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:

Apply log properties and use 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	$lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$
Root	$lo g_{b} (r x) = \frac{1}{r} \cdot lo g_{b} (x)$

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, and treat the denominator as a product:

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

Now use the power and root properties to bring exponents out 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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} y^{'}$ .

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

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 increase the difficulty:

Find the derivative of

$y = x^{x^{x^{2}}}$

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Use log properties

Use the power property to bring the exponent down:

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

The right-hand side still contains a variable in an exponent, so take the natural log of both sides again:

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

Use the product property to separate the right-hand side:

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

Then use the power property again:

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

3. Differentiate implicitly

Differentiate both sides with respect to $x$ .

Left side (chain rule twice):

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

Right side: differentiate $x^{2} ln (x)$ using the product rule, and differentiate $ln (ln (x))$ using the chain rule.

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

Simplifying,

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

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

Multiply both sides by $y ln (y)$ :

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

Finally, replace $y$ with the original function.

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

Examples

The total resistance in a circuit is modeled by the function

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

Find $R^{'} (V)$ (or in other words, $\frac{d R}{d V}$ ) using logarithmic differentiation.

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Expand using log properties

Use the product property first:

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

Now use the power and root properties:

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

Finally, use the quotient property:

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

3. Differentiate implicitly

Differentiate both sides with respect to $V$ .

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

Simplifying,

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

4. Isolate $\frac{d R}{d V}$ to find the derivative

Multiply both sides by $R (V)$ and substitute the original expression for $R (V)$ .

$\frac{d R}{d V} = (\frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}) \cdot (\frac{V ^{2} + 1}{V - 2})^{3} \cdot 5 x + 4$

The population of a bacterial culture is modeled by

$P (t) = (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

Find the instantaneous rate of change of the population at time $t = 1$ .

Solution

(spoiler)

1. Take the natural log of both sides

$ln (P) = ln ((t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1))$

2. Expand using log properties

Use the product property to split the logarithm into a sum:

$ln (P) = ln ((t^{2} - t)^{3}) + ln (e^{2 t}) + ln (ln (t + 1))$

Now use the power property on the first two terms:

$ln (P) = 3 ln (t^{2} - t) + 2 t ln (e) + ln (ln (t + 1))$

Since $ln (e) = 1$ , this becomes

$ln (P) = 3 ln (t^{2} - t) + 2 t + ln (ln (t + 1))$

3. Differentiate implicitly

Differentiate both sides with respect to $t$ .

$\frac{1}{P} \cdot P^{'} (t) = 3 \cdot \frac{1}{t ^{2} - t} \cdot (2 t - 1) + 2 + \frac{1}{ln ( t + 1 )} \cdot \frac{1}{t + 1} \cdot 1$

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

Multiply both sides by $P$ and substitute the original expression for $P (t)$ .

$P^{'} (t) = [\frac{6 t - 3}{t ^{2} - t} + 2 + \frac{1}{( t + 1 ) ln ( t + 1 )}] \cdot (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

At $t = 1$ , the instantaneous rate of change of the population is

$P^{'} (1) = [\frac{6 ( 1 ) - 3}{( 1 ) ^{2} - 1} + 2 + \frac{1}{( 1 + 1 ) ln ( 1 + 1 )}] \cdot (1^{2} - 1)^{3} \cdot e^{2 (1)} \cdot ln (1 + 1)$

$= 0$

Logarithmic differentiation overview

Technique for differentiating complicated functions, especially with variables in both base and exponent
Steps:
- Take natural log of both sides
- Expand using log properties
- Differentiate implicitly
- Isolate derivative and substitute original function

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: $lo g_{b} (x^{p}) = p lo g_{b} (x)$
Root: $lo g_{b} (r x) = \frac{1}{r} lo g_{b} (x)$

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

Expand with logs: $2 ln (2 x + 1) - 3 ln (x) - \frac{1}{2} ln (x^{2} + 2 x)$
Differentiate: $\frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}$
Final derivative: Multiply by original $y$

Variable in base and exponent: $y = x^{x}$

Take log: $ln (y) = x ln (x)$
Differentiate: $\frac{1}{y} y^{'} = 1 + ln (x)$
Final derivative: $y^{'} = x^{x} (1 + ln (x))$

Advanced example: $y = x^{x^{x^{2}}}$

Take log: $ln (y) = x^{x^{2}} ln (x)$
Take log again: $ln (ln (y)) = x^{2} ln (x) + ln (ln (x))$
Differentiate: $\frac{1}{y l n ( y )} y^{'} = x + 2 x ln (x) + \frac{1}{x l n ( x )}$
Final derivative: $y^{'} = (x + 2 x ln (x) + \frac{1}{x l n ( x )}) x^{x^{x^{2}}} ln (x^{x^{x^{2}}})$

