Textbook Question
In Exercises 41–58, find dy/dt.
y = ((3t − 4) / (5t + 2))⁻⁵
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 3.7.10
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
xy = cot(xy)
Verified step by step guidance1
Start by recognizing that the equation xy = cot(xy) involves both x and y, and y is a function of x. This means we need to use implicit differentiation.
Differentiate both sides of the equation with respect to x. For the left side, use the product rule: d/dx[xy] = x(dy/dx) + y.
For the right side, differentiate cot(xy) with respect to x. Use the chain rule: d/dx[cot(u)] = -csc^2(u) * du/dx, where u = xy. So, differentiate xy with respect to x, which gives: d/dx[xy] = x(dy/dx) + y.
Set the derivatives equal: x(dy/dx) + y = -csc^2(xy) * (x(dy/dx) + y).
Solve for dy/dx by isolating it on one side of the equation. This involves algebraic manipulation to factor out dy/dx and simplify the expression.
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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 like xy = cot(xy), where y is not isolated.
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05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In implicit differentiation, the chain rule is crucial when differentiating terms involving products or compositions of x and y, such as cot(xy).
Recommended video:
05:02Intro to the Chain Rule
Trigonometric Derivatives
Understanding the derivatives of trigonometric functions is vital for solving problems involving implicit differentiation. For example, the derivative of cot(u) with respect to u is -csc^2(u). When differentiating an equation like xy = cot(xy), knowing these derivatives allows you to correctly apply the chain rule and find dy/dx, especially when trigonometric functions are involved.
Recommended video:
06:35Derivatives of Other Inverse Trigonometric Functions
Related Practice
Textbook Question
In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.
y = (1 / x³), (−2, −1/8)
Textbook Question
Theory and Examples
The equations in Exercises 49 and 50 give the position s = f(t) of a body moving on a coordinate line (s in meters, t in seconds). Find the body’s velocity, speed, acceleration, and jerk at time t = π/4 sec.
s = 2 − 2 sin t
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = x⁵ - 0.125x² + 0.25x
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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 volume V = x³ of a cube when the edge lengths change from x₀ to x₀ + dx
Textbook Question
Graphs
Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d).
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