Derivative of Logarithmic Functions

Logarithmic functions show up everywhere in calculus, from population growth models to sound intensity formulas, so knowing how to differentiate them is essential. The good news is that there are only a couple of core rules to learn, and once you combine them with the chain rule, you can handle almost any log expression you meet.

The most important rule to memorize is the derivative of the natural logarithm:

\( \dfrac{d}{dx}\left[\ln x\right] = \dfrac{1}{x}, \qquad x > 0 \)

This rule comes directly from the fact that \( \ln x \) is the inverse of \( e^x \). If you have already looked at the derivative of exponential functions, you can see the two are closely linked: differentiating an inverse function essentially "undoes" the exponential relationship, leaving the simple expression \( \frac{1}{x} \).

For a logarithm with a different base, use:

\( \dfrac{d}{dx}\left[\log_a x\right] = \dfrac{1}{x\ln a} \)

Notice that when \( a = e \), \( \ln a = \ln e = 1 \), and the formula reduces to the natural log rule above. This is why \( \ln x \) is the preferred base in calculus, all the other log derivatives are just a constant multiple of it.

Most real problems do not ask for the derivative of \( \ln x \) by itself, they ask for the derivative of \( \ln \) of some function \( u(x) \). The chain rule extends the basic rule to:

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

and similarly \( \dfrac{d}{dx}\left[\log_a u\right] = \dfrac{u'}{u \ln a} \). If you want a refresher on why this "outside times inside" pattern always works, see the chain rule lesson.

Example 1: Derivative of ln(2x)

Let \( u = 2x \), so \( u' = 2 \). Then:

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

Notice the constant 2 cancels out completely, this happens for \( \ln(kx) \) with any nonzero constant \( k \), because \( \ln(kx) = \ln k + \ln x \) and the derivative of the constant \( \ln k \) is zero.

Example 2: Derivative of ln(1/x)

Rewrite \( \ln\left(\frac{1}{x}\right) \) using log properties as \( -\ln x \). Differentiating gives:

\( \dfrac{d}{dx}\left[\ln\left(\dfrac{1}{x}\right)\right] = -\dfrac{1}{x} \)

You get the same answer using the chain rule directly with \( u = \frac{1}{x} \), but simplifying with log properties first is usually faster.

Example 3: Derivative of x ln x

This one needs the product rule as well, since it is \( x \) multiplied by \( \ln x \):

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

Example 4: Derivative of ln(x squared) versus (ln x) squared

These two expressions look similar but are not the same function, so their derivatives differ:

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

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

The first uses the chain rule on the inside function \( x^2 \), while the second treats \( \ln x \) itself as the inner function of a squaring operation.

Before differentiating a complicated logarithm, it often pays to expand it using log properties:

\( \ln(ab) = \ln a + \ln b, \qquad \ln\left(\dfrac{a}{b}\right) = \ln a - \ln b, \qquad \ln(a^n) = n\ln a \)

For example, an expression like \( \ln\left(\dfrac{1+x}{1-x}\right) \) becomes \( \ln(1+x) - \ln(1-x) \), which differentiates term by term to:

\( \dfrac{1}{1+x} - \dfrac{-1}{1-x} = \dfrac{1}{1+x} + \dfrac{1}{1-x} \)

This is much cleaner than trying to apply the chain rule to the quotient directly.

Some functions are awkward to differentiate directly, especially products or quotients of several factors, or a variable raised to a variable power. Logarithmic differentiation solves this by taking the natural log of both sides first, then differentiating implicitly. The steps are:

1. Take \( \ln \) of both sides of \( y = f(x) \).
2. Use log properties to expand the right side into sums and differences.
3. Differentiate both sides with respect to \( x \) (the left side gives \( \frac{y'}{y} \)).
4. Multiply both sides by \( y \) and substitute the original \( f(x) \) back in for \( y \).

Example: y = x to the power of x

This function cannot be handled by the power rule or the exponential rule alone, since both the base and the exponent contain \( x \).

\( y = x^x \)

\( \ln y = x \ln x \)

Differentiating both sides:

\( \dfrac{y'}{y} = \ln x + 1 \)

\( y' = y(\ln x + 1) = x^x(\ln x + 1) \)

This technique is also handy any time you need the slope of a log-based curve at a specific point, which connects directly to the idea of the slope and equation of a tangent line.

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