3.4 Logarithmic differentiation

Achievable AP Calculus AB

3. Advanced differentiation

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Logarithmic differentiation

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What you’ll learn

Logarithmic differentiation: Apply logarithmic properties to simplify complex products and quotients before differentiating.

Consider the function

$y = \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

Differentiating this directly can get tedious because you’d need to use several derivative rules (power, product, quotient, and chain), and some of them more than once.

Logarithmic differentiation is a technique that streamlines the work. by taking a logarithm first, then using log properties to rewrite a complicated expression as a sum and difference of simpler terms. After that, differentiation is much more straightforward.

Step-by-step process

Take the natural logarithm of both sides (skip if one side is already a single logarithm with a complicated argument)
Expand using log properties.
Differentiate implicitly
Isolate $y^{'}$ to find the derivative

Use the following logarithmic properties:

Name	Formula
Product	$ln (a \cdot b) = ln (a) + ln (b)$
Quotient	$ln (\frac{a}{b}) = ln (a) - ln (b)$
Power	$ln (a^{b}) = b ln (a)$

Note: The power property doesn’t apply when the exponent is entirely on the outside:

$(ln a)^{b} \neq = b ln (a)$

Examples

Problem 1. Differentiate

$y = \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

Solution

(spoiler)

1. Take the natural log of both sides

$ln (y) = ln (\frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x})$

2. Expand using log properties

First use the quotient property to turn the fraction into a difference:

$ln (y) = ln ((2 x + 1)^{2}) - ln (x^{3} x^{2} + 2 x)$

Apply the product property on the 2nd logarithm and distribute the negative sign:

$ln (y) = ln ((2 x + 1)^{2}) - (ln (x^{3}) + ln (x^{2} + 2 x)) = ln ((2 x + 1)^{2}) - ln (x^{3}) - ln (x^{2} + 2 x)$

Now use the power property to bring exponents to the front:

$ln (y) = 2 ln (2 x + 1) - 3 ln (x) - \frac{1}{2} ln (x^{2} + 2 x)$

At this point, each term is a constant multiple of a simple logarithm, which is much easier to differentiate.

3. Differentiate implicitly

Differentiate both sides with respect to $x$ . On the left, use the chain rule: $\frac{d}{d x} [ln (y)] = \frac{1}{y} \cdot y^{'}$ .

$\frac{1}{y} \cdot y^{'} = (2 \cdot \frac{1}{2 x + 1} \cdot 2) - \frac{3}{x} - (\frac{1}{2} \cdot \frac{1}{x ^{2} + 2 x} \cdot (2 x + 2))$

Simplifying,

$\frac{1}{y} \cdot y^{'} = \frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}$

4. Isolate $y^{'}$ to find the derivative

Multiply both sides by $y$ , then replace $y$ with the original function so the final answer is written in terms of $x$ .

$y^{'} = (\frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}) \cdot \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

If the function is already written as a logarithm, you can skip directly to expanding using logarithmic properties. The last step of multiplying by $y$ is also not necessary.

Problem 2. The total resistance in a circuit is modeled by the function

$R (V) = ln [(\frac{V ^{2} + 1}{V - 2})^{3} \cdot 5 V + 4]$

Find $R^{'} (V)$ .

(spoiler)

1. Expand using log properties

Use the product property first:

$R (V) = ln ((\frac{V ^{2} + 1}{V - 2})^{3}) + ln (5 V + 4)$

Now use the power property:

$R (V) = 3 ln (\frac{V ^{2} + 1}{V - 2}) + \frac{1}{2} ln (5 V + 4)$

Finally, use the quotient property:

$R (V) = 3 ln (V^{2} + 1) - 3 ln (V - 2) + \frac{1}{2} ln (5 V + 4)$

2. Differentiate

Differentiate both sides with respect to $V$ .

$R^{'} (V) = \frac{3}{V ^{2} + 1} \cdot (2 V) - \frac{3}{V - 2} + \frac{1}{2 ( 5 V + 4 )} \cdot 5$

Simplifying,

$R^{'} (V) = \frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 V + 8}$

Logarithmic Differentiation Overview

Streamlines differentiation of complex products, quotients, and variable exponents
Steps: take natural log, expand with log properties, differentiate implicitly, solve for $y^{'}$
Key log properties:
- Product: $lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$
- Quotient: $lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$
- Power: $lo g_{b} (x^{n}) = n lo g_{b} (x)$

Example: $y = \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

Log expansion: $2 ln (2 x + 1) - 3 ln (x) - \frac{1}{2} ln (x^{2} + 2 x)$
Derivative: $y^{'} = (\frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}) \cdot \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

Variable in Both Base and Exponent (e.g., $y = x^{x}$ )