Example: Circuit resistance $R (V) = (\frac{V ^{2} + 1}{V - 2})^{3} 5 x + 4$

Log expansion: $3 ln (V^{2} + 1) - 3 ln (V - 2) + \frac{1}{2} ln (5 x + 4)$
Differentiate: $\frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}$
Final derivative: Multiply by original $R (V)$

Example: Bacterial population $P (t) = (t^{2} - t)^{3} e^{2 t} ln (t + 1)$

Log expansion: $3 ln (t^{2} - t) + 2 t + ln (ln (t + 1))$
Differentiate: $\frac{6 t - 3}{t ^{2} - t} + 2 + \frac{1}{( t + 1 ) l n ( t + 1 )}$
Final derivative: Multiply by original $P (t)$
At $t = 1$ , $P^{'} (1) = 0$

Logarithmic differentiation

What you’ll learn:

Apply log properties and use 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	$lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$
Root	$lo g_{b} (r x) = \frac{1}{r} \cdot lo g_{b} (x)$

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, and treat the denominator as a product:

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

Now use the power and root properties to bring exponents out 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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} y^{'}$ .

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

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 increase the difficulty:

Find the derivative of

$y = x^{x^{x^{2}}}$

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Use log properties

Use the power property to bring the exponent down:

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

The right-hand side still contains a variable in an exponent, so take the natural log of both sides again:

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

Use the product property to separate the right-hand side:

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

Then use the power property again:

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

3. Differentiate implicitly

Differentiate both sides with respect to $x$ .

Left side (chain rule twice):

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

Right side: differentiate $x^{2} ln (x)$ using the product rule, and differentiate $ln (ln (x))$ using the chain rule.

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

Simplifying,

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

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

Multiply both sides by $y ln (y)$ :

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

Finally, replace $y$ with the original function.

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

Examples

The total resistance in a circuit is modeled by the function

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

Find $R^{'} (V)$ (or in other words, $\frac{d R}{d V}$ ) using logarithmic differentiation.

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Expand using log properties

Use the product property first:

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

Now use the power and root properties:

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

Finally, use the quotient property:

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

3. Differentiate implicitly

Differentiate both sides with respect to $V$ .

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

Simplifying,

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

4. Isolate $\frac{d R}{d V}$ to find the derivative

Multiply both sides by $R (V)$ and substitute the original expression for $R (V)$ .

$\frac{d R}{d V} = (\frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}) \cdot (\frac{V ^{2} + 1}{V - 2})^{3} \cdot 5 x + 4$

The population of a bacterial culture is modeled by

$P (t) = (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

Find the instantaneous rate of change of the population at time $t = 1$ .

Solution

(spoiler)

1. Take the natural log of both sides

$ln (P) = ln ((t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1))$

2. Expand using log properties

Use the product property to split the logarithm into a sum:

$ln (P) = ln ((t^{2} - t)^{3}) + ln (e^{2 t}) + ln (ln (t + 1))$

Now use the power property on the first two terms:

$ln (P) = 3 ln (t^{2} - t) + 2 t ln (e) + ln (ln (t + 1))$

Since $ln (e) = 1$ , this becomes

$ln (P) = 3 ln (t^{2} - t) + 2 t + ln (ln (t + 1))$

3. Differentiate implicitly

Differentiate both sides with respect to $t$ .

$\frac{1}{P} \cdot P^{'} (t) = 3 \cdot \frac{1}{t ^{2} - t} \cdot (2 t - 1) + 2 + \frac{1}{ln ( t + 1 )} \cdot \frac{1}{t + 1} \cdot 1$

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

Multiply both sides by $P$ and substitute the original expression for $P (t)$ .

$P^{'} (t) = [\frac{6 t - 3}{t ^{2} - t} + 2 + \frac{1}{( t + 1 ) ln ( t + 1 )}] \cdot (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

At $t = 1$ , the instantaneous rate of change of the population is

$P^{'} (1) = [\frac{6 ( 1 ) - 3}{( 1 ) ^{2} - 1} + 2 + \frac{1}{( 1 + 1 ) ln ( 1 + 1 )}] \cdot (1^{2} - 1)^{3} \cdot e^{2 (1)} \cdot ln (1 + 1)$

$= 0$

Achievable AP Calculus AB

3. Advanced differentiation

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

Logarithmic differentiation

8 min read

Font

Discuss

Feedback

What you’ll learn:

Apply log properties and use 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	$lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$
Root	$lo g_{b} (r x) = \frac{1}{r} \cdot lo g_{b} (x)$

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, and treat the denominator as a product:

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

Now use the power and root properties to bring exponents out 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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} y^{'}$ .

