Problem 25 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \arctan \left(e^{x}\right) $$

Step-by-Step Solution

Answer

The derivative of \( \arctan(e^x) \) with respect to \( x \) is \( \frac{e^x}{1+e^{2x}} \).

Step 1: Identify the structure of the function

The given function is \( f(x) = \arctan(e^x) \). This is a composition of two functions - the outer function \( g(u) = \arctan(u) \) and the inner function \( u(x) = e^x \). To differentiate this, we'll need to apply the chain rule.

Step 2: Differentiate the outer function

The derivative of \( g(u) = \arctan(u) \) with respect to \( u \) is: \[ g'(u) = \frac{1}{1+u^2} \] This gives us the derivative of the arctangent function.

Step 3: Differentiate the inner function

The inner function is \( u(x) = e^x \). Its derivative with respect to \( x \) is: \[ u'(x) = \frac{d}{dx}(e^x) = e^x \]

Step 4: Apply the chain rule

Using the chain rule, \( \frac{df}{dx} = g'(u) \cdot u'(x) \). Substitute the derivatives from the previous steps: \[ \frac{df}{dx} = \frac{1}{1+(e^x)^2} \cdot e^x \] Simplify this expression:

Step 5: Simplify the result

We have \( \frac{1}{1+(e^x)^2} \cdot e^x \). Rewrite \( (e^x)^2 \) as \( e^{2x} \), so the expression becomes: \[ \frac{e^x}{1+e^{2x}} \] This is the derivative of the given expression.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus, widely used for differentiating a function that is composed of other functions. The main idea is to differentiate these functions one after the other, by working from the outermost function inward. Think of it like peeling an onion, layer by layer. In this exercise, the function \( f(x) = \arctan(e^x) \) is split into two parts. The outer function is \( g(u) = \arctan(u) \) and the inner function is \( u(x) = e^x \). First, you differentiate the outer function as if the inner function is just a simple variable \( u \). You get \( g'(u) = \frac{1}{1+u^2} \). Then, differentiate the inner function \( u(x) = e^x \), which results in \( u'(x) = e^x \). The Chain Rule then multiplies these derivatives together, giving:

• \( \frac{df}{dx} = g'(u) \cdot u'(x) \)

For our function, it becomes \( \frac{df}{dx} = \frac{1}{1+(e^x)^2} \cdot e^x \), leading to the final simplified derivative \( \frac{e^x}{1+e^{2x}} \). This method works so well because it allows you to handle complex compositions by breaking them into manageable parts.

Exponential Function

Exponential functions are a type of mathematical function that involve the constant \( e \) (approximately 2.718), which is the base of natural logarithms. They are widely used for their unique properties, such as modelling growth processes and compound interest. The simplest exponential function is \( e^x \), where the variable \( x \) is the exponent. These functions are particularly interesting because their differentiation is straightforward: the derivative of \( e^x \) with respect to \( x \) is exactly \( e^x \). This characteristic makes them convenient to work with, especially in calculus and differential equations.
In our exercise, the exponential function is the inner function in the composition \( f(x) = \arctan(e^x) \). While utilizing the Chain Rule, we first calculate the derivative of the exponential part, \( e^x \), as \( e^x \).
This result is foundational for further calculations, seamlessly integrating into the chain rule to find the derivative of the full expression \( \frac{e^x}{1+e^{2x}} \). Many complex natural processes, like population growth or radioactive decay, are modeled using exponential functions because of their pervasiveness and elegant property when differentiated.

Inverse Trigonometric Functions

Inverse trigonometric functions, like the arctangent function \( \arctan(x) \), provide angles as outputs for given trigonometric ratios. They are essential in numerous applications that require the determination of angles from known sine, cosine, or tangent values. The inverse trigonometric function used in this exercise, \( \arctan(u) \), is the inverse of the tangent function. When differentiating such functions, each has its own unique formulae. Specifically, the derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \) with respect to \( u \), which showcases their distinct nature compared to regular trigonometric functions.
In the context of our exercise, \( \arctan(e^x) \) forms the outer function when applying the Chain Rule. Therefore, part of differentiating the composed function involves using the derivative mentioned, \( \frac{1}{1+(e^x)^2} \).
This derivative describes how the function \( \arctan(u) \) changes with variations in \( u \), and it is especially important for correctly applying the Chain Rule.

• By combining it with the derivative of the inner function, it results in the final expression \( \frac{e^x}{1+e^{2x}} \)

For students or anyone working with these inverse trigonometric derivatives, it is crucial to remember their specific formulas, as they often appear in calculus problems related to compositions of functions.