Problem 28 - University Calculus: Early Transcendentals

Question

Find \(d p / d q\). $$p=(1+\csc q) \cos q$$

Step-by-Step Solution

Answer

The derivative \( \frac{dp}{dq} = -\frac{\cos^2 q}{\sin^2 q} - \sin q - 1 \).

Step 1: Understand the Problem

You need to differentiate the function \(p = (1 + \csc q) \cos q\) with respect to \(q\). This means you will find \( \frac{dp}{dq} \).

Step 2: Use the Product Rule

The function \(p = (1 + \csc q) \cos q\) is a product of two functions: \((1 + \csc q)\) and \(\cos q\). The product rule for differentiation states that \((uv)' = u'v + uv'\). Identify \(u = (1 + \csc q)\) and \(v = \cos q\).

Step 3: Differentiate \(u = 1 + \csc q\)

Differentiate \(u = 1 + \csc q\). The derivative of a constant \(1\) is 0, and the derivative of \(\csc q\) is \(-\csc q \cot q\). Thus, \(u' = 0 - \csc q \cot q = -\csc q \cot q\).

Step 4: Differentiate \(v = \cos q\)

Differentiate \(v = \cos q\). The derivative of \(\cos q\) is \(-\sin q\). Thus, \(v' = -\sin q\).

Step 5: Apply the Product Rule

Using the product rule, \( \frac{dp}{dq} = u'v + uv' = (-\csc q \cot q)(\cos q) + (1 + \csc q)(-\sin q)\).

Step 6: Simplify the Expression

Simplify the expression from Step 5:\[ (-\csc q \cot q)(\cos q) + (1 + \csc q)(-\sin q) \]Distribute each term:\[ -\csc q \cos q \cot q - \sin q - \csc q \sin q \]Use the identity \(\csc q = \frac{1}{\sin q}\) and \(\cos q \cot q = \frac{\cos^2 q}{\sin q}\) to further simplify:\[ -\frac{\cos^2 q}{\sin^2 q} - \sin q - 1 \]Then, combine like terms if possible.

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus that is essential when differentiating products of two functions. It helps you find the derivative of a function that can be expressed as the product of two or more other functions. To apply the product rule, remember the formula:

• If you have two functions, say \( u(q) \) and \( v(q) \), then the derivative of their product \( (uv)' \) is given by: \( u'v + uv' \).

In our original exercise, the function \( p = (1 + \csc q) \cos q \) is clearly a product of two individual functions: \( u(q) = (1 + \csc q) \) and \( v(q) = \cos q \).
This clear separation allows us to individually differentiate \( u \) and \( v \), and then apply the formula.
This method makes handling complex expressions much more manageable.

Trigonometric Identities

Trigonometric identities are equations that are true for all angles and are especially invaluable when simplifying expressions in calculus.
In this particular problem, identities are useful for simplifying terms during the differentiation process.
Some common trigonometric identities that are frequently used include:

• The reciprocal identity: \( \csc q = \frac{1}{\sin q} \).
• The quotient identity: \( \cot q = \frac{\cos q}{\sin q} \).

Using these identities, we can simplify expressions like \( \csc q \cos q \cot q \) easily because:

• \( \cos q \cot q \) becomes \( \frac{\cos^2 q}{\sin q} \).
• Now, \( \csc q \cos q \cot q \) simplifies to \( \frac{\cos^2 q}{\sin^2 q} \) when combined.

Being able to recognize and apply these identities helps you simplify complex derivatives into more manageable forms.

Derivatives

Derivatives are central to understanding rates of change within calculus. They measure how a function changes as its input changes.
A common basic derivative that is vital in trigonometric calculus involves functions like sine, cosine, or tangent.
For instance:

• The derivative of \( \cos q \) is \(-\sin q \), and is a standard result.
• The derivative of \( \csc q \) is \(-\csc q \cot q \), which might be less familiar but equally important.

In differentiating the function \( p = (1 + \csc q) \cos q \), we had to find the derivatives of \( 1 + \csc q \) and \( \cos q \) separately:

• For \( u' \), differentiating \( 1 + \csc q \) yields \( -\csc q \cot q \).
• For \( v' \), differentiating \( \cos q \) yields \(-\sin q \).

These derivatives are then combined using the product rule to find the full derivative of the function. Understanding these operations gives us a deeper insight into the behavior of the function's graph.