Problem 32 - University Calculus: Early Transcendentals

Question

Find \(d p / d q\). $$p=\frac{3 q+\tan q}{q \sec q}$$

Step-by-Step Solution

Answer

The derivative \(\frac{dp}{dq}\) is \(\frac{q \sec^3 q - \tan q \sec q}{(q \sec q)^2}\).

Step 1: Identify the Function Components

The function given in the problem is \(p = \frac{3q + \tan q}{q \sec q}\). We need to differentiate this function with respect to \(q\). First, recognize that this function is a quotient of two functions: the numerator \(u(q) = 3q + \tan q\) and the denominator \(v(q) = q \sec q\).

Step 2: Apply the Quotient Rule

To differentiate a quotient \(\frac{u(q)}{v(q)}\), use the Quotient Rule: \(\frac{d}{dq}\left(\frac{u(q)}{v(q)}\right) = \frac{u'(q)v(q) - u(q)v'(q)}{(v(q))^2}\). Identify \(u'(q)\) and \(v'(q)\) which are needed for this rule.

Step 3: Differentiate the Numerator

Find \(u'(q)\): - The derivative of \(3q\) is 3.- The derivative of \(\tan q\) is \(\sec^2 q\).Thus, \(u'(q) = 3 + \sec^2 q\).

Step 4: Differentiate the Denominator

Find \(v'(q)\):- Derivative of \(q\) is 1.- Derivative of \(\sec q\) is \(\sec q \tan q\).So for \(v(q) = q \sec q\), apply the product rule: \(v'(q) = 1 \cdot \sec q + q \cdot \sec q \tan q = \sec q + q \sec q \tan q\).

Step 5: Substitute into the Quotient Rule

Substitute \(u'(q)\), \(v'(q)\), \(u(q)\), and \(v(q)\) into the Quotient Rule formula:\[\frac{dp}{dq} = \frac{(3 + \sec^2 q)(q \sec q) - (3q + \tan q)(\sec q + q \sec q \tan q)}{(q \sec q)^2}\].

Step 6: Simplify the Expression

Simplify the numerator and denominator:- The numerator after expanding and simplifying is:\[ (3q \sec q + q \sec^3 q) - (3q \sec q + q \sec^3 q \tan q + \tan q \sec q) \].- Simplify to find the final expression for \(\frac{dp}{dq}\).

Step 7: Final Expression

After simplification, the derivative \(\frac{dp}{dq}\) is:\[ \frac{dp}{dq} = \frac{q \sec^3 q - \tan q \sec q}{(q \sec q)^2} \].

Key Concepts

Quotient Rule

When we are tasked with differentiating a function that is written as the division of two other functions, the Quotient Rule comes into play. This technique allows us to systematically find the derivative of such expressions. For a function in the form of \( \frac{u(q)}{v(q)} \), where \( u(q) \) and \( v(q) \) are two separate functions of \( q \), the Quotient Rule states:

\[ \frac{d}{dq}\left(\frac{u(q)}{v(q)}\right) = \frac{u'(q)v(q) - u(q)v'(q)}{(v(q))^2} \]
This formula gives us a structured approach:

• Differentiate the numerator \( u(q) \) to find \( u'(q) \).
• Differentiate the denominator \( v(q) \) to find \( v'(q) \).
• Substitute these derivatives into the Quotient Rule formula.

By applying these steps, we ensure that we correctly capture the changes in both the numerator and denominator as \( q \) changes.

Trigonometric Derivatives

Trigonometric functions, such as sine, cosine, and tangent, have their own specific derivatives. In calculus, knowing these by heart can greatly simplify differentiation problems. Let's focus on the derivatives relevant to our exercise: the derivative of \( \tan q \) and \( \sec q \).

• For \( \tan q \), the derivative is \( \sec^2 q \). This is a fundamental result derived from the identity involving sine and cosine.
• For \( \sec q \), the derivative is \( \sec q \tan q \). This involves applying both the product and chain rules.

In our problem, these derivatives contribute to finding the rate of change of the expression \( p \) with respect to \( q \). By substituting these derivatives into our equations, we can handle more complex expressions involving trigonometric functions.

Simplification Process

After applying differentiation rules, like the Quotient Rule, the resulting expressions can often be quite complex. Simplification is the process of making this expression more manageable and readable. It's like tidying up after a thorough derivation.

In simplification:

• Start by expanding any expressions that result from multiplication.
• Combine like terms where possible. This involves gathering similar terms to reduce clutter.
• Factor out common terms if possible. This can reduce the expression even further and make it neater.

For our derivative, simplifying involved careful manipulation of trigonometric terms and common factor identification, leading to the more elegant expression \( \frac{q \sec^3 q - \tan q \sec q}{(q \sec q)^2} \). Simplification not only makes the result more aesthetically pleasing, but often reveals deeper insights into the behavior of the function.