/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Use logarithmic differentiation ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use logarithmic differentiation to find the derivative of the function. $$y=\frac{\sin ^{2} x \tan ^{4} x}{\left(x^{2}+1\right)^{2}}$$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = \frac{\sin^2 x \tan^4 x}{(x^2+1)^2} \left( 2\cot x + 4\sec^2 x - \frac{4x}{x^2 + 1} \right) \).

Step by step solution

01

Apply the Natural Logarithm

Take the natural logarithm of both sides of the given function to facilitate differentiation. The original function is \[ y = \frac{\sin^2 x \cdot \tan^4 x}{(x^2+1)^2} \]Taking the log of both sides gives:\[ \ln y = \ln \left( \sin^2 x \right) + \ln \left( \tan^4 x \right) - \ln \left( (x^2 + 1)^2 \right) \]
02

Simplify using Logarithmic Properties

Utilize the properties of logarithms to simplify each term on the right side of the equation:\[ \ln \left( \sin^2 x \right) = 2\ln \left( \sin x \right), \ln \left( \tan^4 x \right) = 4\ln \left( \tan x \right), \ln \left( (x^2 + 1)^2 \right) = 2\ln \left( x^2 + 1 \right) \] Thus, the equation simplifies to:\[ \ln y = 2\ln (\sin x) + 4\ln (\tan x) - 2\ln (x^2 + 1) \]
03

Differentiate Both Sides

Differentiate both sides of the equation with respect to \(x\). Use the chain rule and the derivative of \(\ln u\) which is \(\frac{1}{u} \cdot \frac{du}{dx}\).- Differentiating \( \ln y \) gives:\[ \frac{1}{y} \cdot \frac{dy}{dx} \]- Differentiating \(2\ln (\sin x)\) gives:\[ 2 \cdot \frac{1}{\sin x} \cdot \cos x = 2\cot x \]- Differentiating \(4\ln (\tan x)\) gives:\[ 4 \cdot \frac{1}{\tan x} \cdot \sec^2 x = 4\cdot \frac{1}{\tan x} \cdot \frac{1}{\cos^2 x} = 4\sec^2 x \]- Differentiating \(-2\ln (x^2 + 1)\) gives:\[ -2 \cdot \frac{1}{x^2 + 1} \cdot 2x = -\frac{4x}{x^2 + 1} \]Putting it all together, we have:\[ \frac{1}{y} \cdot \frac{dy}{dx} = 2\cot x + 4\sec^2 x - \frac{4x}{x^2 + 1} \]
04

Solve for \( \frac{dy}{dx} \)

Multiply both sides by \(y\) to isolate \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y \cdot \left(2\cot x + 4\sec^2 x - \frac{4x}{x^2 + 1}\right)\] Substitute back the original function for \(y\):\[ = \frac{\sin^2 x \tan^4 x}{(x^2+1)^2} \cdot \left(2\cot x + 4\sec^2 x - \frac{4x}{x^2 + 1}\right)\]
05

Final Expression for the Derivative

The derivative of the given function using logarithmic differentiation is:\[ \frac{dy}{dx} = \frac{\sin^2 x \tan^4 x}{(x^2+1)^2} \left( 2\cot x + 4\sec^2 x - \frac{4x}{x^2 + 1} \right) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a powerful mathematical tool that simplifies complex functions for differentiation purposes. It's based on the constant \( e \), approximately 2.718, which is the base of natural logarithms. By applying the natural logarithm to both sides of an equation, we can transform potentially unwieldy products and quotients of functions into manageable sums and differences.

For example, when dealing with the function \( y = \frac{\sin^2 x \cdot \tan^4 x}{(x^2+1)^2} \), taking the natural logarithm of both sides allows us to simplify using properties like the log of a product \( \ln(ab) = \ln a + \ln b \) and the log of a quotient \( \ln\frac{a}{b} = \ln a - \ln b \). This leads to easier differentiation by converting multiplicative relationships into additive ones.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. When a function is nested inside another, we use the chain rule to take derivatives. It states that the derivative of a composite function \( f(g(x)) \) is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function: \( f'(g(x)) \cdot g'(x) \).

In our derivation, each natural logarithmic term includes a function within a function, necessitating the chain rule. For instance, differentiating \( \ln(\sin^2 x) \) involves applying the derivative of the natural log to \( \sin^2 x \) and then differentiating \( \sin^2 x \) itself, yielding an overall derivative involving both \( \sin x \) and its derivative \( \cos x \).
Thus, effectively using the chain rule helps break down complex derivatives into manageable parts.
Trigonometric Functions
Trigonometric functions, such as sine \( \sin \), cosine \( \cos \), and tangent \( \tan \), are periodic functions that describe angles and relationships in a right triangle. They play a vital role in calculus, especially in differentiation tasks.

