Problem 30 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Find the domains of the function and its derivative. \(f(x)=\arcsin \left(e^{x}\right)\)

Step-by-Step Solution

Answer

The derivative is \( \frac{e^x}{\sqrt{1-e^{2x}}} \). The domain of \( f(x) \) is \( x = 0 \), and the domain of \( f'(x) \) is \( x < 0 \).

Step 1: Find the Derivative

To find the derivative of the function \( f(x) = \arcsin(e^x) \), we need to apply the chain rule. The derivative of \( \arcsin(u) \) is \( \frac{1}{\sqrt{1-u^2}} \), and the derivative of \( e^x \) is \( e^x \). Therefore, the derivative is:\[ f'(x) = \frac{1}{\sqrt{1-(e^x)^2}} \cdot e^x = \frac{e^x}{\sqrt{1-e^{2x}}} \]

Step 2: Determine the Domain of the Original Function

The function \( \arcsin(e^x) \) is defined if \( -1 \leq e^x \leq 1 \). Since \( e^x \) is always positive, you only need to consider \( 0 \leq e^x \leq 1 \), which reduces to the inequality \( 0 \leq x \leq 0 \). Therefore, the domain of \( f(x) \) is \( x = 0 \).

Step 3: Determine the Domain of the Derivative

The derivative \( f'(x) = \frac{e^x}{\sqrt{1-e^{2x}}} \) is defined if the denominator is not zero and the expression under the square root is positive. The function \( \sqrt{1 - e^{2x}} \) is defined and non-zero when \( 1 > e^{2x} \), or equivalently \( x < 0 \). The derivative is defined for \( x < 0 \).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. When you have a function nested inside another function, like in our exercise where we have \(f(x) = \arcsin(e^x)\), the chain rule comes into play.

• First, identify the outer function and the inner function. Here, the outer function is \(\arcsin(u)\), and the inner function is \(u = e^x\).
• The chain rule formula states that the derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\).
• Apply this by taking the derivative of the outer function with respect to the inner function and multiply by the derivative of the inner function.

In the solution, the derivative of \(\arcsin(u)\) is \(\frac{1}{\sqrt{1-u^2}}\), and the derivative of \(e^x\) is \(e^x\). Multiply these together to get \( \frac{e^x}{\sqrt{1-e^{2x}}} \), which is the derivative of the given function using the chain rule.

Function Domain

The domain of a function is the set of all possible input values (usually \(x\)) for which the function is defined. With our function \(f(x) = \arcsin(e^x)\), we need to ensure that what's inside the arcsine function stays between \(-1\) and \(1\), because those are the limits for which the arcsine function is valid.

• First, look at the expression \(e^x\), which can only take positive values.
• This transforms our domain condition to \(0 \leq e^x \leq 1\).
• Solving this inequality for \(x\) gives \(0 \leq x \leq 0\), meaning only \(x = 0\) is valid.

Thus, the domain of the original function is a single point, \(x = 0\).
For the derivative, we must prevent the denominator from being zero and ensure the expression under the square root remains positive. This requires solving \(1 > e^{2x}\), leading us to determine the domain of the derivative as \(x < 0\). This means the derivative exists only for negative \(x\) values.

Exponential Functions

Exponential functions have the general form \(f(x) = a^{x}\) where \(a\) is a positive constant. In our exercise, we have the exponential function \(e^x\), which is one of the most common and important exponential functions due to its natural growth rate properties.

• Exponential growth is characterized by a constant relative growth rate, leading to the function's values rapidly increasing as \(x\) increases.
• The derivative of \(e^x\) is unique because it is equal to itself; \(\frac{d}{dx} e^x = e^x\).

In the problem, \(e^x\) is what makes the function more complex under the arcsine transformation. When solving the exercise, understanding the behavior and properties of exponential functions is key because it influences the derivative and the domain conditions, especially due to \(e^x\)'s range being \((0, \infty)\). This property dictated the allowed values for \(x\) resulting in specific domains for both the function and its derivative.