Problem 29 - University Calculus: Early Transcendentals

Question

Find \(d p / d q\). $$p=\frac{\sin q+\cos q}{\cos q}$$

Step-by-Step Solution

Answer

\( \frac{d p}{d q} = \sec^2 q \).

Step 1: Simplify the Expression

First, simplify the given expression for \( p \). The original expression is \( p = \frac{\sin q + \cos q}{\cos q} \). We can simplify it to \( p = \frac{\sin q}{\cos q} + 1 \). This can further be written as \( p = \tan q + 1 \).

Step 2: Differentiate with Respect to \(q\)

Differentiate \( p = \tan q + 1 \) with respect to \( q \). Remember that the derivative of \( \tan q \) is \( \sec^2 q \), and the derivative of a constant is zero. Therefore, we get: \[ \frac{d p}{d q} = \sec^2 q \].

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in both differential calculus and many real-world applications. They relate the angles and sides of triangles, primarily in right-angled triangles. The most common trigonometric functions include sine (\(\sin\theta\)), cosine (\(\cos\theta\)), and tangent (\(\tan\theta\)). These functions help us determine the ratios derived from the angles in a triangle and are periodic, meaning they repeat their values in consistent intervals.

In the context of the problem, the goal was to differentiate an expression involving sine and cosine. To simplify the expression, we can utilize the identity between these functions. Notice how both sine and cosine are parts of the ratio \(\tanq = \frac{\sinq}{\cosq}\). Recognizing this identity allows for the simplification of expressions containing multiple trigonometric functions, which is a common step in calculus problems.

Derivative Rules

Derivative rules are essential tools in calculus that help compute the rate of change for functions. When dealing with basic functions, like trigonometric functions, specific derivative rules apply. For trigonometric functions, these rules are built upon the basic principles of differentiation. Knowing these derivative formulas is crucial for solving calculus problems efficiently.

In the given exercise, we need to differentiate \(p = \tanq + 1\) with respect to \(q\). The derivative of the tangent function is \(\sec^2q\). Remembering that the derivative of a constant, such as \(1\), is \(0\), we apply the derivative rule for tangents and obtain \(\frac{dp}{dq} = \sec^2q\).

This differentiation illustrates how applying basic rules can yield the derivative, showcasing how straightforward calculus can be when rules are mastered.

Simplification of Expressions

Simplification of expressions is a key step in solving many calculus problems. Simplifying makes expressions easier to handle and differentiate. It involves using algebraic identities, like trigonometric identities, to rewrite complex expressions in a more manageable form.

For the problem at hand, we started with \(p = \frac{\sinq + \cosq}{\cosq}\). By separating the terms under the common denominator and identifying the trigonometric identity \(\tanq = \frac{\sinq}{\cosq}\), we transformed the expression into \(p = \tanq + 1\). This transformation simplifies the differentiation process significantly.

Understanding simplification is vital as it often reduces complex expressions to basic forms, making it easier to apply derivative rules and solve calculus problems efficiently.