Finding the Derivative of $y = x^x$

To find the derivative $\frac{dy}{dx}$ of the function $y = x^x$, we use the method of logarithmic differentiation.

Step 1: Take the Natural Logarithm

Start with the given function:

$ y = x^x $

Take the natural logarithm ($\ln$) of both sides:

$ \ln y = \ln(x^x) $

Step 2: Simplify using Logarithm Properties

Use the logarithm property $\ln(a^b) = b \ln a$ to simplify the right side:

$ \ln y = x \ln x $

Step 3: Differentiate Implicitly

Differentiate both sides of the equation with respect to $x$. Remember to use the chain rule for $\ln y$ and the product rule for $x \ln x$.

Derivative of the left side:

$ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} $

Derivative of the right side (using product rule: $\frac{d}{dx}(uv) = u'v + uv'$):

$ \frac{d}{dx}(x \ln x) = \left( \frac{d}{dx}(x) \right) \ln x + x \left( \frac{d}{dx}(\ln x) \right) $

$ = (1) \ln x + x \left( \frac{1}{x} \right) $

$ = \ln x + 1 $

Step 4: Equate and Solve for $\frac{dy}{dx}$

Set the derivatives of both sides equal:

$ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 $

Multiply both sides by $y$ to isolate $\frac{dy}{dx}$:

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

Step 5: Substitute Back the Original Function

Substitute $y = x^x$ back into the equation:

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

Conclusion

The derivative of $y = x^x$ is $x^x(\ln x + 1)$.