3.3 Logarithmic differentiation

Achievable AP Calculus AB

3. Advanced differentiation

Our AP Calculus AB course is currently in development and is a work-in-progress.

Logarithmic differentiation

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What you’ll learn

Applying log properties and using logarithmic differentiation to simplify expressions.
Differentiating functions with variables in both the base and exponent (e.g. $y = x^{x}$ )

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 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.

This technique is also useful when the variable appears in both the base and the exponent, such as $y = x^{x}$ .

Here’s how to do this:

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

The log properties to remember are:

Name	Formula
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	$lo g_{b} (x^{n}) = n \cdot lo g_{b} (x)$

Domain note and common final-step error

Logarithmic differentiation technically requires $y > 0$ . In practice, we write $ln ∣ y ∣$ instead of $ln (y)$ ; the resulting derivative formula is exactly the same, so the technique works for functions that may be negative.

Also watch out for the last step: after differentiating, you must multiply both sides by $y$ and replace $y$ with the original expression so the answer is fully in terms of $x$ . Forgetting to substitute back is one of the most common mistakes with this technique.

Examples

Let’s use the function at the start as an example:

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

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:

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

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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} \cdot y^{'}$ .

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

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 $x$ .

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

Logarithmic differentiation is also useful when the function appears in an exponent (so the exponent is not a constant).

Here’s a classic example:

Differentiate $y = x^{x}$

Solution

1. Take the natural log of both sides

$ln (y) = ln (x^{x})$

2. Use log properties

Apply the power property $ln (a^{b}) = b ln (a)$ :

$ln (y) = x ln (x)$

3. Differentiate implicitly

Differentiate both sides. On the right, use the product rule.

$\frac{1}{y} \cdot y^{'} = x \cdot \frac{1}{x} + ln (x) \cdot 1$

$\frac{1}{y} \cdot y^{'} = 1 + ln (x)$

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

Multiply both sides by $y$ , then substitute $y = x^{x}$ .

$y^{'} = x^{x} (1 + ln (x))$

Let’s try another example with a variable in the exponent:

Find the derivative of $y = x^{l n x}$

Solution

(spoiler)

1. Take the natural log of both sides

$ln (y) = ln (x^{l n x})$

2. Use log properties

Use the power property to bring the exponent down:

$ln (y) = ln (x) \cdot ln (x) = (ln x)^{2}$

3. Differentiate implicitly

Differentiate both sides with respect to $x$ . On the right, use the chain rule.

$\frac{1}{y} \cdot y^{'} = 2 ln (x) \cdot \frac{1}{x}$

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

Multiply both sides by $y$ , then substitute $y = x^{l n x}$ .

$y^{'} = x^{l n x} \cdot \frac{2 ln ( x )}{x}$

If the function is already written as a logarithm, you can skip directly to expanding using logarithmic properties.

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 $R^{'} (V)$ .

Solution

(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 and root properties:

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

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

Applying log properties and using logarithmic differentiation to simplify expressions.
Differentiating functions with variables in both the base and exponent (e.g. $y = x^{x}$ )

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.

This technique is also useful when the variable appears in both the base and the exponent, such as $y = x^{x}$ .

Here’s how to do this:

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

The log properties to remember are:

Name	Formula
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	$lo g_{b} (x^{n}) = n \cdot lo g_{b} (x)$

Domain note and common final-step error

Examples

Let’s use the function at the start as an example:

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

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:

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

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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} \cdot y^{'}$ .

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

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 $x$ .

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

Logarithmic differentiation is also useful when the function appears in an exponent (so the exponent is not a constant).

Here’s a classic example:

Differentiate $y = x^{x}$

Solution

1. Take the natural log of both sides

$ln (y) = ln (x^{x})$

2. Use log properties

Apply the power property $ln (a^{b}) = b ln (a)$ :

$ln (y) = x ln (x)$

3. Differentiate implicitly

Differentiate both sides. On the right, use the product rule.

$\frac{1}{y} \cdot y^{'} = x \cdot \frac{1}{x} + ln (x) \cdot 1$

$\frac{1}{y} \cdot y^{'} = 1 + ln (x)$

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

Multiply both sides by $y$ , then substitute $y = x^{x}$ .

$y^{'} = x^{x} (1 + ln (x))$

Let’s try another example with a variable in the exponent:

Find the derivative of $y = x^{l n x}$

Solution

(spoiler)

1. Take the natural log of both sides

$ln (y) = ln (x^{l n x})$

2. Use log properties

Use the power property to bring the exponent down:

$ln (y) = ln (x) \cdot ln (x) = (ln x)^{2}$

3. Differentiate implicitly

Differentiate both sides with respect to $x$ . On the right, use the chain rule.

$\frac{1}{y} \cdot y^{'} = 2 ln (x) \cdot \frac{1}{x}$

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

Multiply both sides by $y$ , then substitute $y = x^{l n x}$ .

