/*! 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 31 Find the derivatives of the func... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivatives of the functions. $$y=\frac{1}{18}(3 x-2)^{6}+\left(4-\frac{1}{2 x^{2}}\right)^{-1}$$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = (3x-2)^5 - \left(4-\frac{1}{2x^2}\right)^{-2} \cdot \frac{1}{x^3} \).

Step by step solution

01

Understand the Problem

We need to find the derivative of the function \( y = \frac{1}{18}(3x-2)^6 + (4-\frac{1}{2x^2})^{-1} \) with respect to \( x \). This involves using differentiation rules such as the power rule and chain rule.
02

Differentiate the First Term

Differentiate \( y_1 = \frac{1}{18}(3x-2)^6 \) using the chain rule. Define \( u = 3x-2 \), then \( y_1 = \frac{1}{18}u^6 \). The derivative is \( \frac{dy_1}{dx} = \frac{1}{18} \cdot 6u^5 \cdot \frac{du}{dx} \). Since \( \frac{du}{dx} = 3 \), then \( \frac{dy_1}{dx} = \frac{1}{18} \cdot 6 \cdot 3(3x-2)^5 = \frac{1}{18} \cdot 18(3x-2)^5 = (3x-2)^5 \).
03

Differentiate the Second Term

Differentiate \( y_2 = \left(4-\frac{1}{2x^2}\right)^{-1} \). Let \( v = 4-\frac{1}{2x^2} \), then \( y_2 = v^{-1} \). The derivative using the chain rule is \( \frac{dy_2}{dx} = -v^{-2} \cdot \frac{dv}{dx} \). Find \( \frac{dv}{dx} = \frac{d}{dx}\left(4-\frac{1}{2x^2}\right) = \frac{d}{dx}(4) - \frac{d}{dx}\left(\frac{1}{2x^2}\right) = 0 + 2x^{-3} = \frac{1}{x^3} \), so \( \frac{dy_2}{dx} = -\left(4-\frac{1}{2x^2}\right)^{-2} \cdot \frac{1}{x^3} \).
04

Combine the Results

Add the results of the derivatives found in Step 2 and Step 3 to get the overall derivative: \( \frac{dy}{dx} = (3x-2)^5 - \left(4-\frac{1}{2x^2}\right)^{-2} \cdot \frac{1}{x^3} \).

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 91Ó°ÊÓ!

Key Concepts

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

Derivatives
Derivatives are a fundamental concept in calculus. They measure the rate at which a function changes as its input changes. Simply put, the derivative of a function represents its slope or steepness at any given point. When we talk about finding the derivative of a function with respect to a variable, we're looking at how the function changes as that variable changes.

In this exercise, we aimed to find the derivative of the function \( y = \frac{1}{18}(3x-2)^6 + (4-\frac{1}{2x^2})^{-1} \). To do this, we applied different rules, including the chain rule and the power rule, to differentiate each term in the function. By doing so, we could determine the rate of change of the entire function with respect to \( x \).

Understanding derivatives is crucial because they help us analyze and predict the behavior of functions, which is fundamental in fields such as physics, engineering, and economics.
Chain Rule
The chain rule is a powerful tool in calculus used to find the derivative of composite functions. A composite function is one where a function is applied inside another function, like \( f(g(x)) \). The chain rule helps us differentiate such functions by handling the "inside" and "outside" functions separately.

To apply the chain rule, we differentiate the "outside" function first, leaving the inside function unchanged. Then, we multiply that result by the derivative of the "inside" function. This is exactly what we did with the first term of our original function, \( y_1 = \frac{1}{18}(3x-2)^6 \).

:- We let \( u = 3x-2 \). Thus, \( y_1 = \frac{1}{18}u^6 \). - Differentiate the outside function: \( \frac{dy_1}{du} = \frac{1}{18} \cdot 6u^5 \). - Find \( \frac{du}{dx} = 3 \), so multiply: \( \frac{dy_1}{dx} = \frac{1}{18} \cdot 6u^5 \cdot 3 \).

As you can see, the chain rule allows us to simplify the process of finding derivatives of nested functions, like polynomials, exponentials, and trigonometric functions.
Power Rule
The power rule is one of the most straightforward differentiation rules in calculus. It states that if you have a function of the form \( f(x) = x^n \), its derivative is \( f'(x) = n \cdot x^{n-1} \). This rule makes it easy to find the derivative of any function involving powers of \( x \).

In our exercise, the power rule was applied during the differentiation of the term \( \frac{1}{18}(3x-2)^6 \). Here's how it works:
  • Identify the power term: \( (3x-2)^6 \).
  • Differentiate using the power rule: bring down the exponent (6 in this case), and subtract 1 from the exponent.
Using this method:- We multiplied by the constant: \( \frac{1}{18} \cdot 6 \cdot (3x-2)^5 \). - This gave us a key part of the derivative when combined with the chain rule.

The power rule makes dealing with polynomial functions simple and efficient, avoiding more complex calculations often associated with more intricate rules.

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

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

Particle motion The position of a particle moving along a coordinate line is \(s=\sqrt{1+4 t},\) with \(s\) in meters and \(t\) in seconds. Find the particle's velocity and acceleration at \(t=6 \mathrm{sec}\)

Use logarithmic differentiation or the method in Example 7 to find the derivative of \(y\) with respect to the given independent variable. $$y=x^{\ln x}$$

Use the identity $$\cot ^{-1} u=\frac{\pi}{2}-\tan ^{-1} u$$ to derive the formula for the derivative of \(\cot ^{-1} u\) in Table 3.1 from the formula for the derivative of \(\tan ^{-1} u\).

Write a differential formula that estimates the given change in volume or surface area. The change in the lateral surface area \(S=\pi r \sqrt{r^{2}+h^{2}}\) of a right circular cone when the radius changes from \(r_{0}\) to \(r_{0}+d r\) and the height does not change

See all solutions

Recommended explanations on Math Textbooks

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.