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Chapter 2: Problem 19

Compute the derivative of the given function. $$g(t)=\sin t \cos ^{-1} t$$

Short Answer

Expert verified
The derivative is \( g'(t) = \cos t \cdot \cos^{-1} t - \frac{\sin t}{\sqrt{1-t^2}} \).

Step by step solution

01

Identify the function components

The given function is a product of two functions: the first function is \( \sin t \) and the second function is \( \cos^{-1} t \). We need to use the product rule of differentiation.
02

Recall the Product Rule

The product rule states that if you have a function that is the product of two differentiable functions, such as \( f(t) = u(t) \, v(t) \), then the derivative \( f'(t) \) is given by \( u'(t) \, v(t) + u(t) \, v'(t) \). Here, \( u(t) = \sin t \) and \( v(t) = \cos^{-1} t \).
03

Differentiate \( \sin t \)

The derivative of \( \sin t \) with respect to \( t \) is \( \cos t \). Thus, \( u'(t) = \cos t \).
04

Differentiate \( \cos^{-1} t \)

The derivative of \( \cos^{-1} t \) with respect to \( t \) is \( -\frac{1}{\sqrt{1-t^2}} \). Thus, \( v'(t) = -\frac{1}{\sqrt{1-t^2}} \).
05

Apply the product rule

Substitute \( u(t), u'(t), v(t), \) and \( v'(t) \) into the product rule formula: \[ g'(t) = \cos t \cdot \cos^{-1} t + \sin t \cdot \left(-\frac{1}{\sqrt{1-t^2}}\right) \].
06

Simplify the derivative expression

Simplify the expression: \[ g'(t) = \cos t \cdot \cos^{-1} t - \frac{\sin t}{\sqrt{1-t^2}} \].

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Key Concepts

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

Product Rule
When differentiating products of functions, the product rule is a powerful tool. It allows us to find the derivative of a product of functions without expanding or going through extensive simplifications.
The product rule can be written as:
  • If you have two functions, say \( u(t) \) and \( v(t) \), their product is \( u(t) \, v(t) \).
  • The derivative of their product is given by \( u'(t) \, v(t) + u(t) \, v'(t) \).
This rule helps keep everything organized, especially with complicated expressions.
In our case, \( u(t) = \sin t \) and \( v(t) = \cos^{-1} t \). The steps involve differentiating each function separately and plugging the results into the formula. This makes the differentiation process smoother and more systematic.
Trigonometric Functions
Trigonometric functions like \( \sin t \) and \( \cos t \) relate to angles and sides of triangles. These functions are periodic, meaning they repeat values in a regular pattern.
The well-known derivatives of trigonometric functions include:
  • The derivative of \( \sin t \) is \( \cos t \).
  • The derivative of \( \cos t \) is \( -\sin t \).
These differentiation rules arise from the circular motions tied to angle measures and are foundational in calculus. Understanding these derivatives is crucial for solving a wide range of problems in calculus and physics.
Inverse Trigonometric Functions
Inverse trigonometric functions go the opposite direction of regular trigonometric functions, taking a value and giving the corresponding angle.
For example, \( \cos^{-1} t \) tells you what angle has \( \cos \) value of \( t \).
The derivatives of inverse trigonometric functions are:
  • The derivative of \( \cos^{-1} t \) is \( -\frac{1}{\sqrt{1-t^2}} \).
  • The derivative of \( \sin^{-1} t \) is \( \frac{1}{\sqrt{1-t^2}} \).
These derivatives are essential because they connect trigonometric relationships back to their original angles, offering a complete picture in analysis problems involving arcs and angles.

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Most popular questions from this chapter

T/F: Implicit differentiation can be used to find the derivative of \(y=x^{3 / 4}\).

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