Question

For \( x<0 \), \( \frac{d}{dx} [|x|^x] \) is given by: 

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

Hint:

When differentiating power functions of the form \( f(x)^g(x) \), taking the logarithm first can simplify the differentiation process significantly.

Answer 1

The Correct Option is B

Step 1: Expressing \( |x|^x \) in a Differentiable Form
For \( x<0 \), we rewrite the given function using \( -x \): \[ |x|^x = (-x)^x. \] Taking the natural logarithm on both sides: \[ \ln y = x \ln(-x). \] Step 2: Differentiating Both Sides
Differentiating both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \ln(-x) + x \cdot \frac{1}{-x} \cdot (-1). \] Simplifying: \[ \frac{1}{y} \frac{dy}{dx} = \ln(-x) + 1. \] Step 3: Substituting \( y = (-x)^x \)
\[ \frac{dy}{dx} = (-x)^x [1 + \log(-x)]. \] Final Answer: \( \boxed{(-x)^x [1 + \log(-x)]} \).

Answer 2

Step 1: Simplify the expression for \( x < 0 \) 

For \( x < 0 \), we know that: \[ |x| = -x \Rightarrow |x|^x = (-x)^x \]

Step 2: Take logarithm to simplify

Let \( y = (-x)^x \)
Take natural logarithm on both sides: \[ \ln y = x \ln(-x) \]

Step 3: Differentiate using chain rule

Differentiate both sides with respect to \( x \): \[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} \left( x \ln(-x) \right) \] Use product rule: \[ \frac{d}{dx} \left( x \ln(-x) \right) = \ln(-x) + x \cdot \frac{d}{dx} [\ln(-x)] \] Note: \[ \frac{d}{dx} [\ln(-x)] = \frac{-1}{x} \Rightarrow x \cdot \left( \frac{-1}{x} \right) = -1 \] So: \[ \frac{dy}{dx} \cdot \frac{1}{y} = \ln(-x) - 1 \Rightarrow \frac{dy}{dx} = y [\ln(-x) - 1] \] Recall \( y = (-x)^x \), so: \[ \frac{d}{dx} [|x|^x] = (-x)^x [\ln(-x) - 1] = (-x)^x [1 + \log(-x)] \] since \( \ln a = \log_e a \) and \( \log(-x) \) in base \( e \).

Answer:

\( \boxed{ (-x)^x [1 + \log(-x)] } \)