Logarithmic Differentiation: Videos & Practice Problems

Logarithmic differentiation simplifies finding derivatives of complex functions. By taking the natural log of both sides, we can apply log properties to expand the equation, transforming products into sums and pulling out exponents. Using implicit differentiation, we derive the left side as $dy/dx$ and the right side involves derivatives of natural logs. This method is particularly useful for functions like $x^x$, where traditional rules may not apply directly.

Logarithmic Differentiation

In calculus, finding the derivative of complex functions can often be challenging, especially when multiple rules such as the quotient rule, product rule, and chain rule are involved. However, logarithmic differentiation provides a more straightforward approach for certain functions. This method is particularly useful when dealing with products, quotients, or powers within a function.

To apply logarithmic differentiation, start by taking the natural logarithm of both sides of the equation. For example, if we have a function defined as:

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

Taking the natural logarithm gives:

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

Using properties of logarithms, we can expand this expression. The logarithm of a product can be expressed as the sum of the logarithms, and the logarithm of a power allows us to bring the exponent in front:

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

Next, we differentiate both sides using implicit differentiation. The derivative of the left side, using the chain rule, is:

$$\frac{d}{dx}(\ln(y)) = \frac{1}{y} \frac{dy}{dx}$$

For the right side, we differentiate each term individually:

1. The derivative of $\ln(x + 4)$ is $\frac{1}{x + 4}$.

2. The derivative of $5\ln(x + 2)$ is $5 \cdot \frac{1}{x + 2}$.

3. The derivative of $-\frac{2}{3}\ln(x^3 - 1)$ involves applying the chain rule, resulting in $-\frac{2}{3} \cdot \frac{1}{x^3 - 1} \cdot 3x^2 = -\frac{2x^2}{x^3 - 1}$.

Combining these results gives:

$$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x + 4} + \frac{5}{x + 2} - \frac{2x^2}{x^3 - 1}$$

To isolate $\frac{dy}{dx}$, multiply both sides by $y$:

$$\frac{dy}{dx} = y \left( \frac{1}{x + 4} + \frac{5}{x + 2} - \frac{2x^2}{x^3 - 1} \right)$$

Finally, substitute back the original function for $y$:

$$\frac{dy}{dx} = \frac{(x + 4)(x + 2)^5}{(x^3 - 1)^{\frac{2}{3}}} \left( \frac{1}{x + 4} + \frac{5}{x + 2} - \frac{2x^2}{x^3 - 1} \right)$$

This process illustrates how logarithmic differentiation simplifies the derivative calculation for complex functions. It is particularly advantageous when direct application of differentiation rules becomes cumbersome. Understanding and practicing this technique will enhance your ability to tackle a variety of functions in calculus.

Logarithmic Differentiation Example 1

To find the derivative of the function \( y = x^x \), we utilize logarithmic differentiation, which is particularly effective for tower functions like this one. The first step involves taking the natural logarithm of both sides:

\( \ln(y) = \ln(x^x) \)

Using the properties of logarithms, we can simplify the right side:

\( \ln(y) = x \ln(x) \)

Next, we apply implicit differentiation to both sides. The derivative of \( \ln(y) \) with respect to \( x \) is:

\( \frac{d}{dx}(\ln(y)) = \frac{1}{y} \frac{dy}{dx} \)

For the right side, we need to use the product rule since we have \( x \) multiplied by \( \ln(x) \). The product rule states that if you have two functions \( u \) and \( v \), then:

\( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \)

In our case, let \( u = x \) and \( v = \ln(x) \). Thus, we find:

\( \frac{d}{dx}(x \ln(x)) = x \cdot \frac{1}{x} + \ln(x) \cdot 1 = 1 + \ln(x) \)

Now, we equate the derivatives:

\( \frac{1}{y} \frac{dy}{dx} = 1 + \ln(x) \)

To isolate \( \frac{dy}{dx} \), we multiply both sides by \( y \):

\( \frac{dy}{dx} = y(1 + \ln(x)) \)

Substituting back \( y = x^x \) gives us the final expression for the derivative:

\( \frac{dy}{dx} = x^x(1 + \ln(x)) \)

This result illustrates how logarithmic differentiation simplifies the process of finding derivatives for functions where the variable is also an exponent. Understanding this technique is crucial for tackling more complex functions in calculus.

