Differentiate $x^x$ with logarithmic differentiation

Step 1 - Set up

We want the derivative of x^x.

Step 1 - Set up

Let y = x^x.

Step 1 - Set up

Since x is both the base and the exponent, use logarithmic differentiation.

Step 2 - Take logs

Take natural logarithms on both sides.

Step 2 - Take logs

This brings the exponent down where we can differentiate it.

Step 2 - Take logs

So ln y = ln(x^x).

Step 2 - Take logs

Which simplifies to x ln x.

Step 3 - Differentiate

Differentiate both sides with respect to x.

Step 3 - Differentiate

On the left, chain rule gives (1/y)(dy/dx).

Step 3 - Differentiate

On the right, product rule gives x(1/x) + (ln x)(1).

Step 3 - Differentiate

So the right-hand side simplifies to ln x + 1.

Step 3 - Differentiate

Therefore (1/y)(dy/dx) = ln x + 1.

Step 4 - Substitute back

Now multiply both sides by y.

Step 4 - Substitute back

This gives dy/dx = y(ln x + 1).

Step 4 - Substitute back

Finally, substitute y = x^x.

Step 5 - Final answer

So the final derivative is x^x(ln x + 1).

Step 5 - Final answer

If x appears in both the base and the exponent, use logarithmic differentiation.

Step 5 - Final answer

For real x, assume x > 0 so that ln x is defined.