Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.9.24

Chapter 3, Problem 3.9.24

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = cos(x²)

Verified step by step guidance

Step 1: Identify the function y = cos(x²). This is a composition of functions, where the outer function is cos(u) and the inner function is u = x².

Step 2: Apply the chain rule for differentiation, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

Step 3: Differentiate the outer function with respect to its argument u. The derivative of cos(u) with respect to u is -sin(u).

Step 4: Differentiate the inner function u = x² with respect to x. The derivative of x² with respect to x is 2x.

Step 5: Combine the results from steps 3 and 4 using the chain rule: dy/dx = -sin(x²) * 2x. This is the derivative of y with respect to x.

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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 technique in calculus used to differentiate composite functions. It states that if a function y = f(g(x)) is composed of two functions, the derivative dy/dx is found by multiplying the derivative of the outer function f with respect to its inner function g, by the derivative of the inner function g with respect to x. This is essential for differentiating y = cos(x²), where x² is the inner function.

Derivative of Trigonometric Functions

Understanding the derivatives of trigonometric functions is crucial for solving problems involving these functions. The derivative of cos(x) with respect to x is -sin(x). This knowledge is applied when differentiating y = cos(x²), where the derivative of the outer function cos is needed to apply the chain rule effectively.

Differential Notation

Differential notation involves expressing the derivative of a function in terms of differentials, such as dy and dx. It provides a way to represent the rate of change of y with respect to x. In the context of the problem, finding dy involves using differential notation to express the derivative of y = cos(x²) in terms of dy and dx, which is useful for understanding changes in y as x varies.

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Related Practice

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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.

Textbook Question

If x²y³ = 4/27 and dy/dt = ¹/₂, then what is dx/dt when x = 2?

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For what value of c is the curve y = c/ (x + 1) tangent to the line through the points (0, 3) and (5, -2)?

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Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m/sec. Because the acceleration of gravity at the planet’s surface was gₛ m/sec², the explorers expected the ball bearing to reach a height of s = 15t − (1/2)gₛt² m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gₛ?

Textbook Question

Approximation Error

In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find

a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.

f(x) = x⁴, x₀ = 1, dx = 0.1

Textbook Question

In Exercises 41–58, find dy/dt.

y = (1/6)(1 + cos²(7t))³

Derivatives in Differential FormIn Exercises 17–28, find dy.y = cos(x²)

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Key Concepts

Chain Rule

Derivative of Trigonometric Functions

Differential Notation

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = cos(x²)