Question

The derivative of \(x^x\) with respective to x is

• A. \(x^x(x + \log x)\)
• B. \(x^x(x - \log x)\)
• C. \(x^x(1 - \log x)\)
• D. \(x^x(1 + \log x)\)

Hint:

The derivative of \(x^x\) is a standard result that's good to remember. Whenever you see a function in the form of (variable)(variable), logarithmic differentiation is the method to use.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to find the derivative of the function \(y = x^x\), which is a function raised to the power of a function.

Step 2: Key Formula or Approach:
For functions of the form \(y = [f(x)]^{g(x)}\), we use logarithmic differentiation.
1. Take the natural logarithm (\(\ln\)) of both sides.
2. Use logarithm properties to simplify the expression.
3. Differentiate both sides implicitly with respect to \(x\).
4. Solve for \(\frac{dy}{dx}\).

Step 3: Detailed Explanation:
Let \(y = x^x\).
Take the natural logarithm of both sides:
\[ \ln y = \ln(x^x) \]
Using the logarithm power rule, \(\ln(a^b) = b \ln a\):
\[ \ln y = x \ln x \]
Now, differentiate both sides with respect to \(x\). We use the product rule on the right side.
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x) \]
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \left(\frac{d}{dx}(x)\right) \cdot \ln x + x \cdot \left(\frac{d}{dx}(\ln x)\right) \]
\[ \frac{1}{y} \frac{dy}{dx} = (1) \cdot \ln x + x \cdot \left(\frac{1}{x}\right) \]
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]
To find \(\frac{dy}{dx}\), multiply both sides by \(y\):
\[ \frac{dy}{dx} = y (1 + \ln x) \]
Finally, substitute back \(y = x^x\):
\[ \frac{dy}{dx} = x^x (1 + \ln x) \]
Assuming \(\log x\) in the options represents the natural logarithm, this matches option (D).

Step 4: Final Answer:
The derivative of \(x^x\) is \(x^x(1 + \log x)\).