Logarithmic differentiation | Advanced 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 would be fairly tedious since it uses every standard derivative rule (power, product, quotient, chain) at some point, some of those more than once.

Logarithmic differentiation simplifies this entire process by utilizing log properties to break complex functions apart into simpler ones before differentiating.

The technique is also useful for differentiating functions that have the variable 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

Using the product and quotient properties,

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

Using the power and root properties,

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

Now this is significantly easier to differentiate.

3. Differentiate implicitly

Don’t forget the chain rule!

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

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

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

Just multiply by $y$ on both sides and replace $y$ with the original function to write $y^{'}$ in terms of $x$ only.

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

Logarithmic differentiation is also useful when a function is in the exponent of another function, as opposed to either the base or the exponent being a constant.

Here’s a classic example:

1. Differentiate $y = x^{x}$

Solution

1. Take the natural log of both sides

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

2. Use log properties

Applying the power property of logs,

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

3. Differentiate implicitly

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

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

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

Multiplying both sides by $y$ ,

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

Replace $y$ with the original function so that the derivative is in terms of $x$ :

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

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

Using the power property,

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

Because the right side still has a variable in the exponent of the variable, take the natural log of both sides again:

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

Using the product property to separate the right side,

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

Then using the power property,

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

3. Differentiate implicitly

Don’t forget the chain rule!

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

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

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

Multiplying both sides by $y ln (y)$ ,

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

Lastly, replace every instance of $y$ 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

1. 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

Using the product property,

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

Using the power and root properties,

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

Using 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

Don’t forget the chain rule!

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

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 replace with the original function (that was in terms of $V$ ).

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

2. 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

Using the product property,

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

Then using the power property,

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

$ln (e) = 1$ so the equation simplifies to

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

3. Differentiate implicitly

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

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

Multiply both sides by $P$ and replace with the original function that was in terms of $t$ :

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

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

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

$= 0$

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 would be fairly tedious since it uses every standard derivative rule (power, product, quotient, chain) at some point, some of those more than once.

Logarithmic differentiation simplifies this entire process by utilizing log properties to break complex functions apart into simpler ones before differentiating.

The technique is also useful for differentiating functions that have the variable 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

Using the product and quotient properties,

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

Using the power and root properties,

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

Now this is significantly easier to differentiate.

3. Differentiate implicitly

Don’t forget the chain rule!

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

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

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

Just multiply by $y$ on both sides and replace $y$ with the original function to write $y^{'}$ in terms of $x$ only.

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

Logarithmic differentiation is also useful when a function is in the exponent of another function, as opposed to either the base or the exponent being a constant.

Here’s a classic example:

1. Differentiate $y = x^{x}$

Solution

1. Take the natural log of both sides

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

2. Use log properties

Applying the power property of logs,

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

3. Differentiate implicitly

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

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

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

Multiplying both sides by $y$ ,

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

Replace $y$ with the original function so that the derivative is in terms of $x$ :

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

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

Using the power property,

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

Because the right side still has a variable in the exponent of the variable, take the natural log of both sides again:

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

Using the product property to separate the right side,

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

Then using the power property,

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

3. Differentiate implicitly

Don’t forget the chain rule!

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

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

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

Multiplying both sides by $y ln (y)$ ,

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

Lastly, replace every instance of $y$ 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

1. 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

Using the product property,

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

Using the power and root properties,

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

Using 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

Don’t forget the chain rule!

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

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 replace with the original function (that was in terms of $V$ ).

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

2. 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

Using the product property,

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

Then using the power property,

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

$ln (e) = 1$ so the equation simplifies to

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

3. Differentiate implicitly

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

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

Multiply both sides by $P$ and replace with the original function that was in terms of $t$ :

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

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

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

$= 0$