/*! 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 1 Given \(y=f(u)\) and \(u=g(x),\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 1

Given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(d y / d x=f^{\prime}(g(x)) g^{\prime}(x)\). $$y=6 u-9, \quad u=(1 / 2) x^{4}$$

Short Answer

Expert verified
\(\frac{dy}{dx} = 12x^3\)

Step by step solution

01

Identify the Functions

From the given information, we have two functions: \(y = f(u) = 6u - 9\) and \(u = g(x) = \frac{1}{2}x^4\). We need to find the derivative \(\frac{dy}{dx}\) using the chain rule.
02

Differentiate y with respect to u

To find \(\frac{dy}{du}\), differentiate \(f(u) = 6u - 9\) with respect to \(u\). This results in \(\frac{dy}{du} = f'(u) = 6\).
03

Differentiate u with respect to x

To find \(\frac{du}{dx}\), differentiate \(g(x) = \frac{1}{2}x^4\) with respect to \(x\). This gives us \(\frac{du}{dx} = g'(x) = 2x^3\).
04

Apply the Chain Rule

The chain rule states \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). Substituting the values from previous steps, we have: \[\frac{dy}{dx} = 6 \times 2x^3 = 12x^3\].

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.

Differentiation
Differentiation is a key concept in calculus that involves finding how a function changes at any point. It is the process by which we calculate the rate at which something changes, often times with respect to another variable.
  • This process is central to physics, engineering, economics, and other fields where change is monitored.
  • In mathematics, it's like asking how the output of a function shifts when there's a small tweak in the input.
Ultimately, differentiation brings us to the derivative, another crucial concept.
But before diving into that, remember that understanding differentiation is all about grasping the process of change across a function.
Derivative
A derivative is the crux of calculus. It shows the rate of change of a function as its input changes. In simple terms, it's like finding the slope of a function at any point along its curve.
  • The derivative can tell us how fast something is speeding up or slowing down.
  • It's written as \( f'(x) \) for a function \( f(x) \).
In the original exercise, the derivative \( \frac{dy}{du} \) means the change in \( y \) as \( u \) changes. And \( \frac{du}{dx} \) is the change in \( u \) when \( x \) changes.
Calculating derivatives helps you understand the behavior of functions, and by mastering them, you unlock many possibilities in problem-solving. When integrated with other rules, like the Chain Rule, it showcases the dynamic nature of calculus.
Composite Function
A composite function is created when one function is plugged into another. It's like stacking functions to create something new. For instance, if you have a function \( f(u) \) and another \( u = g(x) \), the composite function becomes \( f(g(x)) \).
  • This technique is handy for breaking down complex tasks into manageable parts.
  • Understanding composite functions is crucial, particularly in advanced mathematical scenarios.
In our problem, we had \( y = f(u) = 6u - 9 \) and \( u = g(x) = \frac{1}{2}x^4 \). By composing these, we could use the chain rule to differentiate easily.
Composite functions require critical thinking to visualize how combining functions affects the output, especially when applying calculus rules like the Chain Rule.

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

The effect of flight maneuvers on the heart The amount of work done by the heart's main pumping chamber, the left ventricle, is given by the equation $$ W=P V+\frac{V \delta v^{2}}{2 g} $$ where \(W\) is the work per unit time, \(P\) is the average blood pressure, \(V\) is the volume of blood pumped out during the unit of time, \(\delta\) ("delta") is the weight density of the blood, \(v\) is the average velocity of the exiting blood, and \(g\) is the acceleration of gravity. When \(P, V, \delta,\) and \(v\) remain constant, \(W\) becomes a function of \(g\), and the equation takes the simplified form $$ W=a+\frac{b}{g}(a, b \text { constant }) $$ As a member of NASA's medical team, you want to know how sensitive \(W\) is to apparent changes in \(g\) caused by flight maneuvers, and this depends on the initial value of \(g\). As part of your investigation, you decide to compare the effect on \(W\) of a given change \(d g\) on the moon, where \(g=5.2 \mathrm{ft} / \mathrm{sec}^{2}\), with the effect the same change \(d_{8}\) would have on Earth, where \(g=32 \mathrm{ft} / \mathrm{sec}^{2}\). Use the simplified equation above to find the ratio of \(d W_{\text {moon }}\) to \(d W_{\text {Barth }}\)

Find \(d y / d x\) $$\ln x y=e^{x+y}$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{-x}$$

Use a CAS to perform the following steps. a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\) c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x \sqrt{1+2 y}+y=x^{2}, \quad P(1,0)$$

The linearization of \(2^{x}\) a. Find the linearization of \(f(x)=2^{x}\) at \(x=0 .\) Then round its coefficients to two decimal places. b. Graph the linearization and function together for \(-3 \leq x \leq 3\) and \(-1 \leq x \leq 1\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

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.