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In Exercises \(27-32,\) find \(d p / d q\) $$ p=(1+\csc q) \cos q $$

Short Answer

Expert verified
\(\frac{dp}{dq} = -\sin q - \csc q \cos q \cot q - \csc q \sin q\)."

Step by step solution

01

Expand the Expression

The problem given is to find the derivative of the function with respect to \( q \). The function is \( p = (1 + \csc q) \cos q \). Let's expand it by applying the distributive property: \( p = \cos q + \csc q \cos q \).
02

Differentiate Each Term

The expression is now \( p = \cos q + \csc q \cos q \). We'll differentiate each term separately with respect to \( q \):- For \( \cos q \), its derivative is \(-\sin q\).- For \( \csc q \cos q \), use the product rule.
03

Apply the Product Rule

Recall the product rule: \( \frac{d}{dq}[u \cdot v] = u'v + uv' \). Here, let \( u = \csc q \) and \( v = \cos q \):- The derivative of \( u = \csc q \) is \( -\csc q \cot q \).- The derivative of \( v = \cos q \) is \(-\sin q\).- Thus, \( \frac{d}{dq}[\csc q \cos q] = (-\csc q \cot q)\cos q + (\csc q)(-\sin q) \).
04

Combine and Simplify

Now combine the derivatives:- The derivative of \( \cos q \) is \(-\sin q\).- From the product rule applied earlier, the derivative of \( \csc q \cos q \) results in \(-\csc q \cot q \cos q - \csc q \sin q \).- Therefore, \( \frac{dp}{dq} = -\sin q - \csc q \cos q \cot q - \csc q \sin q \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Product Rule
In calculus, the product rule is an essential tool when differentiating expressions where two functions are multiplied together. If you have two functions, say \( u \) and \( v \), and you want to find the derivative of their product \( u \cdot v \) with respect to a variable, the product rule provides a solution. The formula is:
  • \( \frac{d}{dq}[u \cdot v] = u'v + uv' \)
Here, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \, \) respectively.
The rule tells us to take the derivative of the first function and multiply it by the second function, then add this to the product of the first function and the derivative of the second function.
When applying this in practice, it's crucial to correctly identify \( u \) and \( v \), such as in our exercise where \( u = \csc q \) and \( v = \cos q \). By substituting each function and its derivative into the product rule, you obtain the desired derivative, which can then be simplified.
Exploring Trigonometric Functions
Trigonometric functions play a significant role in calculus, especially when they come into interplay during differentiation and integration. The most commonly used trigonometric functions include:
  • Sine \( \sin q \)
  • Cosine \( \cos q \)
  • Cosecant \( \csc q \), which is the reciprocal of the sine function \( \csc q = 1/\sin q \)
  • Cotangent \( \cot q \) is another reciprocal function \( \cot q = \cos q / \sin q \)
When differentiating trigonometric functions, it's crucial to remember the specific derivatives:
  • For \( \cos q \), the derivative is \( -\sin q \)
  • For \( \csc q \), the derivative is \( -\csc q \cot q \)
These derivatives can appear complex, but with practice, they become a familiar part of solving calculus problems.
Mastering Calculus Exercises
Solving calculus exercises involves understanding the problem and strategically choosing the right mathematical tools. Here is how you can approach such problems:
  • Identify the function to be differentiated. Break it down into simpler terms when possible. As in the original exercise, expanding the expression makes differentiation easier.
  • Choose the correct derivative rules. Each type of expression (product, quotient, trigonometric, or composite functions) has specific rules that guide differentiation.
  • Combine and simplify your results. After differentiation, gather all terms, simplify if necessary, and verify correctness.
By habitual practice and familiarizing yourself with derivative rules, calculating derivatives becomes less challenging. Starting with simpler exercises before tackling complex problems can boost your confidence and expertise in calculus.

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