Problem 36 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\cos ^{-1}\left(e^{-t}\right)$$

Step-by-Step Solution

Answer

The derivative is \( \frac{e^{-t}}{\sqrt{1-e^{-2t}}} \).

Step 1: Identify the Type of Differentiation

The given function is a composite function, meaning we will need to use the chain rule to find its derivative.

Step 2: Differentiate the Outer Function

The outer function is \( \cos^{-1}(u) \), where \( u = e^{-t} \). The derivative of \( \cos^{-1}(u) \) with respect to \( u \) is \( -\frac{1}{\sqrt{1-u^2}} \).

Step 3: Differentiate the Inner Function

The inner function is \( e^{-t} \). Its derivative with respect to \( t \) is \( -e^{-t} \).

Step 4: Apply the Chain Rule

Using the chain rule, the derivative of the composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This gives: \(-\frac{1}{\sqrt{1-(e^{-t})^2}} \times (-e^{-t})\).

Step 5: Simplify the Expression

Simplify the expression to find the final derivative: \( \frac{e^{-t}}{\sqrt{1-e^{-2t}}} \).

Key Concepts

Composite Function

Understanding the concept of composite functions is key to tackling problems that involve multiple layers of functions within each other. Simply put, a composite function involves two or more functions combined such that the output of one function becomes the input of another. For example, if you have a function \( f(x) \) and another function \( g(x) \), then a composite function can be expressed as \( f(g(x)) \).

When differentiating a composite function, the most effective technique is the Chain Rule. The Chain Rule states that the derivative of a composite function \( f(g(x)) \) is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \):

• First, identify the inner function \( g(x) \) and differentiate it to find \( g'(x) \).
• Then, differentiate the outer function \( f \) with respect to the inner function to get \( f'(g(x)) \).
• Multiply these derivatives together to get the derivative of the composite function.

In our example \( y = \cos^{-1}(e^{-t}) \), \( u = e^{-t} \) is the inner function and \( \cos^{-1}(u) \) is the outer function.

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions can often appear daunting due to their complex forms. Functions like \( \cos^{-1}(x) \), \( \sin^{-1}(x) \), and \( \tan^{-1}(x) \) have specific derivatives that are crucial to remember. For \( \cos^{-1}(x) \), the derivative with respect to \( x \) is:

• \( -\frac{1}{\sqrt{1-x^2}} \)

This formula arises from the relationship between trigonometric identities and their inverses. When differentiating a function like \( y = \cos^{-1}(e^{-t}) \), you would first consider the derivative of the outer function \( \cos^{-1}(u) \), treating \( u = e^{-t} \) as the variable of interest.

The expression \( -\frac{1}{\sqrt{1-u^2}} \) captures how small changes in \( u \) affect \( \cos^{-1}(u) \), and this is a direct application in using the chain rule for our given problem.

Exponential Function Derivative

Exponential functions, particularly those involving \( e \), have unique derivative properties thanks to the constant \( e \approx 2.718 \). When the exponent is a simple variable, like in \( e^x \), the derivative is simply \( e^x \). However, when there's an additional layer, such as \( e^{-t} \), things require a closer look.

For any exponential function of the form \( e^{f(x)} \), the derivative requires you to multiply \( e^{f(x)} \) by the derivative of the exponent \( f(x) \). In this case, the inner function is \( -t \), making the derivative of \( e^{-t} \) equal to:

• \( -e^{-t} \)

This simplification results from applying the Chain Rule, which is necessary when nested functions are involved, like when \( e^{-t} \) serves as the input to another function. For the targeted exercise, calculating this derivative is essential in understanding how the exponential decay affects the overall rate of change of \( y = \cos^{-1}(e^{-t}) \).