$y^{'} = x^{l n x} \cdot \frac{2 ln ( x )}{x}$

If the function is already written as a logarithm, you can skip directly to expanding using logarithmic properties.

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 $R^{'} (V)$ .

Solution

(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 and root properties:

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

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 currently in development and is a work-in-progress.

Logarithmic differentiation

6 min read

Font

Discuss

Feedback

What you’ll learn

Applying log properties and using logarithmic differentiation to simplify expressions.
Differentiating functions with variables in both the base and exponent (e.g. $y = x^{x}$ )

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.

This technique is also useful when the variable appears in both the base and the exponent, such as $y = x^{x}$ .

Here’s how to do this:

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

The log properties to remember are:

Name	Formula
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	$lo g_{b} (x^{n}) = n \cdot lo g_{b} (x)$

Domain note and common final-step error

Examples

Let’s use the function at the start as an example:

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

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:

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

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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} \cdot y^{'}$ .

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

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 $x$ .

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

Logarithmic differentiation is also useful when the function appears in an exponent (so the exponent is not a constant).

Here’s a classic example:

Differentiate $y = x^{x}$

Solution

1. Take the natural log of both sides

$ln (y) = ln (x^{x})$

2. Use log properties

Apply the power property $ln (a^{b}) = b ln (a)$ :

$ln (y) = x ln (x)$

3. Differentiate implicitly

Differentiate both sides. On the right, use the product rule.

$\frac{1}{y} \cdot y^{'} = x \cdot \frac{1}{x} + ln (x) \cdot 1$

$\frac{1}{y} \cdot y^{'} = 1 + ln (x)$

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

Multiply both sides by $y$ , then substitute $y = x^{x}$ .

$y^{'} = x^{x} (1 + ln (x))$

Let’s try another example with a variable in the exponent:

Find the derivative of $y = x^{l n x}$

Solution

(spoiler)

1. Take the natural log of both sides

$ln (y) = ln (x^{l n x})$

2. Use log properties

Use the power property to bring the exponent down:

$ln (y) = ln (x) \cdot ln (x) = (ln x)^{2}$

3. Differentiate implicitly

Differentiate both sides with respect to $x$ . On the right, use the chain rule.

$\frac{1}{y} \cdot y^{'} = 2 ln (x) \cdot \frac{1}{x}$

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

Multiply both sides by $y$ , then substitute $y = x^{l n x}$ .

$y^{'} = x^{l n x} \cdot \frac{2 ln ( x )}{x}$

If the function is already written as a logarithm, you can skip directly to expanding using logarithmic properties.

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 $R^{'} (V)$ .

Solution

(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 and root properties:

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

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

Applying log properties and using logarithmic differentiation to simplify expressions.
Differentiating functions with variables in both the base and exponent (e.g. $y = x^{x}$ )

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.

This technique is also useful when the variable appears in both the base and the exponent, such as $y = x^{x}$ .

Here’s how to do this:

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

The log properties to remember are:

Name	Formula
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	$lo g_{b} (x^{n}) = n \cdot lo g_{b} (x)$

Domain note and common final-step error

Examples

Let’s use the function at the start as an example:

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

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:

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

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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} \cdot y^{'}$ .

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

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 $x$ .

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

Logarithmic differentiation is also useful when the function appears in an exponent (so the exponent is not a constant).

Here’s a classic example:

Differentiate $y = x^{x}$

Solution

1. Take the natural log of both sides

$ln (y) = ln (x^{x})$

2. Use log properties

Apply the power property $ln (a^{b}) = b ln (a)$ :

$ln (y) = x ln (x)$

3. Differentiate implicitly

Differentiate both sides. On the right, use the product rule.

$\frac{1}{y} \cdot y^{'} = x \cdot \frac{1}{x} + ln (x) \cdot 1$

$\frac{1}{y} \cdot y^{'} = 1 + ln (x)$

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

Multiply both sides by $y$ , then substitute $y = x^{x}$ .

$y^{'} = x^{x} (1 + ln (x))$

Let’s try another example with a variable in the exponent:

Find the derivative of $y = x^{l n x}$

Solution

(spoiler)

1. Take the natural log of both sides

$ln (y) = ln (x^{l n x})$

2. Use log properties

Use the power property to bring the exponent down:

$ln (y) = ln (x) \cdot ln (x) = (ln x)^{2}$

3. Differentiate implicitly

Differentiate both sides with respect to $x$ . On the right, use the chain rule.

$\frac{1}{y} \cdot y^{'} = 2 ln (x) \cdot \frac{1}{x}$

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

Multiply both sides by $y$ , then substitute $y = x^{l n x}$ .

$y^{'} = x^{l n x} \cdot \frac{2 ln ( x )}{x}$

If the function is already written as a logarithm, you can skip directly to expanding using logarithmic properties.

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 $R^{'} (V)$ .

Solution

(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 and root properties:

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

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,

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