Explanation

We want to find the derivative of:

\(y = x^x\)

where:

\(x>0\)

This function is special because:

• the base is $x$
• and the exponent is also $x$

So it is not:

• a normal power function like $x^k$
• or a normal exponential function like $a^x$

Because of this, the usual differentiation rules cannot be applied directly.

In this lesson, we solve the derivative of $x^x$ in two ways:

1. Using exponential and logarithm properties
2. Using logarithmic differentiation

Formula / Rule

Logarithm Property

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

Derivative of $x^x$

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

Example

Method 1 — Rewrite Using Exponentials

Rewrite:

\(x^x=e^{ln(x^x)}\)

Use the logarithm property:

\(x^x=e^{x ln(x)}\)

Differentiate:

\( \frac{d}{dx}​(x^x)=\frac{d}{dx} ​(e^{x ln(x)}) \)

Using the chain rule:

\( \frac{d}{dx}​(x^x) =e^{xln(x)} \frac{d}{dx} ​(xln(x)) \)

Using the product rule:

\( \frac{d}{dx} ​(xln(x))=ln(x)+1 \)

So:

\( \frac{d}{dx}​(x^x) = e^{x ln(x)} (ln(x)+1) \)

Since:

\( e^{x ln(x)} = e^{ ln(x^x)} = x^x\)

So:

\( \frac{d}{dx}​(x^x) = x^x (ln(x)+1) \)

Method 2 — Logarithmic Differentiation

Let:

\(y=x^x\)

Take natural logarithm for both sides:

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

Using the logarithm property:
$$$$\( ln(y)=xln(x) \)

Differentiate implicitly:

\( \frac{y′}{y} ​=ln(x)+1 \)

Multiply both sides by \(y\):

\( y′=y(ln(x)+1) \)

Substitute:

\( y=x^x \)

So:

\( y′=x^x (ln(x)+1) \)

Both methods give the same answer.

Video Explanation

Practice

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