Try differentiating this function using the quotient and chain rules:
$$y = \frac{x^{2/3}(x^2 + 9)^{1/4}}{(5x - 1)^3}$$
You can do it, but it's a nightmare of nested applications. Now consider this:
$$y = x^x$$
Here, both the base AND the exponent are variable. The power rule doesn't apply (it needs a constant exponent). The exponential rule doesn't apply (it needs a constant base). What do we do?
Logarithmic differentiation handles both situations elegantly. By taking the natural log of both sides first, we transform:
This simplifies the differentiation dramatically.
Test: Expand $\ln\left(\frac{x^3 \sqrt{y}}{z^2}\right)$.
Answer: $3\ln x + \frac{1}{2}\ln y - 2\ln z$
Remember:
Test: If $\ln y = x^2$, find $\frac{dy}{dx}$.
Answer: Differentiating both sides: $\frac{1}{y}\frac{dy}{dx} = 2x$, so $\frac{dy}{dx} = 2xy$.
The Four Steps of Logarithmic Differentiation:
| Step | Action | Example |
|---|---|---|
| 1 | Take $\ln$ of both sides | $\ln y = \ln[x^x]$ |
| 2 | Expand using log rules | $\ln y = x \ln x$ |
| 3 | Differentiate implicitly | $\frac{1}{y}\frac{dy}{dx} = \ln x + 1$ |
| 4 | Solve for $\frac{dy}{dx}$ | $\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$ |
When to use it:
Given: $y = f(x)$ (some complicated function)
Step 1: Take natural log of both sides $$\ln y = \ln f(x)$$
(If $f(x)$ can be negative, use $\ln\vert y\vert = \ln\vert f(x)\vert $ instead.)
Step 2: Simplify using logarithm properties $$\ln(ab) = \ln a + \ln b$$ $$\ln(a/b) = \ln a - \ln b$$ $$\ln(a^n) = n \ln a$$
Step 3: Differentiate both sides implicitly
The left side always becomes $\frac{1}{y}\frac{dy}{dx}$.
Step 4: Solve for $\frac{dy}{dx}$ $$\frac{dy}{dx} = y \cdot (\text{what you got from differentiating})$$
Then substitute back $y = f(x)$.
| Case | Form | Rule | Example |
|---|---|---|---|
| 1 | $b^n$ (both constant) | Derivative is 0 | $\frac{d}{dx}(5^3) = 0$ |
| 2 | $[f(x)]^n$ (variable base, constant exponent) | Power rule | $\frac{d}{dx}[(\sin x)^3] = 3(\sin x)^2 \cos x$ |
| 3 | $b^{g(x)}$ (constant base, variable exponent) | Exponential rule | $\frac{d}{dx}(2^x) = 2^x \ln 2$ |
| 4 | $[f(x)]^{g(x)}$ (both variable) | Logarithmic differentiation | $\frac{d}{dx}(x^x) = x^x(\ln x + 1)$ |
Case 4 is the new territory. Neither the power rule nor the exponential rule applies when both base and exponent vary.
Logarithmic differentiation exploits the fact that $\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}$.
This means: the derivative of a function divided by the function itself equals the derivative of its logarithm.
For $y = f(x)$: $$\frac{y'}{y} = \frac{d}{dx}(\ln y)$$
Rearranging: $y' = y \cdot \frac{d}{dx}(\ln y)$
Find $\frac{dy}{dx}$ if $y = x^x$ for $x > 0$.
Find $\frac{dy}{dx}$ if $y = \frac{x^{2/3}(x^2 + 9)^{1/4}}{(5x - 1)^3}$.
Differentiate $y = (\cos x)^x$ for $x$ where $\cos x > 0$.
Find $\frac{dy}{dx}$ if $y = x^{1/x}$ for $x > 0$.
Differentiate $y = x^{\ln x}$ for $x > 0$ using BOTH: (a) Logarithmic differentiation (b) Rewriting as $e^{(\ln x)^2}$
Verify you get the same answer.
Use logarithmic differentiation to prove that $\frac{d}{dx}(x^n) = nx^{n-1}$ for ANY real number $n$ (not just integers).
Use it when:
Don't use it when:
| Mistake | Why It's Wrong | Correct Approach |
|---|---|---|
| Using on $y = x^2 + x$ | $\ln(x^2 + x) \neq \ln(x^2) + \ln(x)$ | Log rules only work for products/quotients |
| Forgetting to substitute back | Answer has $y$ in it | Replace $y$ with original function |
| Missing the $\frac{1}{y}$ on the left | Implicit differentiation of $\ln y$ gives $\frac{1}{y} \cdot \frac{dy}{dx}$ | Always include the $\frac{1}{y}$ factor |
| Applying to $f(x) < 0$ without care | $\ln$ of negative number undefined | Use $\ln\|y\| = \ln\|f(x)\|$ when needed |
"Log, Expand, Differentiate, Substitute"
The key insight: Logarithms convert multiplication to addition and powers to multiplication, both of which are easier to differentiate.
Looking back:
Looking ahead:
Real-world applications:
| Previous | Up | Next |
|---|---|---|
| e as a Limit | Section 6.4 | Section 6.5 |
Last updated: 2026-01-23