Why simplify a logarithm with the log laws before differentiating?

Rewriting a logarithm of a product, quotient or power as a sum or difference of simpler logarithms turns a hard derivative into an easy one. For example ln of ((x-1)/(x+1)) becomes ln(x-1) minus ln(x+1), which differentiates term by term to 1/(x-1) minus 1/(x+1), equal to 2/(x squared minus 1).

What is the most common mistake when differentiating ln(f(x))?

Forgetting the numerator f prime of x. The derivative of ln(f(x)) is f prime of x over f(x), not 1 over f(x). For ln(2x+3) the answer is 2/(2x+3); writing 1/(2x+3) drops the factor 2 from differentiating the inside.

Why simplify a logarithm with the log laws before differentiating?

Rewriting a logarithm of a product, quotient or power as a sum or difference of simpler logarithms turns a hard derivative into an easy one. For example ln of ((x-1)/(x+1)) becomes ln(x-1) minus ln(x+1), which differentiates term by term to 1/(x-1) minus 1/(x+1), equal to 2/(x squared minus 1).

What is the most common mistake when differentiating ln(f(x))?

Forgetting the numerator f prime of x. The derivative of ln(f(x)) is f prime of x over f(x), not 1 over f(x). For ln(2x+3) the answer is 2/(2x+3); writing 1/(2x+3) drops the factor 2 from differentiating the inside.

Year 12 Methods (Unit 3 & 4) Further differentiation and applications

Differentiation of the natural logarithm function

Q: What is the derivative of ln x?

The derivative of ln x is 1 divided by x. In symbols, d/dx of ln(x) equals 1/x, so the gradient of the curve y = ln x at any point is the reciprocal of its x-coordinate.

Q: How do you differentiate ln of a function?

Use the chain form: d/dx of ln(f(x)) equals f prime of x divided by f(x). Differentiate the inside function to get f prime of x, then divide by the inside function. For example d/dx of ln(2x+3) equals 2 divided by (2x+3).

Q: How do you differentiate x ln x?

Use the product rule with the two factors x and ln x. The derivative is 1 times ln x plus x times 1/x, which simplifies to ln x plus 1.

Q: What is the derivative of ln(5x)?

It is 1/x. Because ln(5x) equals ln 5 plus ln x and ln 5 is a constant, its derivative is 0, leaving the derivative of ln x, which is 1/x. The chain form gives the same result: 5 divided by 5x equals 1/x.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the natural logarithm is differentiated with two rules: \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\) and the chain form \(\dfrac{d}{dx}\ln\big(f(x)\big)=\dfrac{f'(x)}{f(x)}\). Where the logarithm is of a product, quotient or power, simplify with the log laws first, then combine with the product, quotient and chain rules.

The natural logarithm \(y=\ln x=\log_e x\) is the inverse of \(y=e^x\). Its derivative is remarkably simple: \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\). Geometrically, the gradient of the curve \(y=\ln x\) at any point is the reciprocal of that point's \(x\)-coordinate, so the curve rises steeply near \(x=0\) and flattens as \(x\) grows.

For a composite logarithm \(\ln\big(f(x)\big)\), the chain form gives \(\dfrac{d}{dx}\ln\big(f(x)\big)=\dfrac{f'(x)}{f(x)}\): differentiate the inside and divide by the inside. The factor \(f'(x)\) in the numerator is essential — omitting it is the classic error.

Before differentiating, it usually pays to simplify with the logarithm laws: \(\ln(ax)=\ln a+\ln x\), \(\ln(x^{n})=n\ln x\), \(\ln\dfrac{a}{b}=\ln a-\ln b\) and \(\ln\sqrt{u}=\tfrac12\ln u\). A logarithm of a messy product or quotient then breaks into simple terms, each easy to differentiate. Logarithms are also combined with the product, quotient and chain rules.

Key idea. \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\) and \(\dfrac{d}{dx}\ln\big(f(x)\big)=\dfrac{f'(x)}{f(x)}\). Simplify a logarithm with the log laws before you differentiate.

\(y=\ln x\): the gradient is \(\dfrac{1}{x}\), so it equals \(1\) at \(x=1\)

At \(x=2\) the gradient is \(\dfrac{1}{2}\) — the tangent is shallower

The basic rule for the natural logarithm:

\[\frac{d}{dx}\ln x=\frac{1}{x}\]

\frac{d}{d x} \ln x = \frac{1}{x}

The chain form for a composite logarithm (divide the derivative of the inside by the inside):

\[\frac{d}{dx}\ln\big(f(x)\big)=\frac{f'(x)}{f(x)}\]

\frac{d}{d x} \ln (f (x)) = \frac{f^{'} (x)}{f (x)}

The logarithm laws used to simplify before differentiating:

\[\ln(ax)=\ln a+\ln x \qquad \ln(x^{n})=n\ln x \qquad \ln\frac{a}{b}=\ln a-\ln b\]

Combine with the rules. Product: \(\dfrac{d}{dx}\,x\ln x=\ln x+1\). Quotient: \(\dfrac{d}{dx}\dfrac{\ln x}{x}=\dfrac{1-\ln x}{x^{2}}\). Chain: \(\dfrac{d}{dx}(\ln x)^{2}=\dfrac{2\ln x}{x}\).

