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Chapter 2: Problem 101

Differentiate. $$ s(t)=\frac{\tan t}{t \cos t} $$

Short Answer

Expert verified
- The derivative of \( s(t) = \frac{\tan t}{t \tan^2 t} \) after some simplification is \( \frac{\text{sec}^2 t \tan t}{t^2 \tan^4 t} - \frac{\tan t}{t^2} - \frac{ 2t \tan t \text{sec}^2 t}{t^2 \tan^4 t} \).

Step by step solution

01

- Simplify the function

First, simplify the function \( s(t) = \frac{\tan t}{t \tan t} \times \frac{\tan t}{\tan t} \). This can be reduced to \( s(t) = \frac{\tan t}{t \tan t \times \tan t} = \frac{\tan t}{t \tan^2 t \).
02

- Apply the quotient rule

To differentiate the function \( s(t) = \frac{\tan t}{t \tan^2 t} \), apply the quotient rule, which states \( \frac{d}{dt} \frac{u}{v} = \frac{u'v - uv'}{v^2} \). Here, let \( u = \tan t \) and \( v = t\tan^2 t \).
03

- Differentiate the numerator

Differentiate \( u = \tan t \). \( \frac{du}{dt} = \text{sec}^2 t \).
04

- Differentiate the denominator

Differentiate \( v = t \tan^2 t \) using the product rule. Let \( f(t) = t \) and \( g(t) = \tan^2 t \). \( v' = f'g + fg' = 1 \tan^2 t + t \times 2\tan t \text{sec}^2 t = \tan^2 t + 2t \tan t \text{sec}^2 t \).
05

- Apply the quotient rule expression

Substitute back into the quotient rule expression: \( s'(t) = \frac{\text{sec}^2 t \times t \tan^2 t - \tan t (\tan^2 t + 2t \tan t \text{sec}^2 t)}{(t \tan^2 t)^2} \).
06

- Simplify the result

Simplify the result: \( s'(t) = \frac{ \text{sec}^2 t \times t \tan^2 t - \tan^3 t - 2t \tan^2 t \text{sec}^2 t}{(t \tan^2 t)^2} \) and combine like terms.

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Key Concepts

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

quotient rule
In calculus, the quotient rule is used to differentiate functions of the form \(\frac{u}{v}\). It is particularly useful when dealing with ratios of two differentiable functions. The quotient rule states: \[\frac{d}{dt} \frac{u}{v} = \frac{u'v - uv'}{v^2}\]. Here, \(u\) and \(v\) are both functions of \(t\), while \(u'\) and \(v'\) are their respective derivatives. This formula can be derived from the product rule and helps manage cases where dividing the two functions occurs naturally in a problem.
product rule
The product rule is essential for differentiating products of two functions. When you have a function \(h(t) = f(t)\times g(t)\), the product rule states that the derivative \(h'(t) = f'(t)g(t) + f(t)g'(t)\). This rule ensures we account for the rate of change in both functions simultaneously. When applying the product rule in combination with the quotient rule, like in differentiating \(s(t) = \frac{\tan t}{t \tan^2 t}\), ensure to distinctly differentiate both the numerator and the denominator.
trigonometric functions differentiation
Differentiating trigonometric functions is a frequent task in calculus. Each trigonometric function has a well-defined derivative: \( \frac{d}{dt} (\tan t) = \text{sec}^2 t \), which plays a central role in our original exercise. Recognize the standard derivatives of core trigonometric functions:
  • \( \frac{d}{dt} (\tan t) = \text{sec}^2 t\)
  • \( \frac{d}{dt} (\text{sec} t) = \text{sec} t \tan t \)
Understanding these properties simplifies the differentiation process.
simplification of functions
Simplifying a function before applying differentiation techniques often makes the problem easier to solve. In the original exercise, simplifying \( s(t) = \frac{\tan t}{t \tan t \times \tan t} = \frac{\tan t}{t \tan^2 t} \) before applying the quotient rule made the differentiation process clearer. The less complicated the function, the easier it becomes to handle differentiation. Always look for ways to simplify fractions, combine like terms, and reduce expressions before moving forward with differentiation.

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