6. Logarithmic Differentiation

Video coming soon4 worked examples

Theory

Logarithmic differentiation takes the natural log of both sides first, using log laws to turn products, quotients and powers into sums and multiples, and then differentiates implicitly. It is essential when the variable appears in the index, such as $y = x^x$ .

The Golden Rule: take $\ln$ of both sides, simplify with log laws, differentiate implicitly (the left side becomes $\frac{1}{y}\frac{dy}{dx}$ ), then multiply through by $y$ and substitute it back.

⚠️ Common Examiner Traps

The left side: differentiating $\ln y$ gives $\frac{1}{y}\frac{dy}{dx}$ — don't forget the $\frac{dy}{dx}$ .
Not substituting $y$ back: the final answer should be in terms of $x$ .
Log laws: $\ln(ab) = \ln a + \ln b$ and $\ln(a^n) = n\ln a$ — apply them before differentiating.

Worked examples

Example 1

Differentiate $y = x^x$ for $x > 0$ .

Step 1: Take natural logs and simplify:

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

Step 2: Differentiate implicitly (product rule on the right):

\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1

Step 3: Multiply by $y = x^x$ :

\frac{dy}{dx} = x^x(\ln x + 1)

Example 2

Use logarithmic differentiation to find $\dfrac{dy}{dx}$ for $y = \dfrac{(2x+1)^3}{(x-1)^2}$ .

Step 1: Take logs and use the log laws:

\ln y = 3\ln(2x+1) - 2\ln(x-1)

Step 2: Differentiate implicitly:

\frac{1}{y}\frac{dy}{dx} = \frac{6}{2x+1} - \frac{2}{x-1}

Step 3: Multiply by $y$ :

\frac{dy}{dx} = \frac{(2x+1)^3}{(x-1)^2}\left(\frac{6}{2x+1} - \frac{2}{x-1}\right)

Example 3

Show that if $y = a^x$ , where $a$ is a positive constant, then $\dfrac{dy}{dx} = a^x \ln a$ .

Step 1: The variable is in the index, so take natural logs of both sides and use $\ln(a^x) = x\ln a$ :

\ln y = x \ln a

Step 2: Differentiate implicitly. Note that $\ln a$ is a constant, so the right-hand side differentiates to just $\ln a$ :

\frac{1}{y}\frac{dy}{dx} = \ln a

Step 3: Multiply through by $y$ and substitute $y = a^x$ :

\frac{dy}{dx} = y\ln a = a^x \ln a

This is a standard result worth remembering — and it confirms why $e^x$ is special: when $a = e$ , $\ln e = 1$ and the derivative is $e^x$ itself.

Example 4

Differentiate $y = (x^2+1)^x$ .

Step 1: Both the base and the index involve $x$ , so neither the power rule nor the result above applies. Take logs:

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

Step 2: Differentiate implicitly. The right-hand side is a product, and $\ln(x^2+1)$ needs the chain rule:

\frac{1}{y}\frac{dy}{dx} = \ln(x^2+1) + x \times \frac{2x}{x^2+1}

Step 3: Tidy the second term:

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

Step 4: Multiply by $y$ and substitute it back:

\frac{dy}{dx} = (x^2+1)^x\left(\ln(x^2+1) + \frac{2x^2}{x^2+1}\right)