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

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 increase the difficulty:

Find the derivative of

$y = x^{x^{x^{2}}}$

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Use log properties

Use the power property to bring the exponent down:

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

The right-hand side still contains a variable in an exponent, so take the natural log of both sides again:

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

Use the product property to separate the right-hand side:

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

Then use the power property again:

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

3. Differentiate implicitly

Differentiate both sides with respect to $x$ .

Left side (chain rule twice):

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

Right side: differentiate $x^{2} ln (x)$ using the product rule, and differentiate $ln (ln (x))$ using the chain rule.

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

Simplifying,

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

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

Multiply both sides by $y ln (y)$ :

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

Finally, replace $y$ with the original function.

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

Examples

The total resistance in a circuit is modeled by the function

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

Find $R^{'} (V)$ (or in other words, $\frac{d R}{d V}$ ) using logarithmic differentiation.

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Expand using log properties

Use the product property first:

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

Now use the power and root properties:

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

Finally, use the quotient property:

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

3. Differentiate implicitly

Differentiate both sides with respect to $V$ .

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

Simplifying,

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

4. Isolate $\frac{d R}{d V}$ to find the derivative

Multiply both sides by $R (V)$ and substitute the original expression for $R (V)$ .

$\frac{d R}{d V} = (\frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}) \cdot (\frac{V ^{2} + 1}{V - 2})^{3} \cdot 5 x + 4$

The population of a bacterial culture is modeled by

$P (t) = (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

Find the instantaneous rate of change of the population at time $t = 1$ .

Solution

(spoiler)

1. Take the natural log of both sides

$ln (P) = ln ((t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1))$

2. Expand using log properties

Use the product property to split the logarithm into a sum:

$ln (P) = ln ((t^{2} - t)^{3}) + ln (e^{2 t}) + ln (ln (t + 1))$

Now use the power property on the first two terms:

$ln (P) = 3 ln (t^{2} - t) + 2 t ln (e) + ln (ln (t + 1))$

Since $ln (e) = 1$ , this becomes

$ln (P) = 3 ln (t^{2} - t) + 2 t + ln (ln (t + 1))$

3. Differentiate implicitly

Differentiate both sides with respect to $t$ .

$\frac{1}{P} \cdot P^{'} (t) = 3 \cdot \frac{1}{t ^{2} - t} \cdot (2 t - 1) + 2 + \frac{1}{ln ( t + 1 )} \cdot \frac{1}{t + 1} \cdot 1$

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

Multiply both sides by $P$ and substitute the original expression for $P (t)$ .

$P^{'} (t) = [\frac{6 t - 3}{t ^{2} - t} + 2 + \frac{1}{( t + 1 ) ln ( t + 1 )}] \cdot (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

At $t = 1$ , the instantaneous rate of change of the population is

$P^{'} (1) = [\frac{6 ( 1 ) - 3}{( 1 ) ^{2} - 1} + 2 + \frac{1}{( 1 + 1 ) ln ( 1 + 1 )}] \cdot (1^{2} - 1)^{3} \cdot e^{2 (1)} \cdot ln (1 + 1)$

$= 0$

Logarithmic differentiation overview

Technique for differentiating complicated functions, especially with variables in both base and exponent
Steps:
- Take natural log of both sides
- Expand using log properties
- Differentiate implicitly
- Isolate derivative and substitute original function

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: $lo g_{b} (x^{p}) = p lo g_{b} (x)$
Root: $lo g_{b} (r x) = \frac{1}{r} lo g_{b} (x)$

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

Expand with logs: $2 ln (2 x + 1) - 3 ln (x) - \frac{1}{2} ln (x^{2} + 2 x)$
Differentiate: $\frac{4}{2 x + 1} - \frac{3}{x} - \frac{x + 1}{x ^{2} + 2 x}$
Final derivative: Multiply by original $y$

Variable in base and exponent: $y = x^{x}$

Take log: $ln (y) = x ln (x)$
Differentiate: $\frac{1}{y} y^{'} = 1 + ln (x)$
Final derivative: $y^{'} = x^{x} (1 + ln (x))$

Advanced example: $y = x^{x^{x^{2}}}$

Take log: $ln (y) = x^{x^{2}} ln (x)$
Take log again: $ln (ln (y)) = x^{2} ln (x) + ln (ln (x))$
Differentiate: $\frac{1}{y l n ( y )} y^{'} = x + 2 x ln (x) + \frac{1}{x l n ( x )}$
Final derivative: $y^{'} = (x + 2 x ln (x) + \frac{1}{x l n ( x )}) x^{x^{x^{2}}} ln (x^{x^{x^{2}}})$

