Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the function given. \(y=\frac{\arctan \left(e^{x}\right)}{e}\)

Step-by-Step Solution

Answer

The derivative of the function \( y=\frac{\arctan \left(e^{x}\right)}{e} \) is \( y'=\frac{e^x}{1+e^{2x}} \)

Step 1: Identify Function Structure

This function has a structure of a quotient, where the numerator is an inverse tangent function \( \arctan (e^x) \) and the denominator is the constant \( e \)

Step 2: Apply Chain Rule

The derivative of \( \arctan\left(e^{x}\right) \) can be obtained by using the chain rule. The derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \). By letting \( u=e^x \), the derivative becomes \( \frac{1}{1+(e^x)^2} \cdot e^x \) which simplifies to \( \frac{e^x}{1+e^{2x}} \).

Step 3: Quotient Rule

To calculate the derivative of the quotient \( \frac{\arctan(e^{x})}{e} \), the formula for the quotient rule of derivatives will be used: (f/g)' = (g f' - f g') / g^2. In this case, f = \( \arctan(e^{x}) \), g = e, f' = \( \frac{e^x}{1+e^{2x}} \), and g' = 0 as e is a constant. Substituting these into the quotient rule gives the derivative of the function as \( \frac{e(\frac{e^x}{1+e^{2x}})-0}{e^2} \).

Step 4: Simplify the Derivative

The derivative simplifies to \( \frac{e^x}{e(1+e^{2x})} \). Because the denominator \( e \) is a constant, the final derivative becomes \( \frac{e^x}{1+e^{2x}} \).

Key Concepts

Quotient Rule

When we want to differentiate a function that is a quotient, such as the one given by \( \frac{\arctan(e^x)}{e} \), we employ the Quotient Rule. The Quotient Rule provides a way to find the derivative of a division of two functions. It states that if you have a function \( f(x) \) divided by \( g(x) \), the derivative is expressed as:

• \( (\frac{f}{g})' = \frac{g \cdot f' - f \cdot g'}{g^2} \)

In our problem, the numerator \( f(x) = \arctan(e^x) \) and the denominator \( g(x) = e \). With this rule:

• \( f'(x) \) is found using the Chain Rule, which we'll explore next.
• Since \( g \) is a constant, \( g'(x) = 0 \).

By substituting these into the quotient formula, we simplify by multiplying only the non-zero terms. This makes the calculation easier and avoids unnecessary steps, especially when dealing with constants. With the basic setup, the remainder of the work is handled by the individual derivatives.

Chain Rule

The Chain Rule is incredibly useful for differentiating composite functions. Our function \( \arctan(e^x) \) is a composition of the inverse tangent function and the exponential function. The Chain Rule allows us to differentiate this by taking the derivative of the outside function and then multiplying it by the derivative of the inside function. In formula terms:

• If \( y = f(g(x)) \), then the derivative \( \, \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

For \( \arctan \), the derivative is \( \frac{1}{1+u^2} \), and we let \( u = e^x \). We then find \( u' = e^x \), the derivative of the inside function, and multiply it by the outside derivative:

• \( \frac{1}{1+(e^x)^2} \cdot e^x = \frac{e^x}{1+e^{2x}} \).

This product gives us the necessary form to plug into the Quotient Rule, making it a crucial step in our solution.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arctan \), have unique derivatives that are essential to know. These functions are the opposites of the regular trigonometric functions but applied inversely. The derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \), a pattern distinct from their trigonometric counterparts which is useful in various calculus problems.
When combined with other functions, such as exponential ones, these derivatives require careful application of rules like the Chain Rule to differentiate them correctly. In our exercise, \( \arctan(e^x) \) demonstrated how combining it with exponentials or other non-trivial expressions showed up as part of composite functions.
Learning these derivatives not only helps in calculus but is foundational for understanding how inverse trigonometric principles apply when breaking down complex functions into simpler derivatives.