Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.15

Chapter 3, Problem 3.15

Find the derivatives of the functions in Exercises 1–42.

s = (sec t + tan t)⁵

Verified step by step guidance

Identify the function: We have s = (sec t + tan t)⁵. This is a composite function where the outer function is u⁵ and the inner function is u = sec t + tan t.

Apply the chain rule: The chain rule states that if you have a composite function f(g(x)), the derivative is f'(g(x)) * g'(x). Here, f(u) = u⁵ and g(t) = sec t + tan t.

Differentiate the outer function: The derivative of u⁵ with respect to u is 5u⁴. So, f'(u) = 5(sec t + tan t)⁴.

Differentiate the inner function: The derivative of sec t is sec t tan t, and the derivative of tan t is sec² t. Therefore, g'(t) = sec t tan t + sec² t.

Combine the derivatives using the chain rule: Multiply the derivative of the outer function by the derivative of the inner function: s'(t) = 5(sec t + tan t)⁴ * (sec t tan t + sec² t).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus that provides the slope of the tangent line to the curve at any given point. The derivative can be computed using various rules, such as the power rule, product rule, and chain rule, depending on the form of the function.

Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions. It states that if you have a function that is composed of another function, the derivative can be found by multiplying the derivative of the outer function by the derivative of the inner function. This is particularly useful when dealing with functions raised to a power, as seen in the given problem.

Trigonometric Functions

Trigonometric functions, such as secant (sec) and tangent (tan), are fundamental functions in calculus that relate angles to ratios of sides in right triangles. Their derivatives are essential for solving problems involving rates of change in contexts like physics and engineering. Understanding the derivatives of these functions is crucial for applying the chain rule effectively in the given exercise.

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Related Practice

Textbook Question

Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 cm/min. At what rate is the plate’s area increasing when the radius is 50 cm?

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 lateral surface area S = 2πrh of a right circular cylinder when the height changes from h₀ to h₀ + dh and the radius does not change

Textbook Question

Finding Derivative Values

In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.

f(u) = ((u − 1) / (u + 1))², u = g(x) = (1 / x²) − 1, x = −1

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Textbook Question

In Exercises 41–58, find dy/dt.

y = sin²(πt − 2)

Textbook Question

Using the Alternative Formula for Derivatives

Use the formula

f'(x) = lim (z → x) (f(z) − f(x)) / (z − x)