Textbook Question
Find all points on the curve y = tan x, −π/2 < x < π/2, where the tangent line is parallel to the line y = 2x. Sketch the curve and tangent lines together, labeling each with its equation.
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 3.6.29
Find the derivatives of the functions in Exercises 19–40.
y = (4x + 3)⁴(x + 1)⁻³
Verified step by step guidance1
Identify the function as a product of two functions: \( u(x) = (4x + 3)^4 \) and \( v(x) = (x + 1)^{-3} \). We will use the product rule for differentiation, which states that if \( y = u(x) \cdot v(x) \), then \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).
Differentiate \( u(x) = (4x + 3)^4 \) using the chain rule. Let \( z = 4x + 3 \), then \( u(x) = z^4 \). The derivative \( u'(x) = 4z^3 \cdot \frac{d}{dx}(4x + 3) \). Since \( \frac{d}{dx}(4x + 3) = 4 \), we have \( u'(x) = 16(4x + 3)^3 \).
Differentiate \( v(x) = (x + 1)^{-3} \) using the power rule. The derivative \( v'(x) = -3(x + 1)^{-4} \cdot \frac{d}{dx}(x + 1) \). Since \( \frac{d}{dx}(x + 1) = 1 \), we have \( v'(x) = -3(x + 1)^{-4} \).
Apply the product rule: \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Substitute \( u'(x) = 16(4x + 3)^3 \), \( v(x) = (x + 1)^{-3} \), \( u(x) = (4x + 3)^4 \), and \( v'(x) = -3(x + 1)^{-4} \).
Simplify the expression: \( y' = 16(4x + 3)^3(x + 1)^{-3} - 3(4x + 3)^4(x + 1)^{-4} \). Factor and combine like terms if possible to simplify further.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Product Rule
The product rule is a fundamental technique in calculus used to find the derivative of a product of two functions. If you have two functions, u(x) and v(x), the derivative of their product is given by u'(x)v(x) + u(x)v'(x). This rule is essential when differentiating expressions where two functions are multiplied together, as seen in the given problem.
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The Product Rule
Chain Rule
The chain rule is a method for differentiating composite functions, where one function is nested inside another. If you have a function y = f(g(x)), the derivative is found by taking the derivative of the outer function f with respect to g, and then multiplying it by the derivative of the inner function g with respect to x. This rule is crucial for handling expressions with powers, such as (4x + 3)⁴ in the problem.
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05:02Intro to the Chain Rule
Negative Exponents
Negative exponents indicate reciprocal functions, and understanding how to differentiate them is important. For a function like (x + 1)⁻³, the derivative involves applying the power rule, which states that the derivative of xⁿ is n*xⁿ⁻¹, and then considering the negative exponent. This concept helps in simplifying and differentiating terms with negative powers in the given expression.
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Related Practice
Textbook Question
If x²y³ = 4/27 and dy/dt = ¹/₂, then what is dx/dt when x = 2?
Textbook Question
Derivative of multiples Does knowing that a function g(t) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7? Give reasons for your answer.
Textbook Question
In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.
g(x) = { 2x − x³ − 1, x ≥ 0
x − (1 / (x + 1)), x < 0
Textbook Question
Slopes and Tangent Lines
In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points P₁ and P₂.
Textbook Question
Find the derivatives of the functions in Exercises 1–42.
𝔂 = 2 tan² x - sec² x