Problem 60 - Calculus and its applications

Question

Use the Chain Rule, implicit differentiation, and other techniques to differentiate each function given. $$ y=x^{x}, \text { for } x>0 $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = x^x(\ln(x) + 1) \).

Step 1: Take Natural Logarithm

Start by taking the natural logarithm of both sides of the equation to make differentiation simpler. \[ \ln(y) = \ln(x^x) \]Then apply the logarithmic identity \( \ln(a^b) = b \ln a \) to simplify:\[ \ln(y) = x \ln(x) \]

Step 2: Differentiate Both Sides Implicitly

Differentiate both sides with respect to \( x \). Remember, you need to use the chain rule on the left side since \( y \) is a function of \( x \):\[ \frac{d}{dx}[\ln(y)] = \frac{d}{dx}[x \ln(x)] \]Using the chain rule and implicit differentiation, we get:\[ \frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln(x) + x \cdot \frac{1}{x} \]Simplify the right side:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + 1 \]

Step 3: Solve for \( \frac{dy}{dx} \)

Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y(\ln(x) + 1) \]Since \( y = x^x \), substitute back:\[ \frac{dy}{dx} = x^x(\ln(x) + 1) \]

Step 4: Write Final Derivative

The derivative of \( y = x^x \) is now shown:\[ \frac{dy}{dx} = x^x(\ln(x) + 1) \]

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function that is not solved explicitly for one variable in terms of another. In our problem, the function is given as \( y = x^x \) and should be differentiated with respect to \( x \).

To start, we take the natural logarithm of both sides, which allows us to use the properties of logarithms to simplify the differentiation process. After taking the natural logarithm of both sides, we get \( \ln(y) = x \ln(x) \).

This is where implicit differentiation becomes handy. When differentiating \( \ln(y) \), we apply the chain rule, acknowledging that \( y \) is a function of \( x \). Hence, the derivative of \( \ln(y) \) with respect to \( x \) is \( \frac{1}{y} \frac{dy}{dx} \).

Using implicit differentiation allows students to analyze complex relationships between variables as we treat both sides of an equation explicitly, helping to track dependencies that direct differentiation wouldn't manage easily.

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are logarithms with base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. They are essential in calculus due to their unique property of being the inverse function of exponential equations.

In the given problem, we use natural logarithms as a tool to simplify differentiation. Initially, \( y = x^x \) doesn't easily lend itself to straightforward differentiation. By taking the natural logarithm of both sides, \( \ln(y) = \ln(x^x) \), we simplify it to \( \ln(y) = x \ln(x) \), which separates the exponent \( x \) from the base \( x \).

This use of logarithms is crucial because it reduces the exponentiation problem into a form where standard differentiation techniques, like the product rule, can be applied.

• It allows us to handle products and powers more conveniently.
• Through properties like \( \ln(a^b) = b \ln a \), they break down more intricate structures into manageable expressions.

Understanding the utility of natural logarithms in calculus makes solving complex differentiation problems far more accessible.

Derivatives

Derivatives represent the rate at which a function changes as its input changes. They are a core concept in calculus, used to determine slopes of curves, optimize functions, and understand dynamic behaviors of mathematical models.

In tackling the problem \( y = x^x \), our end goal is to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \). After manipulating the equation with logarithms and implicit differentiation, we simplify the complex expression.

Ultimately, we solve:

• Start with \( \ln(y) = x \ln(x) \) and differentiate both sides.
• Apply the chain rule to \( \ln(y) \) which gives \( \frac{1}{y} \frac{dy}{dx} \) on the left side.
• Differentiate \( x \ln(x) \) using the product rule: \( 1 \cdot \ln(x) + x \cdot \frac{1}{x} \), giving \( \ln(x) + 1 \).

After isolating \( \frac{dy}{dx} \), we find \( \frac{dy}{dx} = x^x(\ln(x) + 1) \). Hence, the derivative of the original function illustrates how both the base and exponent impact the rate of change of the function. Derivatives in this manner allow us to quantify and predict the behavior of functions, which is especially vital in applied mathematics and real-world problem-solving.