Logarithmic Differentiation

Learning Outcomes

Use logarithmic differentiation to determine the derivative of a function

At this point, we can take derivatives of functions of the form $y=(g(x))^n$ for certain values of $n$ , as well as functions of the form $y=b^{g(x)}$ , where $b>0$ and $b\ne 1$ . Unfortunately, we still do not know the derivatives of functions such as $y=x^x$ or $y=x^{\pi}$ . These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form $h(x)=g(x)^{f(x)}$ . It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of $y=\frac{x\sqrt{2x+1}}{e^x \sin^3 x}$ . We outline this technique in the following problem-solving strategy.

Problem-Solving Strategy: Using Logarithmic Differentiation

To differentiate $y=h(x)$ using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain $\ln y=\ln (h(x))$ .
Use properties of logarithms to expand $\ln (h(x))$ as much as possible.
Differentiate both sides of the equation. On the left we will have $\frac{1}{y}\frac{dy}{dx}$ .
Multiply both sides of the equation by $y$ to solve for $\frac{dy}{dx}$ .
Replace $y$ by $h(x)$ .

It may be useful to review your properties of logarithms. These will help us in step 2 to expand our logarithmic function.

Recall: Properties of logarithms

The Product Rule for Logarithms	${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$
The Quotient Rule for Logarithms	${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$
The Power Rule for Logarithms	${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$
The Change-of-Base Formula	${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\text{ }n>0,n\ne 1,b\ne 1$

Example: Using Logarithmic Differentiation

Find the derivative of $y=(2x^4+1)^{\tan x}$

Show Solution

Use logarithmic differentiation to find this derivative.

\begin{array}{lllll} \ln y & = \ln (2 x^{4} + 1)^{\tan x} & Step 1. Take the natural logarithm of both sides. \\ \ln y & = \tan x \ln (2 x^{4} + 1) & Step 2. Expand using properties of logarithms. \\ \frac{1}{y} \frac{d y}{d x} & = \sec^{2} x \ln (2 x^{4} + 1) + \frac{8 x^{3}}{2 x^{4} + 1} \cdot \tan x & \begin{array}{l} Step 3. Differentiate both sides. Use the \\ product rule on the right. \end{array} \\ \frac{d y}{d x} & = y \cdot (\sec^{2} x \ln (2 x^{4} + 1) + \frac{8 x^{3}}{2 x^{4} + 1} \cdot \tan x) & Step 4. Multiply by y on both sides. \\ \frac{d y}{d x} & = (2 x^{4} + 1)^{\tan x} (\sec^{2} x \ln (2 x^{4} + 1) + \frac{8 x^{3}}{2 x^{4} + 1} \cdot \tan x) & Step 5. Substitute y = (2 x^{4} + 1)^{\tan x} . \end{array}

$\begin{array}{lllll} \ln y & = \ln (2x^4+1)^{\tan x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = \tan x \ln (2x^4+1) & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = \sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x & & & \begin{array}{l}\text{Step 3. Differentiate both sides. Use the} \\ \text{product rule on the right.} \end{array} \\ \frac{dy}{dx} & =y \cdot (\sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x) & & & \text{Step 4. Multiply by} \, y \, \text{on both sides.} \\ \frac{dy}{dx} & = (2x^4+1)^{\tan x}(\sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x) & & & \text{Step 5. Substitute} \, y=(2x^4+1)^{\tan x}.\end{array}$

Watch the following video to see the worked solution to Example: Using Logarithmic Differentiation.

Closed Captioning and Transcript Information for Video

Example: Using Logarithmic Differentiation

Find the derivative of $y=\large \frac{x\sqrt{2x+1}}{e^x \sin^3 x}$

Show Solution

This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.

\begin{array}{lllll} \ln y & = \ln \frac{x \sqrt{2 x + 1}}{e^{x} \sin^{3} x} & Step 1. Take the natural logarithm of both sides. \\ \ln y & = \ln x + \frac{1}{2} \ln (2 x + 1) - x \ln e - 3 \ln \sin x & Step 2. Expand using properties of logarithms. \\ \frac{1}{y} \frac{d y}{d x} & = \frac{1}{x} + \frac{1}{2 x + 1} - 1 - 3 (\frac{\cos x}{\sin x}) & Step 3. Differentiate both sides. \\ \frac{d y}{d x} & = y (\frac{1}{x} + \frac{1}{2 x + 1} - 1 - 3 \cot x) & Step 4. Multiply by y on both sides and simplify. \\ \frac{d y}{d x} & = \frac{x \sqrt{2 x + 1}}{e^{x} \sin^{3} x} (\frac{1}{x} + \frac{1}{2 x + 1} - 1 - 3 \cot x) & Step 5. Substitute y = \frac{x \sqrt{2 x + 1}}{e^{x} \sin^{3} x} . \end{array}

$\begin{array}{lllll} \ln y & = \ln \large \frac{x\sqrt{2x+1}}{e^x \sin^3 x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = \ln x+\frac{1}{2} \ln (2x+1)-x \ln e-3 \ln \sin x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = \frac{1}{x}+\frac{1}{2x+1}-1-3\big(\frac{\cos x}{\sin x}\big) & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = y (\frac{1}{x}+\frac{1}{2x+1}-1-3 \cot x) & & & \text{Step 4. Multiply by} \, y \, \text{on both sides and simplify.} \\ \frac{dy}{dx} & = \large \frac{x\sqrt{2x+1}}{e^x \sin^3 x} \normalsize (\frac{1}{x}+\frac{1}{2x+1}-1-3 \cot x) & & & \text{Step 5. Substitute} \, y=\large \frac{x\sqrt{2x+1}}{e^x \sin^3 x}. \end{array}$

Example: Extending the Power Rule

Find the derivative of $y=x^r$ where $r$ is an arbitrary real number.

Show Solution

The process is the same as in the last example, though with fewer complications.

\begin{array}{lllll} \ln y & = \ln x^{r} & Step 1. Take the natural logarithm of both sides. \\ \ln y & = r \ln x & Step 2. Expand using properties of logarithms. \\ \frac{1}{y} \frac{d y}{d x} & = r \frac{1}{x} & Step 3. Differentiate both sides. \\ \frac{d y}{d x} & = y \frac{r}{x} & Step 4. Multiply by y on both sides. \\ \frac{d y}{d x} & = x^{r} \frac{r}{x} & Step 5. Substitute y = x^{r} . \\ \frac{d y}{d x} & = r x^{r - 1} & Simplify. \end{array}

$\begin{array}{lllll} \ln y & = \ln x^r & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = r \ln x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = r \frac{1}{x} & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = y \frac{r}{x} & & & \text{Step 4. Multiply by} \, y \, \text{on both sides.} \\ \frac{dy}{dx} & = x^r \frac{r}{x} & & & \text{Step 5. Substitute} \, y=x^r. \\ \frac{dy}{dx} & = rx^{r-1} & & & \text{Simplify.} \end{array}$

Try It

Use logarithmic differentiation to find the derivative of $y=x^x$ .

Show Solution

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

Hint

Follow the problem solving strategy.

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

Try It

Find the derivative of $y=(\tan x)^{\pi}$ .

Show Solution

$y^{\prime}=\pi (\tan x)^{\pi -1} \sec^2 x$

Hint

Use the result from the last example.

Module 3: Derivatives