/*! 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 56 Find the second derivative of ea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 56

Find the second derivative of each function. $$ f(x)=e^{-x^{6} / 6} $$

Short Answer

Expert verified
The second derivative is \( f''(x) = e^{-x^6 / 6} (x^{10} - 5x^4) \).

Step by step solution

01

Find the First Derivative

To find the first derivative of the function, we use the chain rule. The function is given by \( f(x) = e^{-x^6 / 6} \). Let \( u = -x^6 / 6 \), so \( f(x) = e^u \). Then, \( \frac{du}{dx} = -x^5 \) and the first derivative is: \( f'(x) = e^u \cdot \frac{du}{dx} = e^{-x^6 / 6} \cdot (-x^5) \). Therefore, \( f'(x) = -x^5 e^{-x^6 / 6} \).
02

Find the Second Derivative

To find the second derivative, we need to differentiate the first derivative \( f'(x) = -x^5 e^{-x^6 / 6} \). Again, we use the product rule and chain rule. For \( u = -x^6 / 6 \) and \( v = -x^5 \), we have: \( \frac{du}{dx} = -x^5 \) and \( \frac{dv}{dx} = -5x^4 \). The product rule gives: \( f''(x) = \left( \frac{dv}{dx} \cdot e^u + v \cdot e^u \cdot \frac{du}{dx} \right) \). Substitute to get: \( f''(x) = (-5x^4)e^{-x^6 / 6} + (-x^5)(e^{-x^6 / 6})(-x^5) \). Simplifying, \( f''(x) = -5x^4 e^{-x^6 / 6} + x^{10} e^{-x^6 / 6} \). Therefore, \( f''(x) = e^{-x^6 / 6} (x^{10} - 5x^4) \).

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 differentiation technique used when dealing with composite functions. Essentially, it helps us find the derivative of a function that is nested within another function. For instance, consider our function \( f(x) = e^{-x^6 / 6} \). Here, the exponent \(-x^6 / 6\) can be viewed as a function of \(x\), let's call it \(u\).
  • In this scenario, \( f(x) = e^u \) and \( u = -x^6 / 6 \).
  • First, we differentiate \( u \) with respect to \( x \), giving us \( \frac{du}{dx} = -x^5 \).
  • Then, the chain rule states that the derivative of \( f \) with respect to \( x \), is given by \( \frac{df}{du} \cdot \frac{du}{dx} \).
Applying this, we find the derivative of \( e^u \) as \( e^u \), overall calculating the first derivative as \( f'(x) = e^{-x^6 / 6} \cdot (-x^5) \). The chain rule simplifies the process of tackling nested functions, making it indispensable in calculus.
Product Rule
When you need to differentiate a product of two functions, the product rule comes into play. It's particularly handy when each part of the expression can itself be a product of different functions. Let's look at the first derivative we found: \( f'(x) = -x^5 e^{-x^6 / 6} \).
  • Here, we have two functions of \(x\): \( v = -x^5 \) and \( u = e^{-x^6 / 6} \).
  • The product rule tells us that the derivative of the product is \( \frac{dv}{dx} \cdot u + v \cdot \frac{du}{dx} \).
  • First, find \( \frac{dv}{dx} = -5x^4 \).
Now, we combine these results with the chain rule (for \( u \)) to differentiate \( u \). Applying the product rule ensures we account for both components correctly, leading us to \( f''(x) = e^{-x^6 / 6} (x^{10} - 5x^4) \), neatly giving us the second derivative.
Differentiation Techniques
Differentiation is the process of finding the derivative of a function, which is essential for understanding how functions change. The techniques used in differentiation, like the chain rule and product rule, serve different purposes depending on the problem.
  • Each technique is structured to handle specific forms of function equally smoothly.
  • The chain rule is best for nested functions, where we unravel the layers step by step.
  • The product rule is specifically tailored for cases where two functions multiply.
These tools combined can tackle even the most complex differentiation problems efficiently. Understanding each rule's purpose deepens comprehension and ensures accurate calculations, especially when calculating higher-order derivatives like the second derivative. By layering products with one method and nested functions with another, students can approach differentiation with confidence.

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

Choose the correct answer: \(\frac{d}{d x} e^{5}=\) a. \(5 e^{4}\) b. \(e^{5}\) c. 0

A supply function \(S(p)\) gives the total amount of a product that producers are willing to supply at a given price \(p\). The elasticity of supply is defined as $$ E_{s}(p)=\frac{p \cdot S^{\prime}(p)}{S(p)} $$ Elasticity of supply measures the relative increase in supply resulting from a small relative increase in price. It is less useful than elasticity of demand, however, since it is not related to total revenue. Use the preceding formula to find the elasticity of supply for a supply function of the form \(S(p)=a p^{n}\), where \(a\) and \(n\) are positive constants.

For each function: a. Find the relative rate of change. b. Evaluate the relative rate of change at the given value(s) of \(t\). $$ f(t)=100 \sqrt[3]{t+2}, \quad t=8 $$

For each demand function \(D(p)\) : a. Find the elasticity of demand \(E(p)\). b. Determine whether the demand is elastic, inelastic, or unit-elastic at the given price \(p\). $$ D(p)=4000 e^{-0.01 p}, \quad p=200 $$

For each function: a. Find \(f^{\prime}(x)\). b. Evaluate the given expression and approximate it to three decimal places. \(f(x)=\frac{e^{x}}{x}\), find and approximate \(f^{\prime}(3) .\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.