Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.7.22

Chapter 3, Problem 3.7.22

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

y² – 2x = 1 – 2y

Verified step by step guidance

Start by differentiating both sides of the equation y² - 2x = 1 - 2y with respect to x. Remember to apply implicit differentiation, which means treating y as a function of x.

Differentiate the left side: The derivative of y² with respect to x is 2y(dy/dx) using the chain rule, and the derivative of -2x is -2.

Differentiate the right side: The derivative of 1 is 0, and the derivative of -2y with respect to x is -2(dy/dx) using the chain rule.

Set the derivatives equal: 2y(dy/dx) - 2 = -2(dy/dx). Solve this equation for dy/dx to find the first derivative.

To find the second derivative d²y/dx², differentiate the expression for dy/dx with respect to x again. Use implicit differentiation and substitute dy/dx from the previous step where necessary. Simplify the expression to write d²y/dx² in terms of x and y.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

12m

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. It involves differentiating both sides of an equation with respect to a variable, typically x, while treating other variables as implicit functions of x. This method is essential for equations where y is not isolated.

First Derivative (dy/dx)

The first derivative, dy/dx, represents the rate of change of y with respect to x. In the context of implicit differentiation, it involves applying the chain rule to differentiate terms involving y, treating y as a function of x. Solving for dy/dx is crucial for understanding the slope of the curve defined by the implicit equation.

Second Derivative (d²y/dx²)

The second derivative, d²y/dx², provides information about the curvature or concavity of the graph of a function. After finding dy/dx using implicit differentiation, d²y/dx² is obtained by differentiating dy/dx again with respect to x, applying the chain rule and product rule as necessary. This derivative helps in analyzing the behavior of the function's graph, such as identifying points of inflection.

Recommended video:

06:02

The Second Derivative Test: Finding Local Extrema

Related Practice

Textbook Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

x²/³ + y²/³ = 1

220

views

Textbook Question

Find the first and second derivatives of the functions in Exercises 33–38.

w = ((1 + 3z) / 3z) (3 − z)

196

views

Textbook Question

A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?

235

views

Textbook Question

Find the derivatives of the functions in Exercises 1–42.

_____

𝔂 = / x² + x

√ x²

360

views

Textbook Question

Derivative Calculations

In Exercises 1–12, find the first and second derivatives.