Use logarithmic differentiation to find the derivative of the given function.
$y=(\sqrt{x+1})^{x}$

• A. $x\left(\sqrt{x+1}\right)^{x-1}$
• B. $\frac{x}{2}\left(\sqrt{x+1}\right)^{\frac{x}{2}-1}$
• C. $[\frac{x}{2(x+1)}+\frac{\ln (x+1)}{2}]{(x+1)}^{\frac{x}{2}}$
• D. $\left\lbrack\frac{\ln\left(x+1\right)+x}{2\left(x+1\right)}\right\rbrack\left(\sqrt{x+1}\right)^{x}$

Answer: (C)

Use logarithmic differentiation to find the derivative of the given function.
$y=\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$

• A. $\left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$
• B. $\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4}$
• C. $\left\lbrack\frac{2}{3\left(x+2\right)}-\frac{2}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$
• D. $\left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{1}{2\left(x^2-4\right)}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$

Answer: (A)

Use logarithmic differentiation to find the derivative of the given function.
$y=\tan^{x}x$

• A. $\left(\tan x\right)^{x}\cdot\ln\left(\tan x\right)\cdot\sec^2x$
• B. $\left(\tan x\right)^{x-1}\cdot\sec^2x$
• C. $\ln\left(\tan x\right)+\frac{x\sec^2x}{\tan x}$
• D. $\left(\ln\left(\tan x\right)+\frac{x\sec^2x}{\tan x}\right)\cdot\tan^{x}x$

Answer: (D)

Here’s what students ask on this topic:

What is logarithmic differentiation and when is it used?

Logarithmic differentiation is a technique used to find the derivative of complex functions by taking the natural log of both sides of the equation. This method simplifies the differentiation process by transforming products into sums and pulling out exponents using log properties. It is particularly useful for functions where traditional differentiation rules are cumbersome or inapplicable, such as functions of the form x^x.

How do you apply logarithmic differentiation to a function?

To apply logarithmic differentiation, follow these steps: 1) Take the natural log of both sides of the function. 2) Use log properties to expand the equation, transforming products into sums and pulling out exponents. 3) Apply implicit differentiation to both sides. 4) Solve for dy/dx by isolating it on one side of the equation. This method simplifies the differentiation process for complex functions.

Can you provide an example of logarithmic differentiation?

Sure! Consider the function y = (x + 4)(x + 2)⁵ / (x³ - 1)^2/3. First, take the natural log of both sides: ln(y) = ln((x + 4)(x + 2)⁵ / (x³ - 1)^2/3). Expand using log properties: ln(y) = ln(x + 4) + 5ln(x + 2) - (2/3)ln(x³ - 1). Differentiate implicitly: (1/y)dy/dx = 1/(x + 4) + 5/(x + 2) - (2/3)(3x² / (x³ - 1)). Finally, solve for dy/dx: dy/dx = y[1/(x + 4) + 5/(x + 2) - 2x² / (x³ - 1)].

Why is logarithmic differentiation useful for functions like x^x?

Logarithmic differentiation is useful for functions like x^x because traditional differentiation rules do not apply directly. By taking the natural log of both sides, ln(y) = ln(x^x) = xln(x), we can differentiate implicitly: (1/y)dy/dx = ln(x) + 1. Solving for dy/dx gives dy/dx = y(ln(x) + 1) = x^x(ln(x) + 1), which simplifies the differentiation process.

What are the steps to expand a logarithmic function using log properties?

To expand a logarithmic function using log properties, follow these steps: 1) Use the product rule: ln(ab) = ln(a) + ln(b). 2) Use the quotient rule: ln(a/b) = ln(a) - ln(b). 3) Use the power rule: ln(a^b) = b ln(a). These properties help transform complex expressions into simpler sums and differences, making differentiation easier.