To differentiate $y = x^x$, we'll proceed as follows:

Step 1: Rewrite the function using logarithms

Given: $y = x^x$

Take the natural logarithm on both sides: $\ln(y) = \ln(x^x)$

Using the logarithmic identity $\ln(a^b) = b \ln(a)$, we get: $\ln(y) = x \ln(x)$

Step 2: Differentiate implicitly

Differentiate both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln(x))$

Now, apply the product rule to $x \ln(x)$: $\frac{d}{dx}(x \ln(x)) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$

Therefore: $\frac{1}{y} \frac{dy}{dx} = \ln(x) + 1$

Step 3: Solve for $\frac{dy}{dx}$

Multiply both sides by $y$: $\frac{dy}{dx} = y(\ln(x) + 1)$

Substituting $y = x^x$: $\frac{dy}{dx} = x^x (\ln(x) + 1)$

Final answer

$\frac{dy}{dx} = x^x (\ln(x) + 1)$

Would you like more details or have any questions?

Related Questions

1. How do you differentiate $y = x^{x^2}$?
2. How would you integrate $x^x$?
3. What is the derivative of $y = x^a$ where $a$ is a constant?
4. How do you find the second derivative of $y = x^x$?
5. How does the differentiation change for $y = a^x$ where $a$ is a constant?

Tip

Always consider using logarithmic differentiation when both the base and exponent involve the variable, as it simplifies the process.