Problem 20 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ g(x)=\sqrt{x} e^{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the function

The given function is \( g(x) = \sqrt{x} e^{x} \). This function is a product of two simpler functions, \( \sqrt{x} \) and \( e^{x} \).

Step 2: Apply the Product Rule

Since \( g(x) = \sqrt{x} e^{x} \) is a product of two functions, we use the product rule to differentiate. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then their product \( u(x)v(x) \) is differentiated as: \( u'(x)v(x) + u(x)v'(x) \). Here, let \( u(x) = \sqrt{x} \) and \( v(x) = e^{x} \).

Step 3: Differentiate \( u(x) = \sqrt{x} \)

To find \( u'(x) \), differentiate \( \sqrt{x} = x^{1/2} \). Using the power rule, \( \frac{d}{dx} x^{n} = nx^{n-1} \), we get: \( u'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).

Step 4: Differentiate \( v(x) = e^{x} \)

The function \( v(x) = e^{x} \) is differentiated as \( v'(x) = e^{x} \) because the derivative of \( e^{x} \) with respect to \( x \) is itself \( e^{x} \).

Step 5: Combine results using the product rule

Substitute back into the product rule: \( g'(x) = u'(x)v(x) + u(x)v'(x) = \frac{1}{2\sqrt{x}}e^{x} + \sqrt{x}e^{x} \).

Step 6: Simplify the expression

Combine and simplify the terms: \( g'(x) = \left( \frac{1}{2\sqrt{x}} + \sqrt{x} \right)e^{x} \). To simplify further, express with a common denominator: \( g'(x) = \left( \frac{1}{2\sqrt{x}} + \frac{x}{\sqrt{x}} \right)e^{x} = \left( \frac{1 + 2x}{2\sqrt{x}} \right)e^{x} \). This is the differentiated function.

Key Concepts

Product Rule

Differentiating functions can seem a bit tricky at first, especially when dealing with products of functions. Luckily, we have a tool called the product rule which makes things easier. The product rule is specifically for cases where you have one function multiplied by another. If you have two functions, say \( u(x) \) and \( v(x) \), and you're looking to differentiate their product, the product rule tells us that the derivative is given by the expression:

• \( u'(x)v(x) + u(x)v'(x) \)

In simple terms, this means that you take the derivative of the first function while leaving the second one alone, and then you take the derivative of the second function while leaving the first one alone, and sum these two results.
It's a great way to break down a complex problem into smaller, more manageable pieces. In our exercise, the function \( g(x) = \sqrt{x} e^{x} \) was differentiated using this very rule, identifying \( \sqrt{x} \) as \( u(x) \) and \( e^{x} \) as \( v(x) \). This key concept allows us to tackle composite functions efficiently by dissecting them into simpler parts.

Power Rule

When we talk about differentiating powers of \( x \), the power rule is your best friend. This rule states that if you have a function \( x^n \), the derivative of this function is \( nx^{n-1} \). It's like a magic formula that helps simplify the process of differentiation. The new power of \( x \) is one less than the original, and you multiply by the original power.

• This is invaluable when dealing with functions like \( \sqrt{x} \).

With \( \sqrt{x} \) being equivalent to \( x^{1/2} \), the power rule tells us that the derivative is \( \frac{1}{2}x^{-1/2} \), which simplifies to \( \frac{1}{2\sqrt{x}} \).
Understanding the power rule is crucial because it not only makes the differentiation process faster but also helps in recognizing how each term in a function is affected by differentiation. This rule is a staple in calculus, underpinning the differentiation of polynomial expressions and more.

Exponential Function

The exponential function, particularly \( e^x \), has a unique property in the realm of differentiation. The derivative of \( e^x \) is itself, \( e^x \). This might seem unusual or even magical at first. However, it is a consequence of the exponential nature of the logarithmic base \( e \), which is approximately 2.718.

• Because \( e^x \) is equal to its derivative, it simplifies a lot of differentiation work.
• This means that no matter how many times you differentiate \( e^x \), you will keep getting \( e^x \).

In our exercise, where \( g(x) = \sqrt{x} e^{x} \), the presence of \( e^x \) allows for a straightforward calculation of its derivative. Understanding this property is important when working with more complex functions because \( e^x \) often appears in various contexts from growth rates to natural processes. Recognizing how to handle the exponential function's derivative is fundamental for solving calculus problems efficiently.