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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.5.29
Chapter 3, Problem 3.5.29

Derivatives


In Exercises 27–32, find dp/dq.


p = (sin q + cos q) / cos q

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1
Step 1: Start by identifying the function p in terms of q. Here, p is given as \( p = \frac{\sin q + \cos q}{\cos q} \).
Step 2: Simplify the expression for p. Divide each term in the numerator by the denominator: \( p = \frac{\sin q}{\cos q} + \frac{\cos q}{\cos q} \). This simplifies to \( p = \tan q + 1 \).
Step 3: Differentiate p with respect to q. The derivative of \( \tan q \) with respect to q is \( \sec^2 q \), and the derivative of a constant (1) is 0.
Step 4: Combine the derivatives to find \( \frac{dp}{dq} \). Since \( p = \tan q + 1 \), \( \frac{dp}{dq} = \sec^2 q + 0 \).
Step 5: Conclude that the derivative of p with respect to q is \( \frac{dp}{dq} = \sec^2 q \).

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Key Concepts

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

Derivatives

A derivative represents the rate at which a function changes as its input changes. It is a fundamental concept in calculus that measures how a function's output value varies with respect to changes in its input variable. The derivative of a function can be interpreted as the slope of the tangent line to the function's graph at a given point.
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Derivatives

Chain Rule

The chain rule is a formula for computing the derivative of a composite function. If a function is composed of two or more functions, the chain rule allows us to differentiate it by multiplying the derivative of the outer function by the derivative of the inner function. This is essential when dealing with functions that are expressed in terms of other functions, as in the case of p = (sin q + cos q) / cos q.
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Intro to the Chain Rule

Trigonometric Derivatives

Trigonometric derivatives involve the differentiation of functions that include trigonometric functions such as sine and cosine. The derivatives of these functions are well-defined: the derivative of sin(q) is cos(q), and the derivative of cos(q) is -sin(q). Understanding these derivatives is crucial for solving problems involving trigonometric functions, especially when applying the quotient rule in this context.
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Derivatives of Other Inverse Trigonometric Functions
Related Practice
Textbook Question

Second Derivatives


Find y'' in Exercises 59–64.


y = (1 − √x)⁻¹

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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

In Exercises 9–18, write the function in the form y = f(u) and u = g(x). Then find dy/dx as a function of x.


y = (2x + 1)⁵

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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 = (4/3)πr³ of a sphere when the radius changes from r₀ to r₀ + dr

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Textbook Question

Derivatives


In Exercises 23–26, find dr/dθ.


r = θ sin θ + cos θ

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Textbook Question

Differentiating Implicitly


Use implicit differentiation to find dy/dx in Exercises 1–14.


x = sec y

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