/*! 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 20 Find \(d y / d x .\) Assume \(a,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 20

Find \(d y / d x .\) Assume \(a, b, c\) are constants. $$x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$$

Short Answer

Expert verified
The derivative \(\frac{dy}{dx}\) is \(-\left(\frac{y}{x}\right)^{1/3}\).

Step by step solution

01

Define Implicit Differentiation

The equation involves both x and y, so we'll use implicit differentiation to find \(\frac{dy}{dx}\). This means we differentiate both sides of the equation with respect to x, treating y as a function of x.
02

Differentiate Both Sides of the Equation

Differentiate the left side: The derivative of \(x^{2/3}\) with respect to x is \(\frac{2}{3}x^{-1/3}\). The derivative of \(y^{2/3}\) with respect to x is \(\frac{2}{3}y^{-1/3}\cdot\frac{dy}{dx}\) by chain rule. The right side, a constant \(a^{2/3}\), has a derivative of zero.
03

Write the Derivative Equation

After differentiating, the equation becomes: \[\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\cdot \frac{dy}{dx} = 0\]
04

Solve for \(\frac{dy}{dx}\)

Rearrange the equation to solve for \(\frac{dy}{dx}\): Subtract \(\frac{2}{3}x^{-1/3}\) from both sides: \[\frac{2}{3}y^{-1/3}\cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3}\] Divide both sides by \(\frac{2}{3}y^{-1/3}\): \[\frac{dy}{dx} = - \frac{\frac{2}{3}x^{-1/3}}{\frac{2}{3}y^{-1/3}} = -\left(\frac{x}{y}\right)^{-1/3}\]
05

Simplify the Expression

Simplify the result: Since \(\left(\frac{x}{y}\right)^{-1/3} = \left(\frac{y}{x}\right)^{1/3}\), the solution is: \[\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3}\]

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.

Chain Rule
The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. In this exercise, we deal with expressions where variables are raised to a power, like \[ y^{2/3} \] which is dependent on \( y \), itself a function of \( x \) in implicit contexts.

To apply the Chain Rule, we view \( y^{2/3} \) as a composition of the power function and \( y(x) \). When differentiating, we proceed as follows:
  • Differentiating the outer function, treating the inside as a fixed object, gives us \( \frac{2}{3}y^{-1/3} \).
  • The inner derivative of \( y(x) \) concerning \( x \) is \( \frac{dy}{dx} \).
Combining these results, the derivative of \( y^{2/3} \) is \( \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} \).

This approach is pivotal when dealing with implicitly defined functions, allowing us to find derivatives even when the function isn't expressed explicitly.
Differentiation
Differentiation is a core concept in calculus, representing the process of finding the rate at which a function changes at any point. It is fundamental to understanding changes in quantities in various fields.

In implicit differentiation, like in our exercise, both \( x \) and \( y \) are treated as variables that depend on each other. The process requires us to:
  • Differentiate the given equation with respect to one variable, typically \( x \).
  • Use the Chain Rule when appropriate, especially when differentiating terms involving \( y \).
  • Solve the resulting equation to express \( \frac{dy}{dx} \), representing the slope of the tangent line to the curve at any given point.
This calculus technique is essential because it allows us to understand complex relationships within functions that are not easily expressed in a single variable form.
Implicit Function Theorem
The Implicit Function Theorem is a significant result in mathematical analysis that relates functions implicitly defined by an equation. This theorem assures us that under certain conditions, a relation \( F(x, y) = 0 \) can be thought of as defining \( y \) as a function of \( x \).

In our exercise, \( x^{2/3} + y^{2/3} = a^{2/3} \), the theorem implies that locally around a point, \( y \) can be expressed as a function of \( x \) as long as certain conditions are met (such as \( F_y eq 0 \)).
  • This is crucial because it allows us to use implicit differentiation with confidence.
  • It also helps us understand that implicit differentiation not only gives the derivative but confirms the existence of \( y \) as a function of \( x \) at least locally.
By converting implicit equations into forms more amenable to differentiation, we can effectively analyze and solve complex calculus problems.

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

find a formula for the error \(E(x)\) in the tangent line approximation to the function near \(x=a\). Using a table of values for \(E(x) /(x-a)\) near \(x=a\), find a value of \(k\) such that \(E(x) /(x-a) \approx k(x-a) .\) Check that, approximately, \(k=f^{\prime \prime}(a) / 2\) and that \(E(x) \approx\left(f^{\prime \prime}(a) / 2\right)(x-a)^{2}\) $$f(x)=\ln x, a=1$$

Give an example of: A hyperbolic function which is concave up.

Suppose that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\) and that \(m \leq f^{\prime}(x) \leq M\) on \((a, b) .\) Use the Racetrack Principle to prove that \(f(x)-f(a) \leq M(x-a)\) for all \(x\) in \([a, b],\) and that \(m(x-a) \leq f(x)-f(a)\) for all \(x\) in \([a, b] .\) Conclude that \(m \leq(f(b)-f(a)) /(b-a) \leq\) \(M .\) This is called the Mean Value Inequality. In words: If the instantaneous rate of change of \(f\) is between \(m\) and \(M\) on an interval, so is the average rate of change of \(f\) over the interval.

Let \(f\) be a differentiable function and let \(L\) be the linear function \(L(x)=f(a)+k(x-a)\) for some constant a. Decide whether the following statements are true or false for all constants \(k\). Explain your answer. (a) \(L\) is the local linearization for \(f\) near \(x=a\) (b) If \(\lim _{x \rightarrow a}(f(x)-L(x))=0,\) then \(L\) is the local linearization for \(f\) near \(x=a\)

In Problems \(96-97,\) give an example of: A function involving a sine and an exponential that requires the chain rule to differentiate.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Calculus

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.