3.8: Differentiation Techniques - Logarithmic Differentiation (2024)

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This page is a draft and is under active development.

Gilbert Strang & Edwin “Jed” Herman
OpenStax

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Learning Objectives

Use logarithmic differentiation to determine the derivative of products andratios of functions.
Use logarithmic differentiation to determine the derivative of functions to functional powers.

So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and hyperbolic functions. In this section, we explore using logarithms to simplify the differentiation process and to take derivatives of functions raised to powers containing functions.

Logarithmic Differentiation

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 \(b \neq 1\). Unfortunately, we still do not know the derivatives of functions such as \(y=x^x\) or \(y=x^{\sin{(x)}}\). 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 \(y=\frac{x\sqrt{2x+1}}{e^x\sin^3 x}\).

Example \(\PageIndex{1}\): Using Logarithmic Differentiation

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

Solution

Use logarithmic differentiation to find this derivative.

\[ \begin{array}{rrclr}
& \ln{(y)} & = & \ln{(2x^4+1)^{\tan{(x)}}} & \left( \text{Take the natural logarithm of both sides.} \right) \\
\implies & \ln{(y)} & = & \tan{(x)} \ln{(2x^4+1)} & \left( \text{Expand using properties of logarithms.} \right) \\
\implies & \dfrac{1}{y}\dfrac{dy}{dx} & = & \sec^2{(x)} \ln{(2x^4+1)} + \dfrac{8x^3}{2x^4+1} \cdot \tan{(x)}& \left( \text{Implicitly differentiate both sides.} \right) \\
\implies & \dfrac{dy}{dx} & = & y \cdot \left(\sec^2{(x)} \ln{(2x^4+1)} + \dfrac{8x^3}{2x^4+1} \cdot \tan{(x)} \right) & \\
\implies & \dfrac{dy}{dx} & = & (2x^4+1)^{\tan{(x)}} \left( \sec^2{(x)} \ln{(2x^4+1)} + \dfrac{8x^3}{2x^4+1} \cdot \tan{(x)} \right) &\left( \text{Substitute }y=(2x^4+1)^{\tan{(x)}}. \right) \\
\end{array} \nonumber \]

Example \(\PageIndex{2}\): Simplifying a "Tough" Differentiation Problem

Find the derivative of \(y=\frac{x\sqrt{2x+1}}{e^x\sin^3 x}\).

Solution

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

\[ \begin{array}{rrclr}
& \ln y & = & \ln\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x} & \left( \text{Take the natural logarithm of both sides.} \right) \\
\implies &\ln y & = & \ln x+\frac{1}{2}\ln(2x+1)−x\ln e−3\ln \sin x & \left( \text{Expand using properties of logarithms.} \right) \\
\implies &\dfrac{1}{y}\dfrac{dy}{dx} & = & \dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\dfrac{\cos x}{\sin x} & \left( \text{Differentiate both sides.} \right) \\
\implies &\dfrac{dy}{dx} & = & y\left(\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\cot x\right) & \\
\implies &\dfrac{dy}{dx} & = & \dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}\left(\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\cot x\right) & \left( \text{Substitute }y=\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}.\right) \\
\end{array} \nonumber \]

Exercise \(\PageIndex{2A}\)

Use logarithmic differentiation to find the derivative of \(y=x^x\).

Hint: Follow the problem solving strategy.

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

Exercise \(\PageIndex{2B}\)

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

Hint: Use the Power Rule (since the exponent \(\pi\) is a constant) and the Chain Rule.

Answer: \(y^{\prime}= \pi (\tan x)^{ \pi −1}\sec^2 x\)

Key Concepts

Logarithmic differentiation allows us to differentiate functions of the form \(y=g(x)^{f(x)}\) or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.

Glossary

logarithmic differentiation: is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly