Textbook Question
Assume that y = 5x and dx/dt = 2. Find dy/dt
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.7.23
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.
2√y = x – y
Verified step by step guidance1
Start by differentiating both sides of the equation 2√y = x - y with respect to x. Remember that y is a function of x, so use implicit differentiation.
Differentiate the left side: The derivative of 2√y with respect to x is 2 * (1/2) * y^(-1/2) * (dy/dx) = (dy/dx) / √y.
Differentiate the right side: The derivative of x is 1, and the derivative of -y is -dy/dx.
Set the derivatives equal: (dy/dx) / √y = 1 - dy/dx. Solve this equation for dy/dx in terms of x and y.
To find the second derivative d²y/dx², differentiate the expression for dy/dx with respect to x again, using implicit differentiation. Substitute dy/dx from the previous step into this new expression to express d²y/dx² in terms of x and y.
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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, often x, while treating other variables as implicit functions of x. This method is essential when dealing with equations where y is not isolated.
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Finding The Implicit Derivative
First Derivative (dy/dx)
The first derivative, dy/dx, represents the rate of change of y with respect to x. It is found by applying implicit differentiation to the given equation. In this context, it involves differentiating each term of the equation with respect to x and solving for dy/dx, which provides the slope of the tangent line to the curve at any point (x, y).
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07:09The First Derivative Test: Finding Local Extrema
Second Derivative (d²y/dx²)
The second derivative, d²y/dx², measures the rate of change of the first derivative, providing information about the concavity of the function. To find it, differentiate the expression for dy/dx implicitly with respect to x again. This step requires applying the chain rule and product rule, and the result helps determine the behavior of the function, such as identifying points of inflection.
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06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.
lim (x → 1) (x⁵⁰ − 1) / (x − 1)
Textbook Question
Derivatives in Differential Form
In Exercises 17–28, find dy.
y = sin(5√x)
Textbook Question
In Exercises 41–58, find dy/dt.
y = (1 + cos(2t))⁻⁴
Textbook Question
Find the first and second derivatives of the functions in Exercises 33–38.
s = (t² + 5t − 1) / t²
Textbook Question
In Exercises 43–50, find by implicit differentiation.
y² = x .
x + 1