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)
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 3.6.48
In Exercises 41–58, find dy/dt.
y = ((3t − 4) / (5t + 2))⁻⁵
Verified step by step guidance1
Identify the function y = ((3t - 4) / (5t + 2))^(-5) and recognize that it is a composition of functions, which suggests the use of the chain rule for differentiation.
Apply the chain rule: If y = u^n, where u is a function of t, then dy/dt = n * u^(n-1) * du/dt. Here, u = (3t - 4) / (5t + 2) and n = -5.
Differentiate u = (3t - 4) / (5t + 2) with respect to t using the quotient rule: If u = v/w, then du/dt = (w * dv/dt - v * dw/dt) / w^2. Here, v = 3t - 4 and w = 5t + 2.
Calculate dv/dt and dw/dt: dv/dt = 3 and dw/dt = 5. Substitute these into the quotient rule to find du/dt.
Substitute u, n, and du/dt into the chain rule expression to find dy/dt. Simplify the expression to complete the differentiation process.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chain Rule
The chain rule is a fundamental differentiation technique used when dealing with composite functions. It states that the derivative of a composite function f(g(x)) is f'(g(x)) * g'(x). In this problem, the chain rule helps differentiate the outer function raised to a power and the inner rational function separately.
Recommended video:
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Intro to the Chain Rule
Quotient Rule
The quotient rule is used to differentiate functions that are expressed as a quotient of two functions, u(t)/v(t). It states that the derivative is (v(t)u'(t) - u(t)v'(t)) / (v(t))². This rule is essential for finding the derivative of the inner function (3t - 4)/(5t + 2) in the given problem.
Recommended video:
06:43The Quotient Rule
Negative Exponent Rule
The negative exponent rule states that a term with a negative exponent can be rewritten as the reciprocal of the term with a positive exponent. In this problem, y = ((3t − 4) / (5t + 2))⁻⁵ implies that the function is the reciprocal of ((3t − 4) / (5t + 2)) raised to the fifth power, which affects how the derivative is calculated.
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Guided course
6:37Zero and Negative Rules
Related Practice
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
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
Normal lines parallel to a line Find the normal lines to the curve xy + 2x – y = 0 that are parallel to the line 2x + y = 0.
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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Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
xy = cot(xy)