Question

Find \( \frac{d}{dx}(x^x) \) and evaluate whether this result is true \( \forall x \in \mathbb{R} \).

Hint:

Whenever a function is of the form \( f(x)^{g(x)} \), always use logarithmic differentiation. Remember that logarithmic functions restrict the domain to positive values.

Answer 1

Step 1: Understanding the Concept:
The function \( x^x \) has both a variable base and a variable exponent. This requires logarithmic differentiation to simplify the expression before taking the derivative.
Step 2: Key Formula or Approach:
Use the identity \( y = x^x \implies \log y = x \log x \).
Step 3: Detailed Explanation:
Let \( y = x^x \).
Taking natural logarithm on both sides:
\[ \log y = \log(x^x) \] \[ \log y = x \log x \] Differentiating both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \log x) \] Using the product rule \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \):
\[ \frac{1}{y} \frac{dy}{dx} = x \left( \frac{1}{x} \right) + \log x (1) \] \[ \frac{1}{y} \frac{dy}{dx} = 1 + \log x \] \[ \frac{dy}{dx} = y(1 + \log x) \] Substituting \( y = x^x \):
\[ \frac{dy}{dx} = x^x(1 + \log x) \] Evaluation of Validity:
The result is obtained using the function \( \log x \). The natural logarithm \( \log x \) is only defined for \( x>0 \).
Therefore, the function \( x^x \) as treated here is primarily defined for positive real numbers. For negative \( x \), \( x^x \) might result in complex values (e.g., \( (-2)^{-2} \) is real, but \( (-2)^{1/2} \) is imaginary).
Hence, the formula \( \frac{d}{dx}(x^x) = x^x(1 + \log x) \) is NOT true for all \( x \in \mathbb{R} \). It is valid only for \( x>0 \).
Step 4: Final Answer:
The derivative is \( x^x(1 + \log x) \) and it is only valid for \( x \in (0, \infty) \), not for all \( x \in \mathbb{R} \).