How to differentiate a natural logarithm

Simplify with the log laws. Rewrite a product, quotient, power or root as a sum or difference of simple logarithms: \(\ln(ax)=\ln a+\ln x\), \(\ln(x^n)=n\ln x\), \(\ln\dfrac{a}{b}=\ln a-\ln b\), \(\ln\sqrt{u}=\tfrac12\ln u\).
Apply the rule. Use \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\); for a composite that cannot be simplified away, use the chain form \(\dfrac{f'(x)}{f(x)}\) — differentiate the inside, divide by the inside.
Use the product / quotient / chain rule where needed. For \(x\ln x\) use the product rule; for \(\dfrac{\ln x}{x}\) the quotient rule; for \((\ln x)^2\) or \(\sqrt{\ln x}\) the chain rule.
Simplify. Combine fractions over a common denominator and leave the derivative in simplest (often factorised) form.

Application. The derivative is a gradient function: substitute an \(x\)-value for the gradient of a tangent, set \(\dfrac{dy}{dx}=0\) for stationary points, and use the sign of \(\dfrac{d^{2}y}{dx^{2}}\) to classify them.

Example 1 — basic and chain form

Differentiate (a) \(y=\ln x\) and (b) \(y=\ln(2x+3)\).

Solution

(a) Apply the basic rule. (b) Use the chain form with \(f(x)=2x+3\), \(f'(x)=2\).

(a) \(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{x}\)
(b) \(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{2}{2x+3}\)

\frac{2}{2 x + 3}

Example 2 — simplify with log laws

Differentiate \(y=\ln\dfrac{x-1}{x+1}\).

Solution

Split with the quotient law, then differentiate each term.

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

\frac{2}{x^{2} - 1}

Example 3 — product rule

Differentiate \(y=x\ln x\).

Solution

Product rule on the factors \(x\) and \(\ln x\).

\(\dfrac{dy}{dx}\)	\(=\)	\((1)\ln x + x\cdot\dfrac{1}{x}\)
	\(=\)	\(\ln x + 1\)

\ln x + 1

Example 4 — gradient of a tangent

Find the gradient of the tangent to \(y=\ln x\) at the point where \(x=2\).

Solution

The gradient function is \(\dfrac{dy}{dx}=\dfrac{1}{x}\); substitute \(x=2\).

\(\dfrac{dy}{dx}\)	\(=\)	\(\dfrac{1}{x}\)
at \(x=2\)	\(=\)	\(\dfrac{1}{2}\)

\frac{1}{2}

Common pitfalls

Do not drop the \(f'(x)\). For a composite, \(\dfrac{d}{dx}\ln(2x+3)=\dfrac{2}{2x+3}\), not \(\dfrac{1}{2x+3}\). The numerator is the derivative of the inside.

Simplify with the log laws first. \(\dfrac{d}{dx}\ln\dfrac{x-1}{x+1}\) is quick once written as \(\ln(x-1)-\ln(x+1)\); differentiating the quotient directly is far messier.

A constant inside vanishes. \(\ln(5x)=\ln 5+\ln x\), and \(\ln 5\) is constant, so \(\dfrac{d}{dx}\ln(5x)=\dfrac{1}{x}\) — the same as \(\ln x\).

Frequently asked questions

What is the derivative of ln x?

\(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\). The gradient of \(y=\ln x\) at any point is the reciprocal of its \(x\)-coordinate.

How do you differentiate ln of a function?

Use the chain form \(\dfrac{d}{dx}\ln\big(f(x)\big)=\dfrac{f'(x)}{f(x)}\): differentiate the inside, divide by the inside. E.g. \(\dfrac{d}{dx}\ln(2x+3)=\dfrac{2}{2x+3}\).

Why simplify with the log laws before differentiating?

Splitting a product, quotient or power into simple logarithms turns a hard derivative into an easy one. \(\ln\dfrac{x-1}{x+1}=\ln(x-1)-\ln(x+1)\) differentiates to \(\dfrac{2}{x^{2}-1}\).

What is the most common mistake?

Forgetting the numerator \(f'(x)\). The derivative of \(\ln(f(x))\) is \(\dfrac{f'(x)}{f(x)}\), not \(\dfrac{1}{f(x)}\).

How do you differentiate x ln x?

Product rule: \(\dfrac{d}{dx}\,x\ln x=(1)\ln x+x\cdot\dfrac{1}{x}=\ln x+1\).

What is the derivative of ln(5x)?

\(\dfrac{1}{x}\). Since \(\ln(5x)=\ln 5+\ln x\) and \(\ln 5\) is constant, the derivative is \(\dfrac{1}{x}\); the chain form gives \(\dfrac{5}{5x}=\dfrac{1}{x}\) too.

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