cases to know where the function exists.

2. Integrate $\frac{1}{x}$ from $-6$ to $-2$

The interval $[-6,-2]$ does not contain $0$, so $\ln \left|x\right|$ is an antiderivative on all of it:

and by the quotient law that is

Check the sign: $\frac{1}{x}$ is negative on the whole interval, so the integral has to be negative, and $-\ln 3\approx -1.10$ is. Try the same computation with $\ln x$ in place of $\ln \left|x\right|$ and the very first step asks for $\ln (-2)$, which does not exist.

Check your understanding

No calculator. The first three are quick; the rest ask for a reason as well as an answer.

1. The tangent to $y=e^x$ at $x=3$ has slope $e^3$. Without differentiating $\ln x$, give the slope of the tangent to $y=\ln x$ at the point $(e^3,3)$.
2. Find the slope of $y=\ln \left|x\right|$ at $x=-5$, and say whether the graph is rising or falling there.
3. Differentiate $g(x)=\ln |4-x|$, and state where your derivative is valid.
4. True or false, with a reason: on the interval $(2,9)$, $\ln x+C$ is just as good an antiderivative of $\frac{1}{x}$ as $\ln \left|x\right|+C$.
5. Differentiate $\ln \left|\cos x\right|$. What antiderivative of $\tan x$ does your answer give you, and why does it need the absolute value?
6. Use the method of this post, writing $y={\log }_{10}x$ as ${10}^y=x$, to find the derivative of ${\log }_{10}x$.
7. Calculus II: evaluate ${\displaystyle {\int }_{-e^2}^{-1}\frac{1}{x}dx}$.
8. Calculus II: find the error. A student writes ${\displaystyle {\int }_{-3}^2\frac{1}{x}dx=\ln \left|2\right|-\ln \left|-3\right|=\ln \frac{2}{3}}$.

Practise it

The free CEGEP Calculus I Derivatives of Transcendental Functions worksheet has logarithmic derivatives with their domains to state, logarithmic differentiation, and the rest of the family: exponentials, trigonometric and inverse trigonometric functions. The questions are on the page, with a printable PDF.

Free worksheets on this topic

• CEGEP Calculus I: Derivatives of Transcendental Functions worksheet
• CEGEP Calculus I: Differentiation Rules worksheet (implicit differentiation)
• CEGEP Calculus II: Antiderivatives and the Indefinite Integral worksheet