Logarithmic differentiation

When a function is a product of many factors, or has the form $u^v$ where both $u$ and $v$ depend on $x$, taking the natural logarithm before differentiating simplifies the calculation dramatically. The trick: $\log(uv) = \log u + \log v$ converts multiplication into addition, and $\log(u^v) = v \log u$ separates the exponent from the base.

The procedure

To differentiate $y = f(x)$ logarithmically:

1. Take $\log$ of both sides: $\log y = \log f(x)$.
2. Differentiate using implicit differentiation: $\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{d}{dx}[\log f(x)]$.
3. Solve: $\dfrac{dy}{dx} = y \cdot \dfrac{d}{dx}[\log f(x)] = f(x) \cdot \dfrac{d}{dx}[\log f(x)]$.

When to use it

The technique is essential when:

• The function is $f(x)^{g(x)}$ , neither pure power rule nor pure exponential applies.
• The function is a long product/quotient: $\log$ turns the product into a sum.

Worked example: $x^x$

Let $y = x^x$. Take $\log$: $\log y = x \log x$. Differentiate: $\dfrac{1}{y} \dfrac{dy}{dx} = \log x + 1$. So $\dfrac{dy}{dx} = x^x (\log x + 1)$.

$(\sin x)^{\cos x}$

$\log y = \cos x \cdot \log(\sin x)$. Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx} = -\sin x \cdot \log(\sin x) + \cos x \cdot \dfrac{\cos x}{\sin x} = -\sin x \log(\sin x) + \cot x \cos x$.

So $\dfrac{dy}{dx} = (\sin x)^{\cos x}\big[-\sin x \log(\sin x) + \cot x \cos x\big]$.

A product example

Differentiate $y = \dfrac{(x + 1)^2 (2x - 1)^3}{(x^2 + 1)^4}$.

$\log y = 2 \log(x + 1) + 3 \log(2x - 1) - 4 \log(x^2 + 1)$.

Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{2}{x + 1} + \dfrac{6}{2x - 1} - \dfrac{8x}{x^2 + 1}$.

So $\dfrac{dy}{dx} = y\left[\dfrac{2}{x + 1} + \dfrac{6}{2x - 1} - \dfrac{8x}{x^2 + 1}\right]$, with $y$ as given.

Worked examples

Example 1. Differentiate $y = x^{\sin x}$.

$\log y = \sin x \log x$. $\dfrac{1}{y}\dfrac{dy}{dx} = \cos x \log x + \dfrac{\sin x}{x}$. So $\dfrac{dy}{dx} = x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right)$.

Example 2. Differentiate $y = (\log x)^x$.

$\log y = x \log(\log x)$. $\dfrac{1}{y}\dfrac{dy}{dx} = \log(\log x) + x \cdot \dfrac{1/x}{\log x} = \log(\log x) + \dfrac{1}{\log x}$. So $\dfrac{dy}{dx} = (\log x)^x\left[\log(\log x) + \dfrac{1}{\log x}\right]$.

Example 3. If $y^x = x^y$, find $\dfrac{dy}{dx}$.

Take $\log$: $x \log y = y \log x$. Differentiate: $\log y + \dfrac{x}{y}\dfrac{dy}{dx} = \dfrac{dy}{dx}\log x + \dfrac{y}{x}$. Solve: $\dfrac{dy}{dx}\left(\dfrac{x}{y} - \log x\right) = \dfrac{y}{x} - \log y$. So $\dfrac{dy}{dx} = \dfrac{y(y - x \log y)}{x(x - y \log x)}$ after multiplying through.

Example 4. Differentiate $y = x^x + x^{1/x}$.

Each term needs logarithmic differentiation.

$u = x^x$: $\dfrac{du}{dx} = x^x(\log x + 1)$.

$v = x^{1/x}$: $\log v = \dfrac{\log x}{x}$. $\dfrac{1}{v}\dfrac{dv}{dx} = \dfrac{1 - \log x}{x^2}$. So $\dfrac{dv}{dx} = x^{1/x} \cdot \dfrac{1 - \log x}{x^2}$.

$\dfrac{dy}{dx} = x^x(\log x + 1) + x^{1/x} \cdot \dfrac{1 - \log x}{x^2}$.

Example 5. Differentiate $y = \sqrt{\dfrac{(x - 1)(x - 2)}{(x - 3)(x - 4)}}$.

$\log y = \tfrac{1}{2}[\log(x - 1) + \log(x - 2) - \log(x - 3) - \log(x - 4)]$.

$\dfrac{1}{y}\dfrac{dy}{dx} = \tfrac{1}{2}\left[\dfrac{1}{x - 1} + \dfrac{1}{x - 2} - \dfrac{1}{x - 3} - \dfrac{1}{x - 4}\right]$.

So $\dfrac{dy}{dx} = \dfrac{y}{2}\left[\dfrac{1}{x - 1} + \dfrac{1}{x - 2} - \dfrac{1}{x - 3} - \dfrac{1}{x - 4}\right]$.

Example 6. Differentiate $y = x^{x^x}$.

$y = x^{u}$ where $u = x^x$. $\log y = u \log x = x^x \log x$.

$\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{d}{dx}(x^x \log x) = (x^x(\log x + 1)) \log x + x^x \cdot \dfrac{1}{x} = x^x[(\log x)^2 + \log x + 1/x]$.

So $\dfrac{dy}{dx} = x^{x^x} \cdot x^x \left[(\log x)^2 + \log x + \dfrac{1}{x}\right]$.

Try it yourself

1. Differentiate $y = x^x$.
2. Differentiate $y = (x^2 + 1)^x$.
3. Differentiate $y = x^{\sin x}$.
4. Differentiate $y = \dfrac{(2x + 1)^3}{(3x - 1)^2 (x + 2)}$ using log differentiation.
5. If $y = (1 + x)(1 + x^2)(1 + x^4)$, differentiate using logs.
6. Differentiate $y = \sqrt{x \sqrt{x \sqrt{x}}}$.
7. Differentiate $y = x^{x^x}$.
8. Differentiate $y = (\sin x)^x$.
9. Differentiate $y = (\sin x)^{\sin x}$.
10. Differentiate $y = x^{\log x}$.
11. Differentiate $y = (\log x)^{\sin x}$.
12. If $x^y \cdot y^x = (x + y)^{x + y}$, set up the log derivative (do not solve).
13. Differentiate $y = (1 + \tan x)^x$.
14. Differentiate $y = \dfrac{e^{2x}(x - 1)^2}{x \sqrt{x + 1}}$ using logs.

Pitfalls / Tricks

• $\log$ of a product becomes a sum , that is the whole point of log differentiation.
• Always restate the original $y$ when writing the final $dy/dx$.
• For functions like $x^x$ where both base and exponent depend on $x$, neither plain power rule nor plain exponential rule works. Log first.
• For products of more than three factors, log differentiation saves enormous time.
• Watch for domain issues: $\log f(x)$ requires $f(x) > 0$.

Next, parametric and higher-order derivatives.