Problem 35 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dr} = -\frac{1}{\sqrt{e^{2r}-1}} \).

Step 1: Recall the Derivative Formula for Inverse Cosecant

The derivative of the inverse cosecant function \( \csc^{-1}(u) \) with respect to \( u \) is given by \( \frac{d}{du}[\csc^{-1}(u)] = -\frac{1}{|u|\sqrt{u^2-1}} \). This formula assumes \( |u|>1 \).

Step 2: Apply the Chain Rule

The function we have is \( y = \csc^{-1}(e^r) \). Here, \( u = e^r \). By the chain rule, the derivative \( \frac{dy}{dr} \) is \( \frac{dy}{du} \cdot \frac{du}{dr} \).

Step 3: Differentiate \( u = e^r \) with Respect to \( r \)

The derivative of \( u = e^r \) with respect to \( r \) is \( \frac{du}{dr} = e^r \).

Step 4: Differentiate \( y = \csc^{-1}(u) \) with Respect to \( u \)

Using the formula from Step 1, the derivative of \( y = \csc^{-1}(u) \) with respect to \( u \) is \( \frac{dy}{du} = -\frac{1}{|e^r|\sqrt{(e^r)^2-1}} \).

Step 5: Combine the Results Using the Chain Rule

Substitute the derivatives from Step 3 and Step 4 into the chain rule formula: \[ \frac{dy}{dr} = \left(-\frac{1}{|e^r|\sqrt{(e^r)^2-1}}\right) \cdot e^r = -\frac{1}{\sqrt{(e^r)^2-1}} \].

Step 6: Simplify the Expression

Since \( e^r \) is always positive, \( |e^r| = e^r \). Thus, the expression for the derivative simplifies to \[ \frac{dy}{dr} = -\frac{1}{\sqrt{e^{2r}-1}} \].

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus used for finding the derivative of composite functions. Basically, it helps us differentiate functions that are "nested" inside each other. To put it simply, when you have a function inside a function, the Chain Rule is your go-to tool.

• Suppose you have a composite function, say \( y = f(g(x)) \). The Chain Rule states that the derivative of \( y \) with respect to \( x \) is \( f'(g(x)) \cdot g'(x) \).
• One way to remember this is: take the derivative of the "outer" function, multiply it by the derivative of the "inner" function.

In our exercise, we used the Chain Rule to derive the function \( y = \csc^{-1}(e^r) \). Here, the inverse cosecant function \( \csc^{-1} \) is the outer function, and \( e^r \) is the inner function, making the Chain Rule necessary to appropriately manage the differentiation process.

Inverse Cosecant Function

The inverse cosecant function, denoted as \( \csc^{-1} \), is the inverse of the cosecant function. The cosecant function itself is the reciprocal of the sine function, making \( \csc(x) = \frac{1}{\sin(x)} \).

• The domain of the inverse cosecant function is \( |x| > 1 \), since \( \csc(x) \) is only defined for values greater than 1 and less than -1.
• Understanding the domain is crucial, as it affects the conditions under which the derivative is valid.
• The derivative of \( \csc^{-1}(u) \) is \( -\frac{1}{|u|\sqrt{u^2-1}} \).

In the given problem, this derivative formula was essential when applying the Chain Rule to find the rate of change with respect to \( r \), helping us break down the differentiation step by step.

Exponential Function

The exponential function, particularly \( e^r \), is a fundamental component in calculus and is characterized by its unique properties.

• One key property of the exponential function is that its derivative is itself, i.e., \( \frac{d}{dr}(e^r) = e^r \). This makes differentiation straightforward.
• Since the base of the exponential function \( e \) (approximately 2.71828) is greater than 1, \( e^r \) is always positive for real values of \( r \).

When using the Chain Rule, the derivative of \( e^r \) was pivotal as it served as the "inner" function's derivative, simplifying the derivative of the composite function \( \csc^{-1}(e^r) \).

Differentiation Techniques

Differentiation, a cornerstone of calculus, involves finding the rate at which a function changes. Various techniques make complex differentiation problems manageable. Let's break down some of these methods:

• **Basic Rules**: Start by mastering the derivatives of common functions like polynomials, trigonometric, and exponential functions.
• **Product Rule**: Use when multiplying two functions. If \( h(x) = f(x) \cdot g(x) \), then the derivative \( h'(x) = f'(x)g(x) + f(x)g'(x) \).
• **Quotient Rule**: Use when one function divides another. If \( h(x) = \frac{f(x)}{g(x)} \), then \( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \).
• **Chain Rule**: Use for composite functions like in our problem, critical for understanding functions nested within each other.

All these tools form the backbone of tackling differentiation problems efficiently, like the exercise we solved using the Chain Rule combined with the rules for inverse trigonometric and exponential functions.