Method guideCalculus IIntermediate9 min read

When logarithmic differentiation is the clean move

Use logarithms to untangle variable exponents and products or quotients with many factors.

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

Method guide

How to recognize the method, run it, and know when it is the wrong choice.

Reviewed July 11, 2026

Use it whenStart here

Choose logarithmic differentiation for variable powers such as f(x)^g(x), or when expanding a product with ordinary rules would create avoidable clutter.

What logarithms simplify

Logarithms move exponents down as factors, turn products into sums, and turn quotients into differences. Those transformations align directly with familiar derivative rules.

The method does not change the function; it rewrites the relationship in a derivative-friendly form.

The hidden implicit step

After taking ln y, differentiation produces y′/y on the left. This is implicit differentiation, so forgetting the inner derivative y′ loses the quantity being solved for.

Multiply by y after differentiating, then replace y with the original expression.

Absolute values and domains

For products that may change sign, ln|y| gives the same derivative y′/y wherever y is nonzero. Domain conditions should be stated instead of silently discarded.

Worked exampleDifferentiate a variable power

Differentiate y = (x² + 1)^x.

\ln y=x\ln(x^2+1)

Bring the variable exponent down.

\frac{y'}y=\ln(x^2+1)+\frac{2x^2}{x^2+1}

Use the product and Chain Rules.

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

Multiply by y and substitute back.

Result

\boxed{(x^2+1)^x\left[\ln(x^2+1)+\dfrac{2x^2}{x^2+1}\right]}

Watch for

Common mistakes

Forgetting y′ when differentiating ln y.
Applying log properties to sums, such as ln(a+b) = ln a + ln b.
Failing to substitute the original y back into the result.

Keep

Three takeaways

Logs turn powers into products.
The left side differentiates to y′/y.
State domain restrictions when logarithms are introduced.