/*! 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 Identify the "inside function" \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 1

Identify the "inside function" \(u=f(x)\) and the "outside function" \(y=g(u) .\) Then find \(d y / d x\) using the Chain Rule $$ y=(2 x+4)^{3} $$

Short Answer

Expert verified
In the composite function \(y=(2x+4)^3\), the inside function is \(u=f(x)=2x+4\) and the outside function is \(y=g(u)=u^3\). To find the derivative \(\frac{dy}{dx}\) using the Chain Rule, we first find the derivatives of the inside and outside functions: \(\frac{du}{dx}=2\) and \(\frac{dy}{du}=3u^2\). Then, apply the Chain Rule: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (3u^2) \cdot 2 = 6(2x+4)^2\).

Step by step solution

01

Identify the inside function (u=f(x)) and the outside function (y = g(u))

To identify the inside function and the outside function from the composite function \(y = (2x+4)^3\), we can see that the function inside the parentheses can be considered as the inside function, while the exponentiation operation outside the parentheses can be considered as the outside function. Therefore, we have the inside function as \(u = f(x) = 2x + 4\) and the outside function as \(y = g(u) = u^3\).
02

Find the derivative of the inside function (u=f(x)) with respect to x

We need to find the derivative of the inside function (\(u = 2x + 4\)) with respect to x. By using the power rule, we can find the derivative: \[ \frac{du}{dx} = \frac{d}{dx}(2x + 4) = 2 \]
03

Find the derivative of the outside function (y=g(u)) with respect to u

We need to find the derivative of the outside function (\(y = g(u) = u^3\)) with respect to u. Again, by using the power rule, we can find the derivative: \[ \frac{dy}{du} = \frac{d}{du}(u^3) = 3u^2 \]
04

Apply the Chain Rule to find the derivative of y with respect to x

The Chain Rule states that if we have a composite function \(y = g(f(x))\), its derivative with respect to x can be found as: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \] Now, substituting the values of the derivatives \(\frac{dy}{du}\) and \(\frac{du}{dx}\) from Steps 2 and 3, and also the expression for the inside function \(u = 2x + 4\), we can find \(\frac{dy}{dx}\) as follows: \[ \frac{dy}{dx} = (3u^2) \cdot 2 = 3(2x+4)^2 \cdot 2 \] So the final answer is, \[ \frac{dy}{dx} = 6(2x+4)^2 \]

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.

Inside Function
When dealing with a composite function like \( y = (2x+4)^3 \), understanding the inside function is crucial. The inside function is the part that is first addressed in the calculation鈥攗sually located within parentheses or deeply nested.

**In practical terms**:
  • Look for expressions inside parentheses or brackets within a function.
  • It is often the piece of the function where the variable \( x \) appears directly.
In our example, the expression \( 2x+4 \) is the inside function, noted as \( u = f(x) = 2x+4 \). This needs to be solved prior to considering the outer layers of the function.

Knowing the inside function is essential for taking derivatives using the Chain Rule. This understanding comes from recognizing it as the component directly influenced by \( x \), and its derivative will be calculated with respect to \( x \).
Outside Function
Let's turn our attention to the outside function, which encompasses the overall operation that affects the inside function. It is applied to the value of the inside function.

In our example, the entire expression \( (2x+4)^3 \) shows how the inside function \( 2x+4 \) is associated with an exponent. Hence, the outside function is \( y = g(u) = u^3 \).

**Identifying the outside function**:
  • It is usually applied after the inside function.
  • This component can include exponents, trigonometric functions, or any other operation acting on the result of the inside function.
Understanding the arrangement of the outside function lets us comprehend how the composite function behaves entirely. This forms an essential part of deriving the composite function by breaking it down part by part.
Composite Function
Now that we have the inside and outside functions, it's time to see how they come together as a composite function. A composite function integrates these two elements, combining them to form a single, unified function. In our case, the composite function looks like this:\[ y = (2x+4)^3\]

**Thinking in terms of composites**:
  • The inside function \( f(x) \) is evaluated first.
  • Then, the result of \( f(x) \) is plugged into the outside function \( g(u) \), replacing \( u \) with \( f(x) \).
They work in tandem: the function within guides the initial value, and then the external operation is applied to the result. By mastering identifying this structure, the Chain Rule's applications will become much clearer. This makes complex derivative calculations far simpler.
Derivative Calculation
Having broken down the functions, let's move to find the derivative using the powerful Chain Rule. The Chain Rule lets us find the derivative of composite functions by taking derivatives step by step.

**Steps to find the derivative**:
  • First, differentiate the inside function: \( \frac{du}{dx} = 2 \).
  • Next, differentiate the outside function with respect to \( u \): \( \frac{dy}{du} = 3u^2 \).
  • Apply the Chain Rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
Fill in the derivatives:\[ \frac{dy}{dx} = (3u^2) \cdot 2 = 3(2x+4)^2 \cdot 2\] So, finally, the derivative of our composite function is\[ \frac{dy}{dx} = 6(2x+4)^2\]Understanding these steps ensures that you can tackle derivatives of more complicated functions confidently.

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

Middle-Distance Race As they round the corner into the final (straight) stretch of the bell lap of a middle-distance race, the positions of the two leaders of the pack, \(A\) and \(B\), are given by $$s_{A}(t)=0.063 t^{2}+23 t+15 \quad t \geq 0$$ and $$s_{B}(t)=0.298 t^{2}+24 t \quad t \geq 0$$ respectively, where the reference point (origin) is taken to be the point located 300 feet from the finish line and \(s\) is measured in feet and \(t\) in seconds. It is known that one of the two runners, \(A\) and \(B\), was the winner of the race and the other was the runner- up. a. Show that \(B\) won the race. b. At what point from the finish line did \(B\) overtake \(A\) ? c. By what distance \(\operatorname{did} B\) beat \(A\) ? d. What was the speed of each runner as he crossed the finish line?

Find the derivative of the function. $$ y=\left(x^{2}+1\right) \tan ^{-1} x $$

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves. In Exercises \(89-92\), show that the curves with the given equations are orthogonal.$$ x^{2}-y^{2}=3, \quad x y=2 $$

Let g denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b) .\) Suppose that \(g\) is the inverse of a function \(f .\) If \(f(2)=4\) and \(f^{\prime}(2)=3\), find \(g^{\prime}(4) .\)

Find an equation of the tangent line to the given curve at the indicated point. $$ 2 y^{2}-x^{3}-x^{2}=0 ; \quad(1,1) $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

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.