Log expansion: $ln (y) = x ln (x)$
Derivative: $y^{'} = x^{x} (1 + ln (x))$

Nested Exponents (e.g., $y = x^{x^{x^{2}}}$ )

Multiple log steps: $ln (ln (y)) = x^{2} ln (x) + ln (ln (x))$
Derivative: $y^{'} = (x + 2 x ln (x) + \frac{1}{x l n ( x )}) \cdot x^{x^{x^{2}}} ln (x^{x^{x^{2}}})$

Functions Already in Logarithmic Form

Expand using log properties before differentiating
Example: $R (V) = ln [(\frac{V ^{2} + 1}{V - 2})^{3} 5 x + 4]$
- Expansion: $3 ln (V^{2} + 1) - 3 ln (V - 2) + \frac{1}{2} ln (5 x + 4)$
- Derivative: $R^{'} (V) = \frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}$

Logarithmic differentiation

What you’ll learn

Logarithmic differentiation: Apply logarithmic properties to simplify complex products and quotients before differentiating.

Consider the function

$y = \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

Differentiating this directly can get tedious because you’d need to use several derivative rules (power, product, quotient, and chain), and some of them more than once.

Step-by-step process

Take the natural logarithm of both sides (skip if one side is already a single logarithm with a complicated argument)
Expand using log properties.
Differentiate implicitly
Isolate $y^{'}$ to find the derivative

Use the following logarithmic properties:

Name	Formula
Product	$ln (a \cdot b) = ln (a) + ln (b)$
Quotient	$ln (\frac{a}{b}) = ln (a) - ln (b)$
Power	$ln (a^{b}) = b ln (a)$

Note: The power property doesn’t apply when the exponent is entirely on the outside:

$(ln a)^{b} \neq = b ln (a)$

Examples

Problem 1. Differentiate

$y = \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

Solution

(spoiler)

1. Take the natural log of both sides

$ln (y) = ln (\frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x})$

2. Expand using log properties

First use the quotient property to turn the fraction into a difference:

$ln (y) = ln ((2 x + 1)^{2}) - ln (x^{3} x^{2} + 2 x)$

Apply the product property on the 2nd logarithm and distribute the negative sign:

$ln (y) = ln ((2 x + 1)^{2}) - (ln (x^{3}) + ln (x^{2} + 2 x)) = ln ((2 x + 1)^{2}) - ln (x^{3}) - ln (x^{2} + 2 x)$

Now use the power property to bring exponents to the front:

$ln (y) = 2 ln (2 x + 1) - 3 ln (x) - \frac{1}{2} ln (x^{2} + 2 x)$

At this point, each term is a constant multiple of a simple logarithm, which is much easier to differentiate.

3. Differentiate implicitly

Differentiate both sides with respect to $x$ . On the left, use the chain rule: $\frac{d}{d x} [ln (y)] = \frac{1}{y} \cdot y^{'}$ .

$\frac{1}{y} \cdot y^{'} = (2 \cdot \frac{1}{2 x + 1} \cdot 2) - \frac{3}{x} - (\frac{1}{2} \cdot \frac{1}{x ^{2} + 2 x} \cdot (2 x + 2))$

Simplifying,

$\frac{1}{y} \cdot y^{'} = \frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}$

4. Isolate $y^{'}$ to find the derivative

Multiply both sides by $y$ , then replace $y$ with the original function so the final answer is written in terms of $x$ .

$y^{'} = (\frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}) \cdot \frac{( 2 x + 1 ) ^{2}}{x ^{3} x ^{2} + 2 x}$

If the function is already written as a logarithm, you can skip directly to expanding using logarithmic properties. The last step of multiplying by $y$ is also not necessary.

Problem 2. The total resistance in a circuit is modeled by the function

$R (V) = ln [(\frac{V ^{2} + 1}{V - 2})^{3} \cdot 5 V + 4]$

Find $R^{'} (V)$ .

(spoiler)

1. Expand using log properties

Use the product property first:

$R (V) = ln ((\frac{V ^{2} + 1}{V - 2})^{3}) + ln (5 V + 4)$

Now use the power property:

$R (V) = 3 ln (\frac{V ^{2} + 1}{V - 2}) + \frac{1}{2} ln (5 V + 4)$

Finally, use the quotient property:

$R (V) = 3 ln (V^{2} + 1) - 3 ln (V - 2) + \frac{1}{2} ln (5 V + 4)$

2. Differentiate

Differentiate both sides with respect to $V$ .

$R^{'} (V) = \frac{3}{V ^{2} + 1} \cdot (2 V) - \frac{3}{V - 2} + \frac{1}{2 ( 5 V + 4 )} \cdot 5$

Simplifying,

$R^{'} (V) = \frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 V + 8}$

Achievable AP Calculus AB

3. Advanced differentiation

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.