Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=\sqrt{e^{2 x}+2 x} $$

Short answer

The derivative is \( f'(x) = \frac{e^{2x} + 1}{\sqrt{e^{2x} + 2x}} \).

Step-by-step solution

Identify the Form of the Function

The function is given by \( f(x) = \sqrt{e^{2x} + 2x} \). We recognize this as a composition of functions: an outer function \( g(u) = \sqrt{u} \) and an inner function \( u(x) = e^{2x} + 2x \).

Differentiate the Outer Function

The derivative of the outer function \( g(u) = \sqrt{u} \) is \( g'(u) = \frac{1}{2\sqrt{u}} \).

Differentiate the Inner Function

The derivative of the inner function \( u(x) = e^{2x} + 2x \) is \( u'(x) = 2e^{2x} + 2 \) using the sum rule and the chain rule for the exponential term.

Apply the Chain Rule

Using the chain rule, we find \( f'(x) = g'(u(x)) \cdot u'(x) \). Substitute the expressions we found: \( f'(x) = \left(\frac{1}{2\sqrt{e^{2x} + 2x}}\right) \cdot (2e^{2x} + 2) \).

Simplify the Expression

Simplifying further, \( f'(x) = \frac{2e^{2x} + 2}{2\sqrt{e^{2x} + 2x}} \), which can be further simplified to \( f'(x) = \frac{e^{2x} + 1}{\sqrt{e^{2x} + 2x}} \).

Key concepts

Chain Rule

The chain rule is fundamental in calculus when differentiating composite functions. It essentially allows us to find the derivative of a function that is the composition of two or more functions. In the context of our exercise, we have an outer and an inner function.
The chain rule states that if a function is written as a composition of two functions, say \( f(x) = g(u(x)) \), then the derivative \( f'(x) \) is the product of the derivative of the outer function \( g \) evaluated at the inner function \( u(x) \) and the derivative of the inner function \( u \).
Let's break this down into a simpler process:

• Identify the outer function \( g(u) \).
• Identify the inner function \( u(x) \).
• Compute the derivative of the outer function, \( g'(u) \), and evaluate it at \( u(x) \).
• Compute the derivative of the inner function, \( u'(x) \).
• Multiply these derivatives to find \( f'(x) \).

This approach is perfect for complex expressions like the one given in the exercise.

Function Composition

Function composition is a powerful concept where the output of one function becomes the input of another. When solving calculus problems, we often deal with functions that are not standalone and are instead combinations or compositions of several simpler functions.
Let's say we have two functions: \( g(u) = \sqrt{u} \) and \( u(x) = e^{2x} + 2x \). The composition \( f(x) = g(u(x)) \) means we are plugging the entire expression of \( u(x) \) into \( g \).

• The key task is to clearly identify what constitutes the outer function and what constitutes the inner function.
• In our exercise, the outer function is \( \sqrt{u} \), which represents the square root of whatever is inside, while the inner function is the expression under the square root \( e^{2x} + 2x \).

Understanding and correctly identifying these components is crucial as it sets the stage for applying the chain rule effectively.

Derivatives of Exponential Functions

Exponential functions have their own set of rules for differentiation. A basic exponential function like \( e^{2x} \) requires the use of both the exponential rule and sometimes the chain rule.
In our exercise, the inner function, \( u(x) = e^{2x} + 2x \), contains an exponential part. Let's see how its derivative is computed:

• The derivative of \( e^{2x} \) involves using the chain rule. The derivative of \( e^{2x} \) is \( 2e^{2x} \) because the derivative of \( e^{k} \) is \( e^{k} \) multiplied by the derivative of the exponent \( k \).
• The derivative of a constant times \( x \), like \( 2x \), is straightforward: it’s simply the constant, in this case, 2.

Combining these gives us the derivative of the inner function, \( u'(x) = 2e^{2x} + 2 \). Understanding the behavior of exponential functions and their derivatives is very useful in calculus.

Simplifying Expressions

After applying calculus rules to find derivatives, the final step is often to simplify the expressions. Simplification helps to make the resulting derivative easier to understand or to use further.
In our example, the expression we got was \( f'(x) = \frac{2e^{2x} + 2}{2\sqrt{e^{2x} + 2x}} \).
Let's go through the steps to simplify:

• Initially, divide all terms in the numerator by 2 to simplify the fraction. This results in \( \frac{e^{2x} + 1}{\sqrt{e^{2x} + 2x}} \).
• By reducing complexity in calculating numerically, we make the expression more compact and manageable while maintaining its equivalence.

Simplifying not only makes the function look neat but also prepares it for subsequent mathematical analyses or solving other problems.