Question

Differentiate \(x^x\) with respect to \(x \log x\).

Hint:

Note that \(x \log x\) is actually the natural logarithm of \(x^x\).
Hence, you are essentially differentiating \(e^v\) with respect to \(v\), which is why the result is simply the original function.

Answer 1

Concept:
• Differentiation of a function with respect to another function: \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\).
• Logarithmic differentiation is used when the base and exponent both contain the variable.

Step 1: Differentiate \(u = x^x\) with respect to \(x\)
Let \(u = x^x\). Taking natural log on both sides: \[ \log u = x \log x \] Differentiating with respect to \(x\): \[ \frac{1}{u} \cdot \frac{du}{dx} = \frac{d}{dx}(x \log x) \] Using product rule: \[ \frac{1}{u} \cdot \frac{du}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \] \[ \frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x) \]

Step 2: Differentiate \(v = x \log x\) with respect to \(x\)
Let \(v = x \log x\). Differentiating with respect to \(x\) using product rule: \[ \frac{dv}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 \] \[ \frac{dv}{dx} = 1 + \log x \]

Step 3: Find the final derivative \(du/dv\)
Using the chain rule in functional form: \[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \] \[ \frac{du}{dv} = \frac{x^x(1 + \log x)}{1 + \log x} \] Cancelling the common term \((1 + \log x)\): \[ \frac{du}{dv} = x^x \]