Problem 16 - Calculus Early Transcendentals

Question

Find the following derivatives. $$\frac{d}{d x}\left(e^{x} \ln x\right)$$

Step-by-Step Solution

Answer

Answer: The derivative of the function \(y = e^x \ln x\) is \(y' = \frac{e^x}{x} + e^x\ln x\).

Step 1: Use the product rule

The product rule states that if \(u(x)\) and \(v(x)\) are two functions of \(x\), then the derivative of their product, \(u(x)v(x)\), is given by $$\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$ In this case, we have \(u(x)=e^x\) and \(v(x)=\ln x\). We will now find their derivatives.

Step 2: Find the derivative of \(u(x)=e^x\)

The derivative of \(e^x\) with respect to \(x\) is simply \(e^x\). Therefore, \(\frac{du}{dx} = e^x\).

Step 3: Find the derivative of \(v(x)=\ln x\)

The derivative of \(\ln x\) with respect to \(x\) is \(\frac{1}{x}\). Therefore, \(\frac{dv}{dx} = \frac{1}{x}\).

Step 4: Apply product rule to find the derivative of \(e^x\ln x\)

Now that we have the derivatives of \(u(x)\) and \(v(x)\), we can apply the product rule: $$\frac{d(e^x \ln x)}{dx} = e^x\left(\frac{1}{x}\right) + \ln x(e^x)$$ Simplifying, we get: $$\frac{d(e^x \ln x)}{dx} = \frac{e^x}{x} + e^x\ln x$$

Key Concepts

Product Rule

When you're working with functions that are multiplied together, the product rule is your go-to tool for differentiation. It helps you find the derivative of two functions being multiplied. The product rule formula is: \[\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\] Here,

• \(u(x)\) and \(v(x)\) are the two functions of \(x\)\
• \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) are their respective derivatives\

To use the product rule, you simply differentiate each function separately and then apply the formula. It's like solving a puzzle where each piece fits together to give the complete picture of the derivative of the product. By applying the product rule, you're essentially saying that the change in the product of two functions depends on how each function changes, independently of their multiplication with the other.

Exponential Functions

Exponential functions, like \(e^x\), have unique characteristics that make them quite special in calculus. The derivative of \(e^x\) is particularly convenient because it remains unchanged; that is, \[\frac{d}{dx} e^x = e^x\] This means that the slope of the tangent to the curve \(e^x\) at any point is the same as the value of the function at that point. This self-replicating property of \(e^x\) is what sets exponential functions apart.

• Exponential growth or decay is modeled exceptionally well by exponential functions, which is one reason they appear so often in real-world applications\
• The constant \(e\) (approximately 2.718) is a mathematical constant that arises naturally in many areas, not just in calculus\

Understanding how to differentiate exponential functions using basic rules can greatly simplify the process of finding derivatives in calculus, especially when they are paired with other types of functions, such as logarithmic functions.

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm \(\ln x\), have unique properties that make their differentiation straightforward. The derivative of \(\ln x\) is: \[\frac{d}{dx} \ln x = \frac{1}{x}\] This tells us that the rate of change of the natural logarithm decreases as \(x\) increases. In other words, the tangent line to the curve \(\ln x\) becomes less steep as \(x\) grows larger.

• \(\ln x\) is the inverse function of the exponential function \(e^x\), providing a powerful relationship in calculus\
• The property \(\ln(ab) = \ln a + \ln b\) simplifies many calculations involving logarithms\

When differentiating products of exponential and logarithmic functions, such as in the example \(e^x \ln x\), utilizing the derivatives of these basic functions and applying the product rule leads to efficient and effective solutions.