Textbook Question
Derivatives in Differential Form
In Exercises 17–28, find dy.
y = x√(1 − x²)
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.7.25
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.
3 + sin y = y – x³
Verified step by step guidance1
Start by differentiating both sides of the equation 3 + sin(y) = y - x³ with respect to x. Remember that y is a function of x, so use implicit differentiation.
Differentiate the left side: The derivative of 3 is 0, and the derivative of sin(y) with respect to x is cos(y) * (dy/dx) using the chain rule.
Differentiate the right side: The derivative of y with respect to x is dy/dx, and the derivative of -x³ is -3x².
Set the derivatives equal: cos(y) * (dy/dx) = dy/dx - 3x². Solve this equation for dy/dx.
To find the second derivative d²y/dx², differentiate the expression for dy/dx with respect to x again. Use implicit differentiation and the chain rule where necessary, and express the result 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:
6mKey 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 differentiate equations where y is not explicitly solved for in terms of x. Instead of isolating y, we differentiate both sides of the equation with respect to x, applying the chain rule when differentiating y. This allows us to find dy/dx in terms of both x and y, which is essential for solving equations that define y implicitly.
Recommended video:
Guided course
05:14
Finding The Implicit Derivative
First Derivative (dy/dx)
The first derivative, denoted as dy/dx, represents the rate of change of y with respect to x. It provides information about the slope of the tangent line to the curve defined by the equation at any point. In the context of implicit differentiation, finding dy/dx allows us to understand how y changes as x varies, which is crucial for further analysis, such as finding the second derivative.
Recommended video:
07:09The First Derivative Test: Finding Local Extrema
Second Derivative (d²y/dx²)
The second derivative, denoted as d²y/dx², measures the rate of change of the first derivative, indicating how the slope of the tangent line is changing. It provides insights into the concavity of the function and can reveal points of inflection. To find d²y/dx² using implicit differentiation, we differentiate dy/dx again, applying the product and chain rules, and express the result in terms of x and y.
Recommended video:
06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
The eight curve Find the slopes of the curve y⁴ = y² – x² at the two points shown here.
Textbook Question
Differential Estimates of Change
In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.
The change in the lateral surface area S = 2πrh of a right circular cylinder when the height changes from h₀ to h₀ + dh and the radius does not change
Textbook Question
Finding Derivative Values
In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.
f(u) = ((u − 1) / (u + 1))², u = g(x) = (1 / x²) − 1, x = −1
1
views
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
s = (sec t + tan t)⁵
Textbook Question
Free-Fall Applications
Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t² on Mars and s = 11.44t² on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m/sec (about 100 km/h) on each planet?