Problem 35 - Thomas Calculus

Question

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\csc ^{-1}\left(e^{t}\right) $$

Step-by-Step Solution

Answer

\(\frac{dy}{dt} = -\frac{1}{\sqrt{(e^t)^2 - 1}}\)

Step 1: Identify the Function

Identify the function given in the problem. Here, the function is \( y = \csc^{-1}(e^t) \), where \( y \) is expressed in terms of \( e^t \).

Step 2: Differentiate Using Inverse Trigonometric Derivative Formula

The derivative of \( \csc^{-1}(x) \) with respect to \( x \) is \( -\frac{1}{|x|\sqrt{x^2 - 1}} \). Since we have \( y = \csc^{-1}(e^t) \), this derivative formula can be applied, treating \( e^t \) as \( x \).

Step 3: Apply Chain Rule

We need to apply the chain rule because \( y \) is a composition of two functions, \( \csc^{-1}(u) \) where \( u = e^t \). The derivative \( \frac{dy}{dt} \) is \( \frac{dy}{du} \cdot \frac{du}{dt} \). Here, \( \frac{du}{dt} = \frac{d}{dt}(e^t) = e^t \).

Step 4: Substitute and Simplify

Substitute \( \frac{dy}{du} = -\frac{1}{|e^t|\sqrt{(e^t)^2 - 1}} \) and \( \frac{du}{dt} = e^t \) into the chain rule formula: \( \frac{dy}{dt} = -\frac{1}{|e^t|\sqrt{(e^t)^2 - 1}} \cdot e^t \). Simplify to get \( \frac{dy}{dt} = -\frac{1}{\sqrt{(e^t)^2 - 1}} \).

Step 5: Ensure Correct Notational Usage

Ensure that the absolute value notation and other parts are handled correctly. The expression \( -\frac{1}{\sqrt{(e^t)^2 - 1}} \) is correctly simplified assuming \( e^t > 1 \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial when dealing with angles and rates of change in calculus. When given a function like \( y = \csc^{-1}(e^{t}) \), it represents the inverse of the cosecant function. The inverse cosecant function, \( \csc^{-1}(x) \), provides the angle whose cosecant is \( x \). Understanding this property helps us determine the derivatives of these functions when combined with other types, like exponential functions.

• \( \csc^{-1}(x) \) covers specific angles where \( x \) is not between \(-1\) and \(1\), as a consequence of the cosecant function's behavior.
• The derivative of \( \csc^{-1}(x) \) with respect to \( x \) is \( -\frac{1}{|x|\sqrt{x^2 - 1}} \). This derivative will play a key role in finding the rate of change of the original compound function.

Learning how to differentiate inverse trigonometric functions enhances your problem-solving toolkit in calculus by allowing you to tackle complex scenarios where these functions are involved.

Chain Rule

The chain rule is an indispensable tool in calculus for differentiating compositions of functions. When a function is nested, or one function is inside another, the chain rule provides a way to differentiate it with respect to a particular variable. In our given exercise, the function \( y = \csc^{-1}(e^t) \) involves the exponential function \( e^t \) nested inside the inverse trigonometric function \( \csc^{-1}(x) \).

• The fundamental idea of the chain rule is to find the derivative of the outer function multiplied by the derivative of the inner function.
• Mathematically, if \( y = f(g(t)) \), then the derivative, \( \frac{dy}{dt} \), is given by \( \frac{dy}{dg} \times \frac{dg}{dt} \).

In this exercise:

• \( f(g) = \csc^{-1}(g) \) and \( g(t) = e^t \).
• Thus, differentiate \( y \) using \( \frac{dy}{du} = -\frac{1}{|e^t|\sqrt{(e^t)^2 - 1}} \) and \( \frac{du}{dt} = e^t \).

Utilizing the chain rule effectively simplifies the calculation and gives a clear path to find the derivative.

Exponential Functions

Exponential functions such as \( e^t \) play significant roles in calculus due to their unique properties. The function \( e^t \) denotes exponential growth and is distinguished by its property that the rate of change (or derivative) is itself, \( e^t \).

• When differentiating \( e^t \) with respect to \( t \), you get \( \frac{d}{dt}(e^t) = e^t \), which highlights its self-replicating nature.
• This characteristic simplifies many problems, including the one in our exercise, where it seamlessly integrates into the chain rule framework without complicating the derivative calculation.

Exponential functions are also essential in solving differential equations and modeling various natural phenomena, making them a staple concept in calculus.

Calculus Problem-Solving

Effective calculus problem-solving involves a step-by-step approach to breaking down complex problems into more manageable parts. In the current exercise, we began by recognizing that our function, \( y = \csc^{-1}(e^t) \), is a composition of an inverse trigonometric and an exponential function.
Here’s how a calculated approach helps:

• **Identification of functions and derivatives** - Recognizing the types of functions involved sets the stage for applying the correct rules and formulas.
• **Use of differentiation rules like the chain rule** - Identifying when to use the chain rule is crucial for correctly differentiating composite functions.
• **Simplification** - After applying the differentiation rules, simplify the expression to make the final derivative clearer.

Follow these strategies to tackle calculus exercises efficiently:

• Break down each function into its components before starting differentiation.
• Check each step carefully for arithmetic or algebraic errors.
• Practice consistently with varied problems to strengthen your understanding.

By mastering these problems step-by-step, you get better prepared to take on increasingly challenging calculus questions with confidence.