/*! 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 117 Use logarithmic differentiation ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 117

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sin x^{x}$$

Short Answer

Expert verified
The derivative is \(x^x(\ln x + 1)\).

Step by step solution

01

Apply Natural Logarithm

Take the natural logarithm on both sides of the equation to make differentiation manageable. This gives us: \( \ln(y) = \ln(\sin(x^x)) \).
02

Differentiate Both Sides

Apply the chain rule to differentiate the left side and the properties of logarithms for the right side. The left becomes \( \frac{1}{y} \frac{dy}{dx} \) and the right becomes \( \csc(x^x) \cdot (x^x)' \).
03

Differentiate the Inner Function

Find the derivative of \(x^x\). First, rewrite \(x^x = e^{x \ln x}\). Differentiating using the chain rule, we have \( (x^x)' = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1) \).
04

Use Product Rule for the Sine Argument

Given that \( \ln(\sin(x^x)) = \ln(\sin u) \) where \( u = x^x \), use the chain rule for differentiation: \( \cot u \cdot u' = \csc(x^x) \cdot x^x (\ln x + 1) \).
05

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

Multiplay both sides by \(y\) to express \(\frac{dy}{dx}\): \( \frac{dy}{dx} = y \cdot \csc(x^x) \cdot x^x (\ln x + 1) \).
06

Substitute Original \(y\)

Since \(y = \sin(x^x)\), substitute back into the derived expression: \( \frac{dy}{dx} = \sin(x^x) \cdot \csc(x^x) \cdot x^x (\ln x + 1) \).
07

Simplify the Derivative Expression

Simplify the expression by recognizing that \( \csc(x^x) = \frac{1}{\sin(x^x)} \), which gives \( \frac{dy}{dx} = x^x (\ln x + 1) \).

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 91影视!

Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus that allows us to differentiate compositions of functions. When you have a function within another function, the chain rule helps you differentiate it by breaking it down into simpler parts.

To apply the chain rule, follow this pattern:
  • Identify the outer and inner functions. The outer function is operated on first, while the inner function is operated on second.
  • Differentiate the outer function with respect to the inner function.
  • Multiply it by the derivative of the inner function.
For example, if you have a function like \(f(g(x))\), the derivative using the chain rule is \(f'(g(x)) \cdot g'(x)\).

In our exercise, we applied the chain rule to differentiate \(x^x\) hidden inside \((\sin(x^x))\). It enabled us to unravel the layers to solve \(\ln(y) = \ln(\sin(x^x))\).
Derivative of Exponential Functions
Exponential functions often appear in calculus problems and have distinct differentiation rules. The basic rule for differentiating expressions like \(e^{u}\) is that the derivative is itself multiplied by the derivative of the exponent.

Exponential differentiation is crucial in our problem because when addressing \(x^x\), we reformulate it using the transformation \(x^x = e^{x \ln x}\). This allows us to take advantage of the properties of exponential functions to find the derivative efficiently.

The key steps include:
  • Rewrite the power function as an exponential through the natural exponential constant.
  • Differentiate using the rule for exponential functions, which in this case, involves applying the chain rule again for the exponent \(x \ln x\).
  • Multiply back by the original term if needed to revert to the standard power form.
Understanding and mastering these techniques are essential for handling complex exponential differentiations.
Trigonometric Functions
Trigonometric functions are central to calculus, and knowing how to differentiate them is a common task. Functions like sine, cosine, tangent, and their reciprocals have specific differentiation rules.

To differentiate \(\sin(x)\), you use the familiar rule that gives \(\cos(x)\). However, when dealing with \(\sin(x^x)\), you apply the chain rule since it鈥檚 a composition of \(\sin\) and \(x^x\).

Here are key points to remember:
  • The derivative of \sin(u)\ is \operatorname{cos}(u) \cdot u'\, where \(u\) itself can be a complex function.
  • Use reciprocal identities whenever \(\csc\), \(\sec\), or other trigonometric reciprocals appear.
  • Combining the initial differentiation step with applying these identities simplifies finding derivatives of composite trigonometric forms.
In our solution, we demonstrated the use of trigonometric identities and differentiation rules to express and simplify the derivative expressions.
Differentiation Techniques
Differentiation techniques encompass a range of rules and strategies to tackle derivatives in various functions. From simple polynomials to intricate combinations involving exponentials and trigonometric functions, these methods are vital.

Some differentiation techniques include:
  • Logarithmic differentiation: This involves taking the natural log of complex functions to simplify the differentiation process.
  • The product rule: Used when differentiating a product of two functions, \( (uv)' = u'v + uv' \).
  • Quotient Rule: Applies to the division of functions, defined by \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
Logarithmic differentiation was particularly useful in our problem as it simplified dealing with the power function embedded in the sine. This technique is particularly valuable when dealing with products, quotients, or powers of functions that don't differentiate neatly through basic applications.

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

Hanging cables Imagine a cable, like a telephone line or TV cable, strung from one support to another and hanging freely. The cable's weight per unit length is a constant \(w\) and the horizontal tension at its lowest point is a vector of length \(H\). If we choose a coordinate system for the plane of the cable in which the \(x\) -axis is horizontal, the force of gravity is straight down, the positive \(y\) -axis points straight up, and the lowest point of the cable lies at the point \(y=H / w\) on the \(y\) -axis (see accompanying figure), then it can be shown that the cable lies along the graph of the hyperbolic cosine Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning "chain." a. Let \(P(x, y)\) denote an arbitrary point on the cable. The next accompanying figure displays the tension at \(P\) as a vector of length (magnitude) \(T\), as well as the tension \(H\) at the lowest point \(A .\) Show that the cable's slope at \(P\) is b. Using the result from part (a) and the fact that the horizontal tension at \(P\) must equal \(H\) (the cable is not moving), show that \(T=w y .\) Hence, the magnitude of the tension at \(P(x, y)\) is exactly equal to the weight of \(y\) units of cable.

Suppose that a cup of soup cooled from \(90^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) after \(10 \mathrm{min}\) in a room whose temperature was \(20^{\circ} \mathrm{C}\). Use Newton's Law of Cooling to answer the following questions. a. How much longer would it take the soup to cool to \(35^{\circ} \mathrm{C} ?\) b. Instead of being left to stand in the room, the cup of \(90^{\circ} \mathrm{C}\) soup is put in a freezer whose temperature is \(-15^{\circ} \mathrm{C}\). How long will it take the soup to cool from \(90^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C} ?\)

Show that if a function \(f\) is defined on an interval symmetric about the origin (so that \(f\) is defined at \(-x\) whenever it is defined at \(x\) ), then $$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$ Then show that \((f(x)+f(-x)) / 2\) is even and that \((f(x)-\) \(f(-x)) / 2\) is odd.

Use your graphing utility. Graph \(f(x)=\tan ^{-1} x\) together with its first two derivatives. Comment on the behavior of \(f\) and the shape of its graph in relation to the signs and values of \(f^{\prime}\) and \(f^{\prime \prime}\).

Which one is correct, and which one is wrong? Give reasons for your answers. a. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\lim _{x \rightarrow 3} \frac{1}{2 x}=\frac{1}{6}\) b. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\frac{0}{6}=0\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation
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.