Question

The derivative of $(\log x)^x$ with respect to $\log x$ is

• A. $(\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right]$
• B. $(\log x)^x \left[ \log x + \frac{1}{\log (\log x)} \right]$
• C. $x (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right]$
• D. $x (\log x)^x \left[ \log x + \frac{1}{\log (\log x)} \right]$

Hint:

To differentiate $u$ with respect to $v$, always compute $\frac{du}{dx}$ and $\frac{dv}{dx}$ separately and divide them. Don't forget the Chain Rule!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the derivative of $u = (\log x)^x$ with respect to $v = \log x$, which is $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.

Step 2: Key Formula or Approach:
Use logarithmic differentiation for $u$ and the chain rule for $v$.

Step 3: Detailed Explanation:
Let $u = (\log x)^x$. Then $\log u = x \log (\log x)$.
Differentiating both sides with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log (\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log (\log x) + \frac{1}{\log x}$.
So, $\frac{du}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right]$.
Now, let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right] \cdot x$.

Step 4: Final Answer:
The derivative is $x (\log x)^x \left[ \frac{1}{\log x} + \log (\log x) \right]$, which is option (C).