Example: Circuit resistance $R (V) = (\frac{V ^{2} + 1}{V - 2})^{3} 5 x + 4$

Log expansion: $3 ln (V^{2} + 1) - 3 ln (V - 2) + \frac{1}{2} ln (5 x + 4)$
Differentiate: $\frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}$
Final derivative: Multiply by original $R (V)$

Example: Bacterial population $P (t) = (t^{2} - t)^{3} e^{2 t} ln (t + 1)$

Log expansion: $3 ln (t^{2} - t) + 2 t + ln (ln (t + 1))$
Differentiate: $\frac{6 t - 3}{t ^{2} - t} + 2 + \frac{1}{( t + 1 ) l n ( t + 1 )}$
Final derivative: Multiply by original $P (t)$
At $t = 1$ , $P^{'} (1) = 0$

Logarithmic differentiation

What you’ll learn:

Apply log properties and use 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	$lo g_{b} (x^{p}) = p \cdot lo g_{b} (x)$
Root	$lo g_{b} (r x) = \frac{1}{r} \cdot lo g_{b} (x)$

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, and treat the denominator as a product:

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

Now use the power and root properties to bring exponents out 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: $\frac{d}{d x} [ln (y)] = \frac{1}{y} y^{'}$ .

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

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 increase the difficulty:

Find the derivative of

$y = x^{x^{x^{2}}}$

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Use log properties

Use the power property to bring the exponent down:

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

The right-hand side still contains a variable in an exponent, so take the natural log of both sides again:

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

Use the product property to separate the right-hand side:

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

Then use the power property again:

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

3. Differentiate implicitly

Differentiate both sides with respect to $x$ .

Left side (chain rule twice):

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

Right side: differentiate $x^{2} ln (x)$ using the product rule, and differentiate $ln (ln (x))$ using the chain rule.

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

Simplifying,

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

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

Multiply both sides by $y ln (y)$ :

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

Finally, replace $y$ with the original function.

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

Examples

The total resistance in a circuit is modeled by the function

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

Find $R^{'} (V)$ (or in other words, $\frac{d R}{d V}$ ) using logarithmic differentiation.

Solution

(spoiler)

1. Take the natural log of both sides

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

2. Expand using log properties

Use the product property first:

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

Now use the power and root properties:

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

Finally, use the quotient property:

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

3. Differentiate implicitly

Differentiate both sides with respect to $V$ .

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

Simplifying,

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

4. Isolate $\frac{d R}{d V}$ to find the derivative

Multiply both sides by $R (V)$ and substitute the original expression for $R (V)$ .

$\frac{d R}{d V} = (\frac{6 V}{V ^{2} + 1} - \frac{3}{V - 2} + \frac{5}{10 x + 8}) \cdot (\frac{V ^{2} + 1}{V - 2})^{3} \cdot 5 x + 4$

The population of a bacterial culture is modeled by

$P (t) = (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

Find the instantaneous rate of change of the population at time $t = 1$ .

Solution

(spoiler)

1. Take the natural log of both sides

$ln (P) = ln ((t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1))$

2. Expand using log properties

Use the product property to split the logarithm into a sum:

$ln (P) = ln ((t^{2} - t)^{3}) + ln (e^{2 t}) + ln (ln (t + 1))$

Now use the power property on the first two terms:

$ln (P) = 3 ln (t^{2} - t) + 2 t ln (e) + ln (ln (t + 1))$

Since $ln (e) = 1$ , this becomes

$ln (P) = 3 ln (t^{2} - t) + 2 t + ln (ln (t + 1))$

3. Differentiate implicitly

Differentiate both sides with respect to $t$ .

$\frac{1}{P} \cdot P^{'} (t) = 3 \cdot \frac{1}{t ^{2} - t} \cdot (2 t - 1) + 2 + \frac{1}{ln ( t + 1 )} \cdot \frac{1}{t + 1} \cdot 1$

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

Multiply both sides by $P$ and substitute the original expression for $P (t)$ .

$P^{'} (t) = [\frac{6 t - 3}{t ^{2} - t} + 2 + \frac{1}{( t + 1 ) ln ( t + 1 )}] \cdot (t^{2} - t)^{3} \cdot e^{2 t} \cdot ln (t + 1)$

At $t = 1$ , the instantaneous rate of change of the population is

$P^{'} (1) = [\frac{6 ( 1 ) - 3}{( 1 ) ^{2} - 1} + 2 + \frac{1}{( 1 + 1 ) ln ( 1 + 1 )}] \cdot (1^{2} - 1)^{3} \cdot e^{2 (1)} \cdot ln (1 + 1)$

$= 0$