In our example, \( \sin x \) and \( \tan x \) are the two core trigonometric components of the function \( y \). Their derivatives are crucial: \( \sin x \) differentiates to \( \cos x \), and \( \tan x \) differentiates to \( \sec^2 x \). These derivatives form part of the derivative expression in logarithmic form and when simplifying equations. Keeping track of these trigonometric derivatives ensures that transformations and simplifications are correctly handled throughout the differentiation process.
Derivative Calculation
Calculating derivatives of complex functions often requires a combination of several rules and principles to arrive at a correct and simplified result.

In the case of logarithmic differentiation, after using logarithms to simplify the function, we apply differentiation rules, namely the chain rule, and track each component's individual derivative. As demonstrated, the setup transitions from logs to derivatives, creating terms like \(2 \cot x\) and \(4 \sec^2 x\).
The final step involves multiplying each side of the equation by the original function, \( y \), capturing the original structural relationships in the expression. The complete derivative, \( \frac{dy}{dx} = \frac{\sin^2 x \tan^4 x}{(x^2+1)^2} (2\cot x + 4\sec^2 x - \frac{4x}{x^2 + 1}) \), signifies the clean integration of all earlier work steps. This approach provides a clear path from manipulated logarithms to concise, accurate derivatives.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\begin{array}{c}{\text { (a) If } g \text { is differentiable, the Reciprocal Rule says that }} \\ {\frac{d}{d x}\left[\frac{1}{g(x)}\right]=-\frac{g^{\prime}(x)}{[g(x)]^{2}}} \\ {\text { Use the Reciprocal Rule to prove the Reciprocal Rule. }} \\ {\text { (b) Use the Reciprocal Rule to differentiate the function. }}\\\ {y=1 /\left(x^{4}+x^{2}+1\right)} \\ {\text { (c) Use the Reciprocal Rule to verify that the Power Rule }} \\ {\text { is valid for negative integers, that is, }}\\\ \frac{d}{d x}\left(x^{-n}\right)=-n x^{-n-1}\\\ {\text { for all positive integers \(n\). }} \end{array}

Sensitivity of the eye to brightness If \(R\) denotes the reaction of the body to some stimulus of strength \(x\) , the sensitivity \(S\) is defined to be the rate of change of the reaction with respect to \(x .\) A particular example is that when the brightness \(x\) of a light source is increased, the eye reacts by decreasing the area \(R\) of the pupil. The experimental formula $$R=\frac{40+24 x^{0.4}}{1+4 x^{0.4}}$$ has been used to model the dependence of \(R\) on \(x\) when \(R\) is measured in square millimeters and \(x\) is measured in appropriate units of brightness. \(\begin{array}{l}{\text { (a) Find the sensitivity. }} \\ {\text { (b) Illustrate part (a) by graphing both } R \text { and } S \text { as functions }} \\ {\text { of } x \text { . Comment on the values of } R \text { and } S \text { at low levels of }} \\ {\text { brightness. Is this what you would expect? }}\end{array}\)

Bone mass In Example 1.1 .6 we found an expression for the mass \(m\) of a human femur of length \(L\) in terms of the outer radius \(r,\) the inner radius \(r_{\text { in }},\) and their ratio \(k=r_{\text { in }} / r .\) More generally, if the bone density is \(\rho,\) measured in \(\mathrm{g} / \mathrm{cm}^{3},\) then bone mass is given by the equation \(m=\pi r^{2} L\left[\rho-(\rho-1) k^{2}\right]\) It may happen that both \(\rho\) and \(k\) change with age, \(t\) (a) If \(\rho\) changes during aging, find an expression for the rate of change of \(m\) with respect to \(t\) . (b) If \(k\) changes during aging, find an expression for the rate of change of \(m\) with respect to \(t\) .

Find the derivative of the function. Simplify where possible. $$f(x)=x \ln (\arctan x)$$

42\. Predator-prey dynamics In Chapter 7 we study a model for the population sizes of a predator and its prey species. If \(u(t)\) and \(v(t)\) denote the prey and predator population sizes at time \(t,\) an equation relating the two is $$v e^{-t} u^{a} e^{-a u}=c$$ where \(c\) and \(\alpha\) are positive constants. Use logarithmic differentiation to obtain an equation relating the relative (per capita) rate of change of predator (that is, \(v^{\prime} / v )\) to that of prey (that is, \(u^{\prime} / u ) .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.