Logarithmic Differentiation

This lesson introduces logarithmic differentiation: the method you use when a function is a power of $x$ or $y$, a complicated product, or a quotient that would be painful to differentiate directly. The idea is to take logarithms of both sides first, simplify with the log laws, and then differentiate.

When to use it

Use logarithmic differentiation when the function involves:

• a power whose base or exponent contains $x$ or $y$,
• a product of more than two functions,
• a complicated quotient.

Laws of logarithms

These are the rules we lean on to simplify before differentiating.

$\log(ab) = \log a + \log b$

$\log\!\left(\tfrac{a}{b}\right) = \log a - \log b$

$\log x^{m} = m\,\log x$

Special cases follow from the power rule:

$\log \sqrt{x} = \log x^{1/2} = \tfrac{1}{2}\log x, \qquad \log \sqrt[n]{a} = \tfrac{1}{n}\log a$

and for products of three factors,

$\log(abc) = \log a + \log b + \log c.$

Also $\log e = 1$, $\log 1 = 0$, and $\log e^{x} = x\log e = x$.

Example 1: a square-root quotient

Differentiate

$y = \sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+3}}.$

Set up. Write the square root as a power $\tfrac{1}{2}$:

$y = \left(\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+3}\right)^{1/2}.$

Take logs on both sides and apply the laws:

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

Differentiate with respect to $x$. The left side gives $\tfrac{1}{y}\tfrac{dy}{dx}$ (implicit differentiation):

$\frac{1}{y}\frac{dy}{dx} = \tfrac{1}{2}\left[\frac{1}{x-3} + \frac{2x}{x^{2}+4} - \frac{6x+4}{3x^{2}+4x+3}\right].$

Solve for the derivative by multiplying through by $y$ and substituting $y$ back:

$\frac{dy}{dx} = \tfrac{1}{2}\sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+3}}\left[\frac{1}{x-3} + \frac{2x}{x^{2}+4} - \frac{6x+4}{3x^{2}+4x+3}\right].$

Example 2: a constant raised to x

Find $\dfrac{d}{dx}\,a^{x}$. Because the exponent contains $x$, take logs.

$y = a^{x} \;\Rightarrow\; \log y = x\log a.$

Differentiating (with $\log a$ a constant):

$\frac{1}{y}\frac{dy}{dx} = \log a \;\Rightarrow\; \frac{dy}{dx} = y\log a = a^{x}\log a.$

So $\dfrac{d}{dx}\,a^{x} = a^{x}\log a$, the standard result.

Example 3: a product of three factors

Differentiate

$y = (x+3)^{2}(x+4)^{3}(2x+5)^{4}.$

Take logs and split the product into a sum, then bring down each power:

$\log y = 2\log(x+3) + 3\log(x+4) + 4\log(2x+5).$

Differentiate term by term, remembering the chain rule on $2x+5$ (its derivative is $2$):

$\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4\cdot 2}{2x+5} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{8}{2x+5}.$

Substitute $y$ back:

$\frac{dy}{dx} = (x+3)^{2}(x+4)^{3}(2x+5)^{4}\left[\frac{2}{x+3} + \frac{3}{x+4} + \frac{8}{2x+5}\right].$

Key takeaways

• Take logarithms of both sides when you have powers, long products, or quotients, then use the log laws to turn them into sums and multiples.
• Differentiating $\log y$ gives $\tfrac{1}{y}\tfrac{dy}{dx}$, so always multiply back by $y$ and substitute the original function for the final answer.
• The standard result $\dfrac{d}{dx}\,a^{x} = a^{x}\log a$ comes straight from this method.