Q. \[ \frac{d}{dx}\left(x^x\right) \]

We want to find the derivative of \(y = x^x\). Because the variable appears both as the base and the exponent, we cannot use the usual power rule directly. Instead, we use logarithmic differentiation.

Step 1: Define the function and take natural logs

Let

\[ y = x^x \]

Take the natural logarithm of both sides:

\[ \ln(y) = \ln(x^x) \]

Using the log rule \(\ln(a^b)=b\ln(a)\), we get:

\[ \ln(y) = x\ln(x) \]

Step 2: Differentiate both sides

Differentiating the left side with respect to \(x\):

\[ \frac{d}{dx}\left(\ln(y)\right) = \frac{1}{y}\frac{dy}{dx} \]

Differentiating the right side:

\[ \frac{d}{dx}\left(x\ln(x)\right) \]

This is a product \(x \cdot \ln(x)\), so we use the product rule:

\[ \frac{d}{dx}\left(x\ln(x)\right) = 1\cdot \ln(x) + x\cdot \frac{1}{x} \]

Simplify:

\[ \frac{d}{dx}\left(x\ln(x)\right) = \ln(x) + 1 \]

Step 3: Set the derivatives equal

So we have:

\[ \frac{1}{y}\frac{dy}{dx} = \ln(x) + 1 \]

Step 4: Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\):

\[ \frac{dy}{dx} = y\left(\ln(x) + 1\right) \]

Recall that \(y = x^x\). Substitute back:

\[ \frac{dy}{dx} = x^x\left(\ln(x) + 1\right) \]

Final Answer

\[ \frac{d}{dx}\left(x^x\right) = x^x\left(\ln(x) + 1\right) \]