Derivative of Logarithmic Functions: A Hard Task Made Easy

The formulas to find the derivative of logarithmic functions:

Logarithm of \( x \)  to the base of \( a \) :

The derivative of \( \log_a x \) : \( \left(\log_a x\right)’ = \frac{1}{x \ln a} \)

Example:

\( \text{Given function: } \log_2 x \)

\( \text{Derivative: } \left(\log_2 x\right)’ = \frac{1}{x \ln 2} \)

Logarithm of \( f(x) \) to the base of \( a \):

The derivative of \( \log_a f(x) \) : \( \left(\log_a f(x)\right)’ = \frac{f'(x)}{f(x) \ln a} \)

Example:

\( \text{Given function: } \log_3 (x^2 + 1) \)

\( \text{Derivative: } \left(\log_3 (x^2 + 1)\right)’ = \frac{2x}{(x^2 + 1) \ln 3} \)

Natural logarithm of \( x \):

The derivative of \( \ln x \) : \( \left(\ln x\right)’ = \frac{1}{x} \)

Example:

\( \text{Given function: } \ln (3x) \)

\( \text{Derivative: } \left(\ln (3x)\right)’ = \frac{1}{3x} \cdot 3 = \frac{1}{x} \)

Natural logarithm of \( f(X) \):

The derivative of \( \ln f(x) \) : \( \left(\ln f(x)\right)’ = \frac{f'(x)}{f(x)} \)

Example:

\( \text{Given function: } \ln (x^3 + 2x) \)

\( \text{Derivative: } \left(\ln (x^3 + 2x)\right)’ = \frac{3x^2 + 2}{x^3 + 2x} \)

Examples:

Let’s consider a complex example:

\( \text{Find the derivative of } h(x) = \ln(x) \cdot \log_2(x^2 + 1) \)

1. Apply the product rule

\( h'(x) = f'(x)g(x) + f(x)g'(x) \)

2. Define \( f'(x) \) and \( g'(x) \)

\( f(x) = \ln(x) \rightarrow f'(x) = \frac{1}{x} \)

\( g(x) = \log_2(x^2 + 1) \rightarrow g'(x) = \frac{2x}{(x^2 + 1) \ln 2} \)

3. Combine using the product rule

\( h'(x) = \frac{1}{x} \cdot \log_2(x^2 + 1) + \ln(x) \cdot \frac{2x}{(x^2 + 1) \ln 2} \)

Here is another example involving radicals:

\( \text{Find the derivative of } h(x) = \sqrt{\ln x} \cdot \log_2(x^3 + 1) \)

1. Apply the product rule

\( h'(x) = f'(x)g(x) + f(x)g'(x) \)

2. Define \( f'(x) \) and \( g'(x) \)

\( f(x) = \sqrt{\ln x} \rightarrow f'(x) = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} \)

\( g(x) = \log_2(x^3 + 1) \rightarrow g'(x) = \frac{3x^2}{(x^3 + 1) \ln 2} \)

3. Combine using the product rule

\( h'(x) = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} \cdot \log_2(x^3 + 1) + \sqrt{\ln x} \cdot \frac{3x^2}{(x^3 + 1) \ln 2} \)

Derivative of exponential functions

Here are the formula for finding the derivative of exponential functions.

Real number \( a \) to the power of \( x \): \( a^x \)

\( \left(a^x\right)’ = a^x \ln a \)

Example:

\( \text{Given function: } 5^{2x + 3} \)

\( \text{Derivative: } \left(5^{2x + 3}\right)’ = 5^{2x + 3} \ln 5 \cdot 2 \)

Real number \( a \) to the power of \( f(x) \):  \( a^x \)

\( \left(a^{f(x)}\right)’ = a^{f(x)} \ln a \cdot f'(x) \)

Example:

\( \text{Given function: } 4^{\sin x} \)

\( \text{Derivative: } \left(4^{\sin x}\right)’ = 4^{\sin x} \ln 4 \cdot \cos x \)

\( e \) to the power of \( x \): \( e^x \)

\( \left(e^x\right)’ = e^x \)

Example:

\( \text{Given function: } e^{3x – 2} \)

\( \text{Derivative: } \left(e^{3x – 2}\right)’ = e^{3x – 2} \cdot 3 \)

\( e \) to the power of \( f(x) \): \( e^{f(x)} \)

\( \left(e^{f(x)}\right)’ = e^{f(x)} \cdot f'(x) \)

