The derivative of x^{x} is equal to x^{x}(1+ln x). That is, d/dx(x^{x})=x^{x}(1+ln x). This can be proved by the logarithmic differentiation. Here, ln=log_{e}. In this post, we will learn how to find the derivative of x to the x.

## Derivative of x^{x} Formula

The formula of x^{x} derivative is given by

d/dx(x^{x})=x^{x}(1+ln x)

## Derivative of x^{x} Proof

Now, we will give a proof of the derivative of x^{x}, that is, d/dx(x^{x}) = x^{x}(1+ln x).

*Proof:*

Let y=x^{x}.

Taking the natural logarithms both sides, we have

ln y = ln x^{x}

⇒ ln y =x ln x.

Taking d/dx on both sides, we have

$\dfrac{d}{dx}(\ln y)=\dfrac{d}{dx}{x \ln x}$

Now, applying the product rule of derivatives, it follows that

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

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=x \cdot \dfrac{1}{x}+\ln x$ as the derivative of lnx is 1/x and d/dx(x)=1.

⇒ $\dfrac{dy}{dx}=y(1+\ln x)$

⇒ $\dfrac{dy}{dx}=x^x(1+\ln x)$ as y=x^{x}.

Thus, the derivative of x^{x} (x to the x) is equal to x^{x}(1+ln x) and this is obtained by the logarithmic differentiation.

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## FAQs

**Q1: What is the derivative of x^x?**

Answer: The derivative of x^x is equal to (1+ln x)x^x.