/*! 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 41 Find the derivative. \(e^{=x^{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 41

Find the derivative. \(e^{=x^{2}} \ln \left(x^{2}\right)\)

Short Answer

Expert verified
The derivative is \( 2x e^{x^2} \ln(x^2) + \frac{2e^{x^2}}{x} \).

Step by step solution

01

Identify the function

The function given is a composition of an exponential function and a logarithmic function: \( f(x) = e^{x^2} \ln(x^2) \).
02

Apply the Product Rule

The function \( f(x) = e^{x^2} \ln(x^2) \) is a product of two functions: \( u = e^{x^2} \) and \( v = \ln(x^2) \). The product rule states that the derivative \( (uv)' \) is \( u'v + uv' \).
03

Differentiate \( e^{x^2} \)

Compute \( u' \). The derivative of \( u = e^{x^2} \) is \( u' = \frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot 2x \) using the chain rule.
04

Differentiate \( \ln(x^2) \)

Compute \( v' \). The derivative of \( v = \ln(x^2) \) is \( v' = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \), again applying the chain rule.
05

Substitute into the Product Rule

Substitute \( u' \), \( u \), \( v \), and \( v' \) into the product rule formula: \( f'(x) = (e^{x^2} \cdot 2x) \ln(x^2) + e^{x^2} \cdot \frac{2}{x} \).
06

Simplify the Expression

Simplify the expression to find the derivative: \( f'(x) = 2x e^{x^2} \ln(x^2) + \frac{2e^{x^2}}{x} \).

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.

Product Rule
The product rule is a fundamental tool in calculus, designed to help us find the derivative of a product of two functions. This is especially important when dealing with complex expressions where functions are multiplied together. The rule is simple, yet powerful, and can be stated as follows:
  • If you have two functions, say \( u(x) \) and \( v(x) \), their product \( uv \) has a derivative given by \( (uv)' = u'v + uv' \).
  • This means you need to first take the derivative of the first function while keeping the second unchanged, and then add the product of the first function with the derivative of the second.
In our problem, we applied the product rule to the functions \( u = e^{x^2} \) and \( v = \ln(x^2) \). By breaking the product apart and handling the derivatives separately, we were able to simplify the differentiation process.
Chain Rule
When derivatives involve composite functions, the chain rule becomes an essential tool. This rule allows us to differentiate the outer function, while factoring in the influence of the inner functions. Essentially, it helps us manage functions nested within each other.
  • The chain rule can be articulated as: if a function \( y \) is a composition of \( u \) and \( g \), such that \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) is \( f'(g(x)) \times g'(x) \).
  • For instance, with \( e^{x^2} \), regard \( e^u \) as the outer function where \( u = x^2 \) and then \( u' = 2x \).
In this exercise, the chain rule was integral in differentiating both \( e^{x^2} \) and \( \ln(x^2) \) effectively. Simply differentiate the outer function first and multiply it by the derivative of the inner function, as we've done to evaluate the respective derivative components.
Exponential Functions
Exponential functions, characterized by the base of the natural logarithm \( e \), are found frequently in calculus due to their unique properties. These functions are expressed in the form \( e^x \) or more generally \( e^{f(x)} \). They grow at a rate proportional to their current value, an attribute that is both fascinating and useful in various fields.
  • The derivative of \( e^x \) itself is quite remarkable as it remains \( e^x \).
  • However, for \( e^{x^2} \), we must apply the chain rule. The derivative becomes \( e^{x^2}\cdot 2x \).
This applies in our problem to find \( u' \) when \( u = e^{x^2} \). Understanding the differentiation of exponential functions, especially using the chain rule, enables us to tackle more complex derivatives involving expressions like \( e^{f(x)} \).

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

Finance Fry \({ }^{15}\) studied 85 developing countries and found that the average percentage growth rate \(y\) in each country was approximated by the equation \(y=g(r)=\) \(-0.033 r^{2}+0.008 r^{3},\) where \(r\) is the real interest rate in the country. Find \(g^{\prime}(r)\), and explain what this term means. Find \(g^{\prime}(-2)\) and \(g^{\prime}(2) .\) Explain in each case what these two numbers mean. What is the significance of the sign of each number?

Take \(x>0\) and find the derivatives. \(\left(x^{3}+2\right) \ln x\)

Elasticity Given the demand curve \(x=10-2 p,\) determine whether the demand is elastic, inelastic, or unit elastic if (a) \(p=2,\) (b) \(p=2.5,\) (c) \(p=3\).

In this section we considered demand \(x\) as a function \(x=f(p)\) of price and then defined the elasticity as \(E=-\frac{p}{x} \frac{d x}{d p} .\) But demand is also a function of other variables. For example, Cotterill and Haller \({ }^{79}\) recently found that the demand \(x\) for the breakfast cereal Shredded Wheat was approximately related to the amount \(a\) of coupons issued by \(x=B a^{0.0229}\), where \(B\) is a constant. Define an elasticity with respect to coupons in a way analogous to what was done for demand with respect to price. Find the elasticity with respect to couponing in this case, and explain in words what it means.

Biology Potter and colleagues \(^{23}\) showed that the percent mortality \(y\) of a New Zealand thrip was approximated by \(\quad y=f(T)=81.12+0.465 T-0.828 T^{2}+0.04 T^{3}\), where \(T\) is the temperature measured in degrees Celsius. Graph on your grapher using a window of dimensions [0,20] by [0,100] a. Estimate the value of the temperature where the tangent line to the curve \(y=f(T)\) is horizontal. b. Check your answer using calculus. (You will need to use the quadratic formula.) c. Did you miss any points in part (a)? d. Is this another example of how your computer or graphing calculator can mislead you? Explain.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Calculus

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.