Example:

\( \text{Given function: } e^{\sqrt{x}} \)

\( \text{Derivative: } \left(e^{\sqrt{x}}\right)’ = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \)

Here is one more example:

\( \text{Find the derivative of } h(x) = e^{2x} \cdot \ln(x^2) \)

1. Apply the product rule

\( h'(x) = e^{2x} \cdot (\ln(x^2))’ + (e^{2x})’ \cdot \ln(x^2) \)

2. Define the derivatives

\( (\ln(x^2))’ = \frac{2}{x} \)

\( (e^{2x})’ = e^{2x} \cdot 2 \)

3. Combine using the product rule

\( h'(x) = e^{2x} \cdot \frac{2}{x} + e^{2x} \cdot 2 \cdot \ln(x^2) \)

Recommended EffortlessMath Books

For a workbook that nails down log and exponential rules before calculus, the Algebra II for Beginners covers logarithms with worked examples. For full precalc-into-calc prep, the Pre-Calculus for Beginners builds the analytic foundation calculus depends on.

Frequently Asked Questions

What’s the derivative of ln x?

\(\frac{d}{dx}[\ln x] = \frac{1}{x}\) for \(x > 0\). This is the foundational log derivative — every other log derivative builds on it. Example: at \(x = 2\), the slope of \(\ln x\) is \(1/2\). At \(x = 10\), the slope is \(1/10\) — flatter, since log grows slowly.

What’s the derivative of log base a of x?

\(\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}\). The \(\ln a\) comes from the change-of-base formula \(\log_a x = \ln x / \ln a\). Differentiating both sides gives the formula. For \(\log_{10}\), divide by \(\ln 10 \approx 2.303\); for \(\log_2\), divide by \(\ln 2 \approx 0.693\).

How do I use the chain rule with log?

\((\ln u)’ = u’/u\). The derivative of \(\ln\) of something is the derivative of that something divided by the something itself. Example: \(d/dx \ln(\sin x) = \cos x / \sin x = \cot x\). Always write \(u\) and \(u’\) clearly before plugging in.

What’s logarithmic differentiation?

A technique for differentiating complicated products, quotients, and functions with variables in both the base and exponent. Take \(\ln\) of both sides, use log properties to simplify, differentiate implicitly, then solve for \(y’\). Example: \(y = x^x\) becomes \(\ln y = x\ln x\), giving \(y’ = x^x(\ln x + 1)\).

When should I use logarithmic differentiation?

When the function is a product or quotient of many factors (taking the log turns it into a sum, much easier), or when the function has the form \(f(x)^{g(x)}\) (variable base and variable exponent). For straightforward functions, plain differentiation rules are faster.

What’s the derivative of ln|x|?

\(\frac{d}{dx}[\ln|x|] = \frac{1}{x}\) for \(x \neq 0\). The absolute value lets the derivative work on both positive and negative \(x\). For negative \(x\), \(\ln|x| = \ln(-x)\), and chain rule gives derivative \((-1)/(-x) = 1/x\) — same answer either way.

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

You can derive it from \((e^x)’ = e^x\) and the fact that \(\ln\) and \(e^x\) are inverses. Let \(y = \ln x\). Then \(e^y = x\). Differentiate implicitly: \(e^y \cdot y’ = 1\), so \(y’ = 1/e^y = 1/x\). The relationship between log and exponential is the source.

Can I differentiate a log inside a log?

Yes — chain rule twice. Example: \(d/dx \ln(\ln x) = (1/\ln x) \cdot (1/x) = 1/(x \ln x)\). The outer derivative gives \(1/\ln x\); chain rule then multiplies by the inner derivative \(1/x\).

What’s the domain issue with log derivatives?

\(\ln x\) is only defined for \(x > 0\), so its derivative is also only valid for \(x > 0\). If you write \((\ln x)’ = 1/x\), it’s understood that the domain is \(x > 0\). For negative \(x\), use \(\ln|x|\) instead, whose derivative \(1/x\) works on the whole nonzero real line.

Where does the derivative of log show up on tests?

Every calc I exam, AP Calc AB and BC, college calculus finals, engineering placement exams, and any standardized test that includes calculus. Expect logarithmic differentiation on harder problems involving exponential growth, optimization, and related rates.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• Product rule
• Exponential and log integrals
• Second derivatives
• Sketching curves with derivatives
• Rules of integration