Problem 96 - Calculus and its applications

Question

Differentiate. $$ f(x)=e^{\sqrt{x}}+\sqrt{e^{x}} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the differentiation rules involved

To differentiate the given function, we need to use the chain rule. For a function of the form \( e^{u(x)} \), the derivative is \( e^{u(x)} \cdot u'(x) \). For \( \sqrt{u(x)} \), expressed as \( (u(x))^{1/2} \), the derivative is \( \frac{1}{2}(u(x))^{-1/2} \cdot u'(x) \).

Step 2: Differentiate the first term

The first term is \( e^{\sqrt{x}} \). Let \( u(x) = \sqrt{x} = x^{1/2} \), then \( u'(x) = \frac{1}{2}x^{-1/2} \). Applying the chain rule, the derivative of \( e^{\sqrt{x}} \) is \( e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2} \).

Step 3: Differentiate the second term

The second term is \( \sqrt{e^{x}} \), which can be written as \( (e^{x})^{1/2} \). Let \( v(x) = e^x \), then \( v'(x) = e^x \). Applying the chain rule, the derivative of \( (e^{x})^{1/2} \) is \( \frac{1}{2}(e^{x})^{-1/2} \cdot e^x \).

Step 4: Simplify the derivatives

For the first term: \( \frac{1}{2} e^{\sqrt{x}} x^{-1/2} \) is already simplified. For the second term: \( \frac{1}{2} e^{x/2} \), as \( (e^x)^{1/2} = e^{x/2} \) means \( e^x (e^x)^{-1/2} = e^{x/2} \).

Step 5: Write the final differentiated function

Combine the simplified derivatives of each term: \( f'(x) = \frac{1}{2} e^{\sqrt{x}} x^{-1/2} + \frac{1}{2} e^{x/2} \).

Key Concepts

Chain Rule

The chain rule is a key tool in calculus used for differentiating composite functions. A composite function is essentially a function within another function. The rule simplifies the process of differentiation when a function is not in its basic form. Specifically, when you have a function like \( f(g(x)) \), and you need to find its derivative, you would do the following:

• First, differentiate the outer function \( f \), treating \( g(x) \) as a single entity, which we call \( f'(g(x)) \).
• Next, differentiate the inner function \( g(x)\), resulting in \( g'(x) \).
• Finally, multiply these two results to get the overall derivative: \( f'(g(x)) \times g'(x) \).

Applying the chain rule is crucial when dealing with functions like \( e^{\sqrt{x}} \) and \( \sqrt{e^x} \). Here, we identify the inner functions \( u(x) = \sqrt{x} \) and \( v(x) = e^x \) respectively, differentiate them, and then apply the chain rule to find their derivatives effectively.

Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent, commonly expressed as \( a^x \), where \( a \) is a constant base. The most frequent base used in calculus is \( e \), approximately equal to 2.718, which makes the function \( e^x \). These functions are unique in their differentiation properties.
Differentiating \( e^x \) is straightforward since its derivative remains \( e^x \). However, when you deal with more complex expressions such as \( e^{u(x)} \), the chain rule is required. The derivative in such a scenario becomes:

• Multiply \( e^{u(x)} \) by the derivative of the inner function \( u'(x) \), leading to the result \( e^{u(x)} \cdot u'(x) \).

In our exercise challenge, the expression \( e^{\sqrt{x}} \) involved using the chain rule to handle the inner function \( \sqrt{x} \) properly. Such steps highlight how exponential functions are managed in differentiation, maintaining respect to their inherent properties.

Square Root Functions

Square root functions involve the radical expression \( \sqrt{x} \) and can be rewritten in an exponential form as \( x^{1/2} \). This conceptualization is crucial as it aids in applying differentiation rules more smoothly. When differentiating these functions, a fundamental rule is key:

• The derivative of \( x^{n} \) for any power \( n \) is given by \( nx^{n-1} \).

Thus, the derivative of a function like \( \sqrt{x} = x^{1/2} \) becomes:

• \( \frac{1}{2} x^{-1/2} \), which is achieved by lowering the power by 1 and multiplying by the original exponent.

This knowledge becomes vital in handling expressions like \( \sqrt{e^x} \). Here, the square root is treated using this power rule, and combined with the chain rule, we effectively differentiate composite functions involving a square root and another function like \( e^x \). These principles ensure clarity and precision in tackling differentiated terms in calculus.