3.8: Differentiation Techniques - Logarithmic Differentiation

Learning Objectives

• Use logarithmic differentiation to determine the derivative of products and ratios of functions.
• Use logarithmic differentiation to determine the derivative of functions to functional powers.

So far, we have learned how to differentiate various 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.\[ \begin{array}{rrclcl}
& \ln{(y)} & = & \ln{(2x^4+1)^{\tan{(x)}}} & \quad & \left( \text{taking the natural logarithm of both sides} \right) \\[16pt]
\implies & \ln{(y)} & = & \tan{(x)} \ln{(2x^4+1)} & \quad & \left( \text{Properties of Logarithms} \right) \\[16pt]
\implies & \dfrac{1}{y}\dfrac{dy}{dx} & = & \sec^2{(x)} \ln{(2x^4+1)} + \dfrac{8x^3}{2x^4+1} \cdot \tan{(x)} & \quad & \left( \text{implicitly differentiating} \right) \\[16pt]
\implies & \dfrac{dy}{dx} & = & y \cdot \left(\sec^2{(x)} \ln{(2x^4+1)} + \dfrac{8x^3}{2x^4+1} \cdot \tan{(x)} \right) & & \\[16pt]
\implies & \dfrac{dy}{dx} & = & (2x^4+1)^{\tan{(x)}} \left( \sec^2{(x)} \ln{(2x^4+1)} + \dfrac{8x^3}{2x^4+1} \cdot \tan{(x)} \right) & \quad & \left( \text{substituting }y=(2x^4+1)^{\tan{(x)}}. \right) \\[16pt]
\end{array} \nonumber \]

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 uses the Properties of Logarithms and the differentiation rules introduced in this chapter.\[ \begin{array}{rrclcl}
& \ln y & = & \ln\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x} & \quad & \left( \text{taking the natural logarithm of both sides} \right) \\[16pt]
\implies & \ln y & = & \ln x+\frac{1}{2}\ln(2x+1)−x\ln e−3\ln \sin x & \quad & \left( \text{Properties of Logarithms} \right) \\[16pt]
\implies & \dfrac{1}{y}\dfrac{dy}{dx} & = & \dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\dfrac{\cos x}{\sin x} & \quad & \left( \text{implicitly differentiating} \right) \\[16pt]
\implies & \dfrac{dy}{dx} & = & y\left(\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\cot x\right) & & \\[16pt]
\implies & \dfrac{dy}{dx} & = & \dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}\left(\dfrac{1}{x}+\dfrac{1}{2x+1}−1−3\cot x\right) & \quad & \left( \text{substituting }y=\dfrac{x\sqrt{2x+1}}{e^x\sin^3 x}. \right) \\[16pt]
\end{array} \nonumber \]

Checkpoint \(\PageIndex{2A}\)

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

Answer

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

Checkpoint \(\PageIndex{2B}\)

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

Answer

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

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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.

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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