Problem 24 - Thomas Calculus

Question

In Exercises \(21-24,\) find \(d p / d q\) $$ p=\frac{\tan q}{1+\tan q} $$

Step-by-Step Solution

Answer

\( \frac{d p}{d q} = \frac{\sec^2 q}{(1 + \tan q)^2} \)

Step 1: Identify the Function and Rewriting

You are given the function \( p = \frac{\tan q}{1 + \tan q} \). Our task is to find \( \frac{d p}{d q} \), which is the derivative of \( p \) with respect to \( q \). We will use the quotient rule for derivatives because \( p \) is a quotient of two functions of \( q \).

Step 2: Recall the Quotient Rule Formula

The quotient rule for derivatives states that if you have a function \( g(q) = \frac{u(q)}{v(q)} \), then its derivative is \( \frac{d}{dq} g(q) = \frac{v(q) u'(q) - u(q) v'(q)}{(v(q))^2} \). Here, \( u(q) = \tan q \) and \( v(q) = 1+\tan q \).

Step 3: Calculate Derivatives of "u" and "v"

Compute the derivatives of the numerator and the denominator in the quotient. The derivative \( u'(q) = \sec^2 q \), because the derivative of \( \tan q \) is \( \sec^2 q \). Next, the derivative \( v'(q) = \sec^2 q \) because the derivative of \( 1 + \tan q \) is also \( \sec^2 q \).

Step 4: Apply the Quotient Rule

Substitute the values of \( u, u', v, \) and \( v' \) into the quotient rule formula: \[ \frac{d}{dq} p = \frac{(1 + \tan q)(\sec^2 q) - \tan q (\sec^2 q)}{(1 + \tan q)^2} \] Simplify the numerator: \[ (1 + \tan q)(\sec^2 q) - \tan q(\sec^2 q) = \sec^2 q + \tan q \cdot \sec^2 q - \tan q \cdot \sec^2 q = \sec^2 q \] Which simplifies to \( \sec^2 q \).

Step 5: Final Simplification

The derivative simplifies to: \[ \frac{d p}{d q} = \frac{\sec^2 q}{(1 + \tan q)^2} \] Nothing more can be combined or simplified from this form, so this is the final answer.

Key Concepts

Quotient Rule

When you come across a function that is a ratio of two other functions, such as fractions like \( \frac{u(q)}{v(q)} \), the Quotient Rule is your best friend. It's a formula specifically designed to help differentiate these types of functions. The rule states that if \( g(q) = \frac{u(q)}{v(q)} \), then the derivative is given by:

• \( \frac{d}{dq} g(q) = \frac{v(q) u'(q) - u(q) v'(q)}{(v(q))^2} \)

To use this formula, you need to know:

• What \( u(q) \) and \( v(q) \) are,
• How to find their derivatives, \( u'(q) \) and \( v'(q) \).

After finding these components, simply plug them into the Quotient Rule formula. It's that easy!
For our exercise, \( u(q) = \tan q \) and \( v(q) = 1 + \tan q \). We computed their derivatives and plugged them into the formula to find the derivative of \( p \).

Trigonometric Functions

Trigonometric functions like \( \tan q \) often appear in math problems because they describe relationships in triangles and circles. When differentiating these functions, there are specific rules to follow. For example:

• The derivative of \( \tan q \) is \( \sec^2 q \).
• \( \sec q \) is the trigonometric identity \( 1/\cos q \), and \( \sec^2 q = (\sec q)^2 \).

Knowing these derivatives is crucial for solving problems that involve trigonometric functions.
In our scenario, both the numerator and the denominator of the given function involve \( \tan q \), and therefore \( \sec^2 q \) appears in the differentiation process. This trigonometric knowledge helped us easily tackle the differentiation task.

Derivatives

The concept of a derivative represents the rate at which a function changes at a particular point. It's a fundamental concept in calculus that gives insight into how variables are correlated. To differentiate a function means to find its derivative. This can involve using specific rules, like the Power Rule or the Quotient Rule, depending on the type of function you have.In our exercise, we are differentiating a function that involves trigonometric and fractional components. By applying the Quotient Rule, we find how the function \( p(q) \) changes as \( q \) changes.
Derivatives give us:

• The slope of the tangent line at any point on a graph,
• Information about increasing or decreasing functions,
• Critical points that can be used to determine local maxima and minima.

In essence, understanding derivatives is like understanding the language of change and